CBSE Class 10 Science Electricity Notes

CBSE Class 10 Science Electricity Notes. There are many more useful educational material which the students can download in pdf format and use them for studies. Study material like concept maps, important and sure shot question banks, quick to learn flash cards, flow charts, mind maps, teacher notes, important formulas, past examinations question bank, important concepts taught by teachers. Students can download these useful educational material free and use them to get better marks in examinations.  Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

CBSE Class 10 Science: Electricity Notes

Electricity is a general term that encompasses a variety of phenomena resulting from the presence and flow of electric charge. These include many easily recognizable phenomena such as lightning and static electricity, but in addition, less familiar concepts such as the electromagnetic field and electromagnetic induction.

Electric charge

Electric charge is a fundamental conserved property of some subatomic particles, which determines their electromagnetic interaction. Electrically charged matter is influenced by, and produces, electromagnetic fields. The interaction between a moving charge and an electromagnetic field is the source of the electromagnetic force, which is one of the four fundamental forces. Electric charge is conserved, additive and quantised. The S.I. unit of electric charge is the coulomb (C).

Any charged body will have a charge

\[Q=ne\]

where \(n\) is the number of electrons and \(e\) is the charge on an electron, \(e=1.6\times10^{-19}\) coulomb.

Electric current

Electric current is a flow of electrons in a conductor such as a metal wire. Electric current is expressed by the amount of charge flowing through a particular area in unit time. In other words, it is the rate of flow of electric charges. In circuits using metallic wires, electrons constitute the flow of charges. However, electrons were not known at the time when the phenomenon of electricity was first observed. So, electric current was considered to be the flow of positive charges, and the direction of flow of positive charges was taken to be the direction of electric current. Conventionally, in an electric circuit the direction of electric current is taken as opposite to the direction of the flow of electrons, which are negative charges.

The magnitude of electric current in a conductor is the amount of electric charge passing through a given point of the conductor in 1 second.

\[I=\frac{Q}{t}\]

The S.I. unit of electric current is the ampere (A), named after the French scientist André-Marie Ampère (1775–1836).

One ampere: when 1 coulomb of charge flows through any cross-section of a conductor in 1 second, the electric current flowing through it is said to be 1 ampere.

Smaller units of current are the milliampere (mA) and the microampere (\(\mu\)A): \(1\text{ mA}=10^{-3}\text{ A}\), \(1\ \mu\text{A}=10^{-6}\text{ A}\).

An instrument called an ammeter measures electric current in a circuit. It is always connected in series in the circuit through which the current is to be measured. The direction of electric current is from the positive terminal to the negative terminal through the electric circuit.

Intext questions (page 200)

What does an electric circuit mean?

Answer: An electric circuit consists of electric devices, switching devices, a source of electricity, etc. that are connected by conducting wires.

Define the unit of current.

Answer: The unit of electric current is the ampere (A). 1 A is defined as the flow of 1 C of charge through a wire in 1 s.

Calculate the number of electrons constituting one coulomb of charge.

Answer: One electron carries a charge of \(1.6\times10^{-19}\) C, so \(1\) C of charge is contained in \(\dfrac{1}{1.6\times10^{-19}}=6.25\times10^{18}\) electrons.
Therefore, \(6.25\times10^{18}\) electrons constitute one coulomb of charge.

Potential difference

Potential difference \(V_A-V_B\) between two points A and B is the work done per unit charge in taking a charge from B to A.

\[V_A-V_B=\frac{\text{work done}}{\text{charge}}\]

where \(V_A\) is the potential at point A and \(V_B\) is the potential at point B. The S.I. unit of potential is the volt (V), named after Alessandro Volta (1745–1827), an Italian physicist.

Electric potential

Electric potential at a point is defined as the work done per unit charge in bringing a charge from infinity to that point.

\[V=\frac{\text{work done}}{\text{charge}}=\frac{W}{Q}\]

The potential difference is measured by an instrument called a voltmeter. The voltmeter is always connected in parallel across the points between which the potential difference is to be measured.

One volt: the potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of electric charge from one point to the other.

Electrons always flow from lower potential to higher potential.

Intext questions (page 202)

Name a device that helps to maintain a potential difference across a conductor.

Answer: A source of electricity such as a cell, battery or power supply helps to maintain a potential difference across a conductor.

What is meant by saying that the potential difference between two points is 1 V?

Answer: If 1 J of work is required to move a charge of 1 C from one point to another, then the potential difference between the two points is 1 V.

How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer: The energy given to each coulomb of charge is equal to the work required to move it. \(\text{Potential difference}=\dfrac{\text{work done}}{\text{charge}}\), so work done = potential difference \(\times\) charge. With charge = 1 C and potential difference = 6 V, work done \(=6\times1=6\) J.
Therefore, 6 J of energy is given to each coulomb of charge passing through a 6 V battery.

Numerical problems

Find the charge if the number of electrons is \(4\times10^{-18}\).

Answer: \(Q=ne=4\times10^{-18}\times1.6\times10^{-19}=6.4\times10^{-37}\text{ C}\)

Find the number of electrons constituting one coulomb of charge.

Answer: \(n=\dfrac{Q}{e}=\dfrac{1}{1.6\times10^{-19}}=6.25\times10^{18}\) electrons

How much work is done in moving a charge of 3 coulombs from a point at 118 V to a point at 128 V?

Answer: Potential difference \(=128-118=10\) V.
\(W=QV=3\times10=30\) J

How much work is done in moving a charge of 2 C across two points having a potential difference of 12 V?

Answer: \(W=QV=2\times12=24\) J

Calculate the amount of work done to carry 4 C from a point at 100 V to a point at 120 V.

Answer: Potential difference \(=120-100=20\) V.
\(W=QV=4\times20=80\) J

How much work will be done in bringing a charge of \(2\times10^{-3}\) coulomb from infinity to a point P at which the potential is 5 V?

Answer: \(W=QV=2\times10^{-3}\times5=0.01\text{ J}\)

How much work will be done in bringing a charge of \(3\times10^{-2}\) coulomb from infinity to a point P at which the potential is 20 V?

Answer: \(W=QV=3\times10^{-2}\times20=0.6\text{ J}\)

How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer: Energy per coulomb \(=\) potential difference \(\times\) 1 C \(=6\times1=6\) J

How much energy is transferred by a 12 V power supply to each coulomb of charge which it moves around a circuit?

Answer: Energy per coulomb \(=12\times1=12\) J

What is the potential difference between the terminals of a battery if 250 joules of work is required to transfer 20 coulombs of charge from one terminal of the battery to the other?

Answer: \(V=\dfrac{W}{Q}=\dfrac{250}{20}=12.5\text{ V}\)

What is the potential difference between the conductors A and B shown in the figure below? If the conductors are connected by a length of wire, which way will electrons flow? When will this flow of electrons stop?

Answer: The potential difference between A and B \(=3000-(-1000)=4000\) V.
Electrons flow from the lower potential to the higher potential, so they flow from A to B. The flow stops when the two conductors reach the same potential.

A particle of charge 2 C is taken from a point at a potential of 100 V to another point at a potential of 150 V. Calculate the work done.

Answer: Potential difference \(=150-100=50\) V.
\(W=QV=2\times50=100\) J

What is the potential difference between the conductors A and B shown in the figure below? If the conductors are connected by a length of wire, which way will electrons flow? When will this flow of electrons stop?

Answer: The potential difference between A and B \(=1200-(-500)=1700\) V.
Electrons flow from A (lower potential) to B (higher potential). The flow stops when the two conductors reach the same potential.

A particle of charge \(5\times10^{-2}\) C is taken from a point at a potential of 50 V to another point at a potential of 250 V. Calculate the work done.

Answer: Potential difference \(=250-50=200\) V.
\(W=QV=5\times10^{-2}\times200=10\text{ J}\)

Three 2 V cells are connected in series and used as a battery in a circuit. (a) What is the potential difference at the terminals of the battery? (b) How many joules of electrical energy does 1 C gain on passing through (i) one cell (ii) all three cells?

Answer: (a) For cells in series the potential differences add: \(2+2+2=6\) V.
(b) (i) Through one cell: \(W=QV=1\times2=2\) J. (ii) Through all three cells: \(W=1\times6=6\) J.

Circuit diagram

A schematic diagram in which the different components of a circuit are represented by symbols is called a circuit diagram. The conventional symbols used to represent some of the most commonly used electrical components are given below.

Ohm's law

According to Ohm's law, "At constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends."

\(I\propto V\) or \(V\propto I\) at constant temperature
\(V=IR\), where \(R\) is the constant of proportionality, which is known as resistance.

Resistance

It is the ratio of the potential difference applied between the ends of a conductor to the current flowing through it. The unit of resistance is the ohm (\(\Omega\)).

\[R=\frac{V}{I}\;\Rightarrow\;I=\frac{V}{R},\qquad 1\text{ ohm}=\frac{1\text{ volt}}{1\text{ ampere}}\]

One ohm: one ohm is the resistance of a conductor such that when a potential difference of 1 volt is applied to its ends, a current of 1 ampere flows through it.

If the resistance is doubled the current gets halved. In many practical cases it is necessary to increase or decrease the current in an electric circuit. A component used to regulate current without changing the voltage source is called a variable resistance. In an electric circuit, a device called a rheostat is often used to change the resistance in the circuit.

Factors on which the resistance of a conductor depends

The resistance of a conductor depends (i) on its length, (ii) on its area of cross-section, and (iii) on the nature of its material.

Resistance depends on the area of cross-section: it is inversely proportional to the area of cross-section \(A\): \(R\propto\dfrac1A\).

Resistance depends on the length of the wire: it is directly proportional to the length \(l\) of the wire: \(R\propto l\).

Combining the above we get

\[R=\rho\frac{l}{A}\]

where \(\rho\) (rho) is a constant of proportionality called the resistivity or specific resistance of the material. If \(l=1\) m and \(A=1\) m², then \(R=\rho\). Resistivity of a material is the resistance of a unit length of the material having unit area of cross-section.

Intext questions (page 209)

On what factors does the resistance of a conductor depend?

Answer: The resistance of a conductor depends upon the following factors: (a) length of the conductor, (b) cross-sectional area of the conductor, (c) material of the conductor, and (d) temperature of the conductor.

Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Answer: Resistance of a wire is \(R=\rho\dfrac{l}{A}\), where \(\rho\) is the resistivity of the material, \(l\) is the length and \(A\) is the area of cross-section. Resistance is inversely proportional to the area of cross-section. The thicker the wire, the lower is its resistance, and vice versa. Therefore, current flows more easily through a thick wire than through a thin wire.

Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Answer: By Ohm’s law \(V=IR\), so \(I=\dfrac{V}{R}\). The new potential difference is \(V/2\) and the resistance is unchanged, so \[I^{\prime}=\frac{V/2}{R}=\frac12\cdot\frac{V}{R}=\frac{I}{2}\] Therefore, the current through the component is reduced to half.

Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Answer: The resistivity of an alloy is higher than that of a pure metal. Moreover, at high temperatures the alloys do not melt readily. Hence the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal.

(a) Which among iron and mercury is a better conductor? (b) Which material is the best conductor?

Answer: (a) Resistivity of iron \(=10.0\times10^{-8}\ \Omega\) m and resistivity of mercury \(=94.0\times10^{-8}\ \Omega\) m. The resistivity of mercury is more than that of iron, so iron is a better conductor than mercury.
(b) From the table of resistivities (Table 12.2 of the textbook) the resistivity of silver is the lowest among the listed materials. Hence silver is the best conductor.

Numerical problems

What current must flow if 0.24 coulomb is to be transferred in 15 ms?

Answer: \(I=\dfrac{Q}{t}=\dfrac{0.24}{15\times10^{-3}}=16\text{ A}\)

If a current of 10 A flows for four minutes, find the quantity of electricity transferred.

Answer: \(t=4\times60=240\) s, so \(Q=It=10\times240=2400\text{ C}\)

An electric bulb draws a current of 0.25 A for 20 minutes. Calculate the electric charge that flows through the circuit.

Answer: \(t=20\times60=1200\) s, so \(Q=It=0.25\times1200=300\text{ C}\)

If the amount of electric charge passing through a conductor in 10 min is 300 C, find the current.

Answer: \(t=10\times60=600\) s, so \(I=\dfrac{Q}{t}=\dfrac{300}{600}=0.5\text{ A}\)

How many electrons are flowing per second past a point in a circuit in which there is a current of 4 A?

Answer: Charge flowing per second \(Q=It=4\times1=4\) C. Number of electrons \(n=\dfrac{Q}{e}=\dfrac{4}{1.6\times10^{-19}}=2.5\times10^{19}\)

A lamp of resistance 80 \(\Omega\) draws a current of 0.75 A. Find the line voltage.

Answer: \(V=IR=0.75\times80=60\text{ V}\)

An electric heater draws a current of 5 A when connected to 220 V mains. Calculate the resistance of its filament.

Answer: \(R=\dfrac{V}{I}=\dfrac{220}{5}=44\ \Omega\)

How much current will an electric bulb draw from a 200 V source, if the resistance of the filament is 1200 \(\Omega\)?

Answer: \(I=\dfrac{V}{R}=\dfrac{200}{1200}=0.1667\text{ A}\)

How much current will an electric heater draw from a 200 V source, if the resistance of the filament is 100 \(\Omega\)?

Answer: \(I=\dfrac{V}{R}=\dfrac{200}{100}=2\text{ A}\)

How much current does an electric heater draw from a 220 V line, if the resistance of the heater (when hot) is 50 \(\Omega\)?

Answer: \(I=\dfrac{V}{R}=\dfrac{220}{50}=4.4\text{ A}\)

A bulb when cold has 1 \(\Omega\) resistance. It draws a current of 0.3 A when glowing from a source of 3 V. Calculate the resistance of the bulb when glowing and explain the reason for the difference in resistance.

Answer: Resistance when glowing \(R=\dfrac{V}{I}=\dfrac{3}{0.3}=10\ \Omega\).
The resistance of the glowing bulb (10 \(\Omega\)) is much more than when it is cold (1 \(\Omega\)), because the filament becomes very hot when current flows and the resistance of a metal increases with increase in temperature.

Calculate the potential difference required across a conductor of resistance 5 \(\Omega\) to make a current of 1.5 A flow through it.

Answer: \(V=IR=1.5\times5=7.5\text{ V}\)

What is the resistance of an electric arc lamp when hot, if the lamp uses 20 A when connected to a 220 V line?

Answer: \(R=\dfrac{V}{I}=\dfrac{220}{20}=11\ \Omega\)

Calculate the amount of work done to draw a current of 8 A from a point at 100 V to a point at 120 V in 2 seconds.

Answer: Charge moved \(Q=It=8\times2=16\) C. Potential difference \(=120-100=20\) V.
\(W=QV=16\times20=320\text{ J}\)

If 200 C of charge pass a point in a circuit in 4 s, what current is flowing?

Answer: \(I=\dfrac{Q}{t}=\dfrac{200}{4}=50\text{ A}\)

A current of 4 A flows around a circuit in 10 seconds. How much charge flows past a point in the circuit in this time? Also find the number of electrons that pass in the circuit.

Answer: \(Q=It=4\times10=40\text{ C}\).
Number of electrons \(n=\dfrac{Q}{e}=\dfrac{40}{1.6\times10^{-19}}=2.5\times10^{20}\)

The current flowing through a resistor is 0.8 A when a p.d. of 20 V is applied. Determine the value of the resistance.

Answer: \(R=\dfrac{V}{I}=\dfrac{20}{0.8}=25\ \Omega\)

Determine the p.d. which must be applied to a 2 k\(\Omega\) resistor in order that a current of 10 mA may flow.

Answer: \(R=2\text{ k}\Omega=2000\ \Omega\), \(I=10\text{ mA}=0.01\) A.
\(V=IR=0.01\times2000=20\text{ V}\)

A coil has a current of 50 mA flowing through it when the applied voltage is 12 V. What is the resistance of the coil?

Answer: \(I=50\text{ mA}=0.05\) A, so \(R=\dfrac{V}{I}=\dfrac{12}{0.05}=240\ \Omega\)

A 100 V battery is connected across a resistor and causes a current of 5 mA to flow. Determine the resistance of the resistor. If the voltage is now reduced to 25 V, what will be the new value of the current flowing?

Answer: \(R=\dfrac{V}{I}=\dfrac{100}{5\times10^{-3}}=20000\ \Omega=20\text{ k}\Omega\).
The resistance stays the same, so at 25 V: \(I=\dfrac{25}{20000}=0.00125\text{ A}=1.25\text{ mA}\)

What is the resistance of a coil which draws a current of (a) 50 mA and (b) 200 \(\mu\)A from a 120 V supply?

Answer: (a) \(R=\dfrac{120}{50\times10^{-3}}=2400\ \Omega=2.4\text{ k}\Omega\)
(b) \(R=\dfrac{120}{200\times10^{-6}}=6\times10^{5}\ \Omega=600\text{ k}\Omega\)

If a current of 5 A flows for 2 minutes, find the quantity of electricity transferred.

Answer: \(Q=It=5\times120=600\text{ C}\)

A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit.

Answer: \(Q=It=0.5\times600=300\text{ C}\)

How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 \(\Omega\)?

Answer: \(I=\dfrac{V}{R}=\dfrac{220}{1200}=0.1833\text{ A}\)

How much current will an electric heater coil draw from a 220 V source, if the resistance of the heater coil is 100 \(\Omega\)?

Answer: \(I=\dfrac{220}{100}=2.2\text{ A}\)

The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V?

Answer: \(R=\dfrac{V}{I}=\dfrac{60}{4}=15\ \Omega\) (constant). New current \(I^{\prime}=\dfrac{120}{15}=8\text{ A}\)

When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Answer: \(R=\dfrac{V}{I}=\dfrac{12}{2.5\times10^{-3}}=4800\ \Omega=4.8\text{ k}\Omega\)

An electric heater is connected to the 230 V mains supply. A current of 8 A flows through the heater. (a) How much charge flows around the circuit each second? (b) How much energy is transferred to the heater each second?

Answer: (a) \(Q=It=8\times1=8\text{ C}\) each second.
(b) \(E=VIt=230\times8\times1=1840\text{ J}\) each second.

How many electrons are flowing per second past a point in a circuit in which there is a current of 5 A?

Answer: \(Q=It=5\times1=5\) C, so \(n=\dfrac{5}{1.6\times10^{-19}}=3.12\times10^{19}\)

An electric iron draws a current of 3.4 A from the 220 V supply line. What current will this electric iron draw when connected to a 110 V supply line?

Answer: \(R=\dfrac{220}{3.4}=64.71\ \Omega\). At 110 V: \(I^{\prime}=\dfrac{110}{64.7}=1.7\text{ A}\) (half of 3.4 A, since the potential difference is halved).

A simple electric circuit has a 24 V battery and a resistor of 60 \(\Omega\). What will be the current in the circuit?

Answer: \(I=\dfrac{V}{R}=\dfrac{24}{60}=0.4\text{ A}\)

When a 4 \(\Omega\) resistor is connected across the terminals of a 12 V battery, find the number of coulombs passing through the resistor per second.

Answer: \(I=\dfrac{12}{4}=3\text{ A}\), i.e. 3 C of charge passes through the resistor every second.

An electric room heater draws a current of 2.4 A from the 120 V supply line. What current will this room heater draw when connected to a 240 V supply line?

Answer: \(R=\dfrac{120}{2.4}=50\ \Omega\). At 240 V: \(I^{\prime}=\dfrac{240}{50}=4.8\text{ A}\)

A current of 200 mA flows through a 4 k\(\Omega\) resistor. What is the p.d. across the resistor?

Answer: \(I=0.2\) A and \(R=4000\ \Omega\), so \(V=IR=0.2\times4000=800\text{ V}\)

A p.d. of 10 V is needed to make a current of 0.02 A flow through a wire. What p.d. is needed to make a current of 250 mA flow through the same wire?

Answer: \(R=\dfrac{10}{0.02}=500\ \Omega\). For \(I=250\text{ mA}=0.25\) A: \(V=IR=0.25\times500=125\text{ V}\)

A TV draws a current of 5 A from the 240 V supply line. What current will this TV draw when it is connected to a 100 V supply line?

Answer: \(R=\dfrac{240}{5}=48\ \Omega\). At 100 V: \(I^{\prime}=\dfrac{100}{48}=2.083\text{ A}\)

The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V?

Answer: \(R=\dfrac{60}{4}=15\ \Omega\). New current \(=\dfrac{120}{15}=8\text{ A}\)

A bulb of resistance 400 \(\Omega\) is connected to 220 V mains. Calculate the magnitude of current.

Answer: \(I=\dfrac{220}{400}=0.55\text{ A}\)

A battery of two cells is used to light a torch bulb of resistance 5 \(\Omega\). The cells maintain a potential difference of 3 V across the bulb. How much current will flow through the bulb?

Answer: \(I=\dfrac{V}{R}=\dfrac{3}{5}=0.6\text{ A}\)

A steady current of 5 A flows through a circuit for 30 minutes. How much charge has circulated through the circuit in this time?

Answer: \(t=30\times60=1800\) s, so \(Q=It=5\times1800=9000\text{ C}\)

Numerical problems on resistivity

Calculate the resistance of a copper wire of length 2 m and area of cross-section \(10^{-6}\) m². Resistivity of copper is \(1.7\times10^{-8}\ \Omega\) m.

Answer: \(R=\rho\dfrac{l}{A}=\dfrac{1.7\times10^{-8}\times2}{10^{-6}}=0.034\ \Omega\)

A copper wire of length 2 m and area of cross-section \(1.7\times10^{-6}\) m² has a resistance of \(2\times10^{-2}\) ohm. Calculate the resistivity of copper.

Answer: \(\rho=\dfrac{RA}{l}=\dfrac{2\times10^{-2}\times1.7\times10^{-6}}{2}=1.7\times10^{-8}\ \Omega\text{ m}\)

The amount of charge passing through a cell in 12 seconds is 3 C. What is the current supplied by the cell?

Answer: \(I=\dfrac{Q}{t}=\dfrac{3}{12}=0.25\text{ A}\)

A 12 V battery of a car is connected across a 4 \(\Omega\) resistor. Calculate the current passing through the resistor.

Answer: \(I=\dfrac{12}{4}=3\text{ A}\)

Resistivity of a given copper wire of length 2 m is \(1.7\times10^{-8}\ \Omega\) m. The wire is stretched so that its length becomes 4 m. Find the new resistivity of the copper wire.

Answer: Resistivity depends only on the nature (and temperature) of the material, not on the length or area. So the resistivity stays the same: \(1.7\times10^{-8}\ \Omega\) m.

The resistance of a given wire of length \(l\) is 3 \(\Omega\). The wire is stretched uniformly such that its length becomes \(2l\). Find the new resistance of the stretched wire.

Answer: On stretching, the volume stays the same, so when the length doubles the area becomes half. \(R^{\prime}=\rho\dfrac{2l}{A/2}=4R=4\times3=12\ \Omega\)

The resistance of a given wire of length \(l\) is 4 \(\Omega\). The wire is stretched uniformly such that its length becomes \(3l\). Find the new resistance of the stretched wire.

Answer: The volume stays the same, so the area becomes one third. \(R^{\prime}=\rho\dfrac{3l}{A/3}=9R=9\times4=36\ \Omega\)

A copper wire has a diameter of 0.5 mm and resistivity of \(1.6\times10^{-8}\ \Omega\) m. What will be the length of this wire to make its resistance 10 \(\Omega\)? How much does the resistance change if the diameter is doubled?

Answer: \(A=\pi\left(\dfrac{0.5\times10^{-3}}{2}\right)^2=1.96\times10^{-7}\text{ m}^2\), so \(l=\dfrac{RA}{\rho}=\dfrac{10\times1.96\times10^{-7}}{1.6\times10^{-8}}=122.7\text{ m}\).
If the diameter is doubled, the area becomes 4 times, so the resistance becomes one fourth: \(R^{\prime}=\dfrac{10}{4}=2.5\ \Omega\). The resistance decreases from 10 \(\Omega\) to 2.5 \(\Omega\).

A 6 \(\Omega\) resistance wire is doubled up by folding. Calculate the new resistance of the wire.

Answer: On folding, the length becomes half and the area of cross-section becomes double. \(R^{\prime}=\rho\dfrac{l/2}{2A}=\dfrac{R}{4}=\dfrac{6}{4}=1.5\ \Omega\)

Calculate the resistance of an aluminium cable of length 10 km and diameter 20 mm if the resistivity of aluminium is \(2.7\times10^{-8}\ \Omega\) m.

Answer: \(l=10^4\) m, radius \(=10\) mm \(=10^{-2}\) m, so \(A=\pi r^2=3.14\times10^{-4}\text{ m}^2\).
\(R=\rho\dfrac{l}{A}=\dfrac{2.7\times10^{-8}\times10^4}{3.14\times10^{-4}}=0.8594\ \Omega\)

Calculate the area of cross-section of a wire if its length is 1.0 m, its resistance is 23 \(\Omega\) and the resistivity of the material of the wire is \(1.84\times10^{-6}\ \Omega\) m.

Answer: \(A=\dfrac{\rho l}{R}=\dfrac{1.84\times10^{-6}\times1.0}{23}=8\times10^{-8}\text{ m}^2\)

A piece of wire of resistance 20 \(\Omega\) is drawn out so that its length is increased to twice its original length. Calculate the resistance of the wire in the new situation.

Answer: Doubling the length (volume constant) halves the area, so the resistance becomes 4 times: \(R^{\prime}=4\times20=80\ \Omega\)

Two cylindrical wires of the same material have their lengths in the ratio of 4 : 9. What should be the ratio of their radii so that their resistances are in the ratio of 4 : 1?

Answer: \(R=\rho\dfrac{l}{\pi r^2}\), so \(\dfrac{R_1}{R_2}=\dfrac{l_1}{l_2}\cdot\dfrac{r_2^2}{r_1^2}\). Then \(\dfrac41=\dfrac49\cdot\dfrac{r_2^2}{r_1^2}\Rightarrow\dfrac{r_2^2}{r_1^2}=9\Rightarrow\dfrac{r_2}{r_1}=3\).
Therefore \(r_1:r_2=1:3\).

Two wires of the same metal have the same area of cross-section but their lengths are in the ratio of 3 : 1. What should be the ratio of current flowing through them respectively, when the same potential difference is applied across each of their lengths?

Answer: For the same material and area, \(R\propto l\), so \(R_1:R_2=3:1\). For the same potential difference \(I=\dfrac{V}{R}\propto\dfrac1R\), so \(I_1:I_2=1:3\).

Two wires A and B of length 30 m and 10 m have radii 2 cm and 1 cm respectively. Compare the resistances of the two wires. Which will have less resistance?

Answer: Assuming the same material, \(\dfrac{R_A}{R_B}=\dfrac{l_A}{l_B}\cdot\dfrac{r_B^2}{r_A^2}=\dfrac{30}{10}\times\dfrac{1^2}{2^2}=\dfrac34\).
So \(R_A:R_B=3:4\), and wire A has the lesser resistance.

Calculate the resistance of a 1 km long copper wire of radius 1 mm. Resistivity of copper is \(1.7\times10^{-8}\ \Omega\) m.

Answer: \(l=1000\) m, \(A=\pi(10^{-3})^2=3.14\times10^{-6}\text{ m}^2\).
\(R=\rho\dfrac{l}{A}=\dfrac{1.7\times10^{-8}\times1000}{3.14\times10^{-6}}=5.411\ \Omega\)

A 4 \(\Omega\) wire is doubled on itself. Calculate the new resistance of the wire.

Answer: Doubling on itself makes the length half and the area double, so \(R^{\prime}=\dfrac{R}{4}=\dfrac{4}{4}=1\ \Omega\)

What should be the length of the nichrome wire of resistance 4.5 \(\Omega\), if the length of a similar wire is 60 cm and resistance 2.5 \(\Omega\)?

Answer: For similar wires (same material and area) \(R\propto l\), so \(\dfrac{l_2}{60}=\dfrac{4.5}{2.5}\Rightarrow l_2=60\times1.8=108\text{ cm}\)

A metal wire of resistivity \(64\times10^{-6}\ \Omega\) m and length 198 cm has a resistance of 7 \(\Omega\). Calculate its radius.

Answer: \(A=\dfrac{\rho l}{R}=\dfrac{64\times10^{-6}\times1.98}{7}=1.81\times10^{-5}\text{ m}^2\).
\(r=\sqrt{\dfrac{A}{\pi}}=\sqrt{\dfrac{1.81\times10^{-5}}{3.14}}=0.0024\text{ m}\approx2.4\text{ mm}\)

Calculate the resistivity of the material of a wire 1.0 m long, 0.4 mm in diameter and having a resistance of 2.0 \(\Omega\).

Answer: \(r=0.2\) mm \(=2\times10^{-4}\) m, so \(A=\pi r^2=1.26\times10^{-7}\text{ m}^2\).
\(\rho=\dfrac{RA}{l}=\dfrac{2.0\times1.26\times10^{-7}}{1.0}=2.51\times10^{-7}\ \Omega\text{ m}\)

Resistance of a system of resistors

Resistors in series

In a series circuit (a) the current \(I\) is the same in all parts of the circuit, and (b) the sum of the voltages \(V_1\), \(V_2\) and \(V_3\) is equal to the total applied voltage \(V\), i.e. \(V=V_1+V_2+V_3\).

From Ohm's law: \(V_1=IR_1\), \(V_2=IR_2\), \(V_3=IR_3\) and \(V=IR\), where \(R\) is the total circuit resistance. Since \(V=V_1+V_2+V_3\), then \(IR=IR_1+IR_2+IR_3\). Dividing throughout by \(I\) gives

\[R=R_1+R_2+R_3\]

Thus for a series circuit, the total resistance is obtained by adding together the values of the separate resistances. When several resistors are connected in series, the resistance of the combination \(R_s\) is equal to the sum of their individual resistances \(R_1,R_2,R_3\), and is thus greater than any individual resistance.

Resistors in parallel

In a parallel circuit (a) the sum of the currents \(I_1\), \(I_2\) and \(I_3\) is equal to the total circuit current \(I\), i.e. \(I=I_1+I_2+I_3\), and (b) the source p.d. \(V\) volts is the same across each of the resistors.

From Ohm's law: \(I_1=\dfrac V{R_1}\), \(I_2=\dfrac V{R_2}\), \(I_3=\dfrac V{R_3}\) and \(I=\dfrac VR\), where \(R\) is the total resistance of the circuit. Since \(I=I_1+I_2+I_3\), then

\[\frac VR=\frac V{R_1}+\frac V{R_2}+\frac V{R_3}\]

Dividing throughout by \(V\), we get

\[\frac1R=\frac1{R_1}+\frac1{R_2}+\frac1{R_3}\]

This equation must be used when finding the total resistance \(R\) of a parallel circuit. Thus the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.

For the special case of two resistors in parallel:

\[\frac1R=\frac1{R_1}+\frac1{R_2}=\frac{R_1+R_2}{R_1R_2}\quad\Rightarrow\quad R=\frac{R_1R_2}{R_1+R_2}\ \left(\text{i.e. }\frac{\text{product}}{\text{sum}}\right)\]

Intext questions (page 213)

Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 \(\Omega\) resistor, an 8 \(\Omega\) resistor, a 12 \(\Omega\) resistor and a plug key, all connected in series.

Answer: Three cells of potential 2 V each connected in series are equivalent to a battery of potential \(2+2+2=6\) V. The following circuit diagram shows the three resistors of 5 \(\Omega\), 8 \(\Omega\) and 12 \(\Omega\) connected in series with a battery of potential 6 V.

Redraw the circuit of the previous question, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 \(\Omega\) resistor. What would be the readings in the ammeter and the voltmeter?

Answer: To measure the current flowing through the resistors, an ammeter is connected in series with the resistors. To measure the potential difference across the 12 \(\Omega\) resistor, a voltmeter is connected in parallel with it, as shown in the figure.

Answer: The resistors are in series, so \(R=5+8+12=25\ \Omega\) and \(V=6\) V. By Ohm’s law \(I=\dfrac{V}{R}=\dfrac{6}{25}=0.24\text{ A}\).
Potential difference across the 12 \(\Omega\) resistor: \(V_1=IR=0.24\times12=2.88\text{ V}\).
Therefore the ammeter reads 0.24 A and the voltmeter reads 2.88 V.

Intext questions (page 216)

Judge the equivalent resistance when the following are connected in parallel: (a) 1 \(\Omega\) and \(10^6\ \Omega\), (b) 1 \(\Omega\), \(10^3\ \Omega\) and \(10^6\ \Omega\).

Answer: (a) \(\dfrac1R=\dfrac11+\dfrac1{10^6}\Rightarrow R=\dfrac{10^6}{10^6+1}\approx1\ \Omega\). Therefore the equivalent resistance is about 1 \(\Omega\).
(b) \(\dfrac1R=\dfrac11+\dfrac1{10^3}+\dfrac1{10^6}=\dfrac{10^6+10^3+1}{10^6}\Rightarrow R=\dfrac{1000000}{1001001}=0.999\ \Omega\). Therefore the equivalent resistance is 0.999 \(\Omega\).

An electric lamp of 100 \(\Omega\), a toaster of resistance 50 \(\Omega\), and a water filter of resistance 500 \(\Omega\) are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Answer: Let \(R\) be the equivalent resistance. \(\dfrac1R=\dfrac1{100}+\dfrac1{50}+\dfrac1{500}=\dfrac{5+10+1}{500}=\dfrac{16}{500}\Rightarrow R=\dfrac{500}{16}=31.25\ \Omega\).
Current drawn by all three appliances \(I=\dfrac VR=\dfrac{220}{31.25}=7.04\text{ A}\).
The electric iron takes the same current, 7.04 A, from the same 220 V source, so its resistance is \(R^{\prime}=\dfrac VI=\dfrac{220}{7.04}=31.25\ \Omega\).
Therefore the resistance of the electric iron is 31.25 \(\Omega\) and the current through it is 7.04 A.

What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Answer: There is no division of voltage among the appliances when they are connected in parallel. The potential difference across each appliance is equal to the supplied voltage. The total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel.

How can three resistors of resistances 2 \(\Omega\), 3 \(\Omega\) and 6 \(\Omega\) be connected to give a total resistance of (a) 4 \(\Omega\), (b) 1 \(\Omega\)?

Answer: (a) Connect the 6 \(\Omega\) and 3 \(\Omega\) resistors in parallel and put this combination in series with the 2 \(\Omega\) resistor. The parallel combination has \(R=\dfrac{6\times3}{6+3}=2\ \Omega\), so the total resistance is \(2+2=4\ \Omega\).

Answer: (b) Connect all three resistors in parallel. \(\dfrac1R=\dfrac12+\dfrac13+\dfrac16=\dfrac{3+2+1}{6}=\dfrac66=1\Rightarrow R=1\ \Omega\).

What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 \(\Omega\), 8 \(\Omega\), 12 \(\Omega\), 24 \(\Omega\)?

Answer: (a) The highest resistance is obtained when the coils are connected in series: \(4+8+12+24=48\ \Omega\).
(b) The lowest resistance is obtained when the coils are connected in parallel: \(\dfrac1R=\dfrac14+\dfrac18+\dfrac1{12}+\dfrac1{24}=\dfrac{6+3+2+1}{24}=\dfrac{12}{24}\Rightarrow R=2\ \Omega\). Therefore 2 \(\Omega\) is the lowest total resistance.

Numerical problems

For the circuit shown in the figure below, determine (a) the battery voltage \(V\), (b) the total resistance of the circuit, and (c) the values of resistance of resistors \(R_1\), \(R_2\) and \(R_3\), given that the p.d.'s across \(R_1\), \(R_2\) and \(R_3\) are 5 V, 2 V and 6 V respectively.

Answer: (a) In a series circuit \(V=V_1+V_2+V_3=5+2+6=13\) V.
(b) \(R=\dfrac VI=\dfrac{13}{4}=3.25\ \Omega\).
(c) \(R_1=\dfrac5{4}=1.25\ \Omega\), \(R_2=\dfrac24=0.5\ \Omega\), \(R_3=\dfrac64=1.5\ \Omega\).

For the circuit shown in the figure below, determine the p.d. across resistor \(R_3\). If the total resistance of the circuit is 100 \(\Omega\), determine the current flowing through resistor \(R_1\). Find also the value of resistor \(R_2\).

Answer: The p.d. across \(R_3\): \(V_3=25-10-4=11\) V.
The current is the same in a series circuit: \(I=\dfrac VR=\dfrac{25}{100}=0.25\) A, which flows through \(R_1\).
\(R_2=\dfrac{V_2}{I}=\dfrac4{0.25}=16\ \Omega\).

A 12 V battery is connected in a circuit having three series-connected resistors having resistances of 4 \(\Omega\), 9 \(\Omega\) and 11 \(\Omega\). Determine the current flowing through, and the p.d. across the 9 \(\Omega\) resistor.

Answer: \(R=4+9+11=24\ \Omega\), so \(I=\dfrac{12}{24}=0.5\) A. The p.d. across the 9 \(\Omega\) resistor is \(V_1=IR=0.5\times9=4.5\) V.

Find the voltage \(V\) in the given figure.

Answer: The 4 \(\Omega\) and 6 \(\Omega\) resistors are in series across the 50 V battery, and \(V\) is the p.d. across the 6 \(\Omega\) resistor. \(I=\dfrac{50}{4+6}=5\) A, so \(V=IR=5\times6=30\) V.

For the circuit shown in the figure, determine (a) the reading on the ammeter, and (b) the value of resistor \(R_2\).

Answer: The p.d. across each branch is the same: \(V=8\times5=40\) V.
(a) The ammeter is in series with \(R_3\): \(I_3=\dfrac{40}{20}=2\) A.
(b) Current in \(R_2\): \(I_2=11-8-2=1\) A, so \(R_2=\dfrac{40}{1}=40\ \Omega\).

Two resistors are connected in series across a 24 V supply and a current of 3 A flows in the circuit. If one of the resistors has a resistance of 2 \(\Omega\), determine (a) the value of the other resistor, and (b) the p.d. across the 2 \(\Omega\) resistor. If the circuit is connected for 50 hours, how much energy is used?

Answer: Total resistance \(R=\dfrac{24}{3}=8\ \Omega\).
(a) Other resistor \(=8-2=6\ \Omega\).
(b) p.d. across the 2 \(\Omega\) resistor \(=3\times2=6\) V.
Energy used \(=VIt=24\times3\times50\ \text{Wh}=3600\ \text{Wh}=3.6\) kWh (\(=3.6\times3.6\times10^6=1.296\times10^7\) J).

Two resistors of resistance 3 \(\Omega\) and 6 \(\Omega\) are connected in parallel across a battery having a voltage of 12 V. Determine (a) the total circuit resistance and (b) the current flowing in the 3 \(\Omega\) resistor.

Answer: (a) \(R=\dfrac{3\times6}{3+6}=2\ \Omega\).
(b) The p.d. across each resistor is 12 V, so \(I_1=\dfrac{12}{3}=4\) A.

Given four 1 \(\Omega\) resistors, state how they must be connected to give an overall resistance of (a) \(\dfrac14\ \Omega\), (b) 1 \(\Omega\), (c) \(1\dfrac13\ \Omega\), (d) \(2\dfrac12\ \Omega\), all four resistors being connected in each case.

Answer: (a) All four in parallel: \(R=\dfrac14\ \Omega\).
(b) Two in series (2 \(\Omega\)) in parallel with the other two in series (2 \(\Omega\)): \(R=\dfrac{2\times2}{2+2}=1\ \Omega\).
(c) Three in parallel (\(\dfrac13\ \Omega\)) in series with the fourth: \(R=\dfrac13+1=1\dfrac13\ \Omega\).
(d) Two in parallel (\(\dfrac12\ \Omega\)) in series with the other two in series (2 \(\Omega\)): \(R=\dfrac12+2=2\dfrac12\ \Omega\).

For the circuit shown in the figure below, find (a) the value of the supply voltage \(V\) and (b) the value of current \(I\).

Answer: (a) \(V=I_2R_2=3\times20=60\) V.
(b) \(I_1=\dfrac{60}{10}=6\) A, \(I_3=\dfrac{60}{60}=1\) A, so \(I=I_1+I_2+I_3=6+3+1=10\) A.

For the series-parallel arrangement shown in the figure below, find (a) the supply current, (b) the current flowing through each resistor and (c) the p.d. across each resistor.

Answer: The 6 \(\Omega\) and 2 \(\Omega\) resistors are in parallel: \(R_{23}=\dfrac{6\times2}{6+2}=1.5\ \Omega\). Total resistance \(R=2.5+1.5+4=8\ \Omega\).
(a) Supply current \(I=\dfrac{200}{8}=25\) A.
(b) \(R_1\) and \(R_4\) carry the full 25 A. The p.d. across the parallel pair is \(25\times1.5=37.5\) V, so \(I_2=\dfrac{37.5}{6}=6.25\) A and \(I_3=\dfrac{37.5}{2}=18.75\) A.
(c) p.d. across \(R_1=25\times2.5=62.5\) V; across \(R_4=25\times4=100\) V; across \(R_2\) and \(R_3\) = 37.5 V each.

For the arrangement shown in the figure below, find the current \(I_x\).

Answer: Start from the far end. The 2 \(\Omega\) and 8 \(\Omega\) resistors are in parallel: \(\dfrac{2\times8}{2+8}=1.6\ \Omega\). In series with 1.4 \(\Omega\): \(1.4+1.6=3\ \Omega\). This is in parallel with 9 \(\Omega\): \(\dfrac{9\times3}{9+3}=2.25\ \Omega\). Total resistance \(=2+2.25=4.25\ \Omega\).
\(I=\dfrac{17}{4.25}=4\) A. The p.d. across the 9 \(\Omega\) resistor is \(4\times2.25=9\) V, so \(I_1=\dfrac{9}{3}=3\) A. The p.d. across the 2 \(\Omega\)–8 \(\Omega\) pair is \(3\times1.6=4.8\) V, so \(I_x=\dfrac{4.8}{8}=0.6\) A.

Four resistances of 16 \(\Omega\) each are connected in parallel. Four such combinations are connected in series. What is the total resistance?

Answer: One parallel combination: \(\dfrac{16}{4}=4\ \Omega\). Four such combinations in series: \(4\times4=16\ \Omega\).

A battery of 9 V is connected in series with resistors of 0.2 \(\Omega\), 0.3 \(\Omega\), 0.4 \(\Omega\), 0.5 \(\Omega\) and 12 \(\Omega\). How much current would flow through the 12 \(\Omega\) resistor?

Answer: \(R=0.2+0.3+0.4+0.5+12=13.4\ \Omega\). In a series circuit the same current flows through every resistor: \(I=\dfrac{9}{13.4}=0.6716\text{ A}\).

An electric bulb of resistance 20 \(\Omega\) and a resistance wire of 4 \(\Omega\) are connected in series with a 6 V battery. Draw the circuit diagram and calculate: (a) total resistance of the circuit (b) current through the circuit (c) potential difference across the electric bulb (d) potential difference across the resistance wire.

Answer: (a) \(R=20+4=24\ \Omega\).
(b) \(I=\dfrac{6}{24}=0.25\) A.
(c) p.d. across the bulb \(=0.25\times20=5\) V.
(d) p.d. across the wire \(=0.25\times4=1\) V.

Find the equivalent resistance of the given circuit.

Answer: Five equal resistors in parallel: \(R=\dfrac{20}{5}=4\ \Omega\).

Find the value of \(V_T\) in the given circuit.

Answer: The resistors are in series: \(R=10+20+30=60\ \Omega\). \(V_T=IR=1\times60=60\) V.

Find the voltage across each resistance in the given circuit.

Answer: \(R=10+10=20\ \Omega\), \(I=\dfrac{30}{20}=1.5\) A. Voltage across each resistor \(=1.5\times10=15\) V.

A potential difference of 4 V is applied to two resistors of 6 \(\Omega\) and 2 \(\Omega\) connected in series. Calculate: (a) the combined resistance (b) the current flowing (c) the potential difference across the 6 \(\Omega\) resistor.

Answer: (a) \(R=6+2=8\ \Omega\).
(b) \(I=\dfrac48=0.5\) A.
(c) \(V=0.5\times6=3\) V.

Resistors of 20 \(\Omega\), 20 \(\Omega\) and 30 \(\Omega\) are connected in parallel. What resistance must be added in series with the combination to obtain a total resistance of 10 \(\Omega\)?

Answer: \(\dfrac1R=\dfrac1{20}+\dfrac1{20}+\dfrac1{30}=\dfrac{3+3+2}{60}=\dfrac8{60}\Rightarrow R=7.5\ \Omega\). The resistance to be added in series \(=10-7.5=2.5\ \Omega\).

If four identical lamps are connected in parallel and the combined resistance is 100 \(\Omega\), find the resistance of one lamp.

Answer: For \(n\) identical resistors in parallel \(R_p=\dfrac Rn\), so \(R=4\times100=400\ \Omega\).

Find the current across each resistance and the total current flowing in the given circuit.

Answer: In parallel each resistor has the full 10 V: \(I_1=\dfrac{10}5=2\) A, \(I_2=\dfrac{10}{10}=1\) A, \(I_3=\dfrac{10}{20}=0.5\) A. Total current \(I=2+1+0.5=3.5\) A.

In the given circuit, the resistances \(R_1\) and \(R_2\) are connected in parallel. (i) Find the value of \(V_T\). (ii) Find the total current and equivalent resistance in the circuit if resistance \(R_2=10\ \Omega\).

Answer: (i) \(V_T=I_1R_1=5\times10=50\) V.
(ii) With \(R_2=10\ \Omega\): \(I_2=\dfrac{50}{10}=5\) A, so the total current \(I=5+5=10\) A and the equivalent resistance \(R=\dfrac{50}{10}=5\ \Omega\) (also \(\dfrac{10\times10}{10+10}=5\ \Omega\)).

In the given circuit, (i) find the equivalent resistance of the circuit and the total current flowing in the circuit. (ii) Find the current flowing through \(R_2\) and \(R_3\). (iii) Find the voltage across each resistance.

Answer: (i) \(R_2\) and \(R_3\) in parallel: \(\dfrac{4\times4}{4+4}=2\ \Omega\). Total \(R=2+2=4\ \Omega\), so \(I=\dfrac{10}4=2.5\) A.
(ii) The two 4 \(\Omega\) resistors are equal, so the current divides equally: \(I_2=I_3=1.25\) A.
(iii) \(V_1=2.5\times2=5\) V; \(V_2=V_3=1.25\times4=5\) V.

In the given circuit, (i) find the equivalent resistance and the total current flowing in the circuit. (ii) Find the voltage and current across each resistance in the circuit.

Answer: (i) \(R_{23}=\dfrac{10\times20}{10+20}=\dfrac{20}{3}\ \Omega\). Total \(R=30+\dfrac{20}3=\dfrac{110}{3}=36.67\ \Omega\). \(I=\dfrac{40}{110/3}=\dfrac{12}{11}=1.09\) A.
(ii) \(R_1\): \(I_1=1.09\) A, \(V_1=1.09\times30=32.7\) V. The p.d. across the parallel pair \(=40-32.7=7.27\) V, so \(V_2=V_3=7.27\) V, \(I_2=\dfrac{7.27}{10}=0.727\) A and \(I_3=\dfrac{7.27}{20}=0.364\) A.

In the given circuit, (i) find the equivalent resistance and the total current flowing in the circuit. (ii) Find the voltage and current across each resistance in the circuit.

Answer: (i) \(R_2\) and \(R_3\) are in series: \(20+30=50\ \Omega\). This is in parallel with \(R_1\): \(R=\dfrac{10\times50}{10+50}=8.33\ \Omega\). \(I=\dfrac{40}{8.33}=4.8\) A.
(ii) \(R_1\): \(V_1=40\) V, \(I_1=\dfrac{40}{10}=4\) A. The series branch carries \(I=\dfrac{40}{50}=0.8\) A: \(V_2=0.8\times20=16\) V and \(V_3=0.8\times30=24\) V.

Find the current through the 10 ohm resistor for the following circuit.

Answer: The 4 A current source forces 4 A into the parallel combination of the other two branches. The 5 \(\Omega\) and 3 \(\Omega\) resistors are in series: \(8\ \Omega\). The 4 A divides between 10 \(\Omega\) and 8 \(\Omega\) in inverse proportion to resistance: \(I_{10}=4\times\dfrac{8}{10+8}=1.78\) A. (The 2 \(\Omega\) resistor in series with the current source does not change this current.)

In the given circuit, (i) find the equivalent resistance and the total current flowing in the circuit. (ii) Find the voltage and current across each resistance in the circuit.

Answer: (i) \(R_3+R_4=20\ \Omega\) in parallel with \(R_2=20\ \Omega\): \(10\ \Omega\). Total \(R=5+10+10=25\ \Omega\). \(I=\dfrac{100}{25}=4\) A.
(ii) \(R_1\): 4 A, 20 V. \(R_5\): 4 A, 40 V. The parallel part has \(100-20-40=40\) V across it: \(R_2\): 40 V, 2 A. \(R_3\) and \(R_4\) (in series, 2 A): 2 A each, 20 V each.

Find the equivalent resistance of the following circuits:

(i)

Answer: \(R_{23}=15\ \Omega\); total \(=10+15=25\ \Omega\).

(ii)

Answer: \(R_2\) and \(R_3\) in series: \(40\ \Omega\); in parallel with \(R_4\): \(\dfrac{20\times40}{20+40}=13.33\ \Omega\). Total \(=30+13.33+25=68.33\ \Omega\).

(iii)

Answer: \(\dfrac1{R_p}=\dfrac1{10}+\dfrac14+\dfrac18=0.475\Rightarrow R_p=2.105\ \Omega\). Total \(=7+2+2.105=11.1\ \Omega\).

(iv)

Answer: \(\dfrac1{R_p}=\dfrac13+\dfrac16+\dfrac1{18}=\dfrac{6+3+1}{18}\Rightarrow R_p=1.8\ \Omega\). Total \(=1+2.2+1.8+4=9\ \Omega\).

(v)

Answer: \(\dfrac{6\times18}{6+18}=4.5\ \Omega\). Total \(=2+4.5+1.5=8\ \Omega\).

(vi)

Answer: First group: \(\dfrac{15}2=7.5\ \Omega\); second group: \(\dfrac{15}3=5\ \Omega\). Total \(=15+7.5+5=27.5\ \Omega\).

For the circuit shown in the figure below, determine the value of \(V_1\). If the total circuit resistance is 36 \(\Omega\), determine the supply current and the value of resistors \(R_1\), \(R_2\) and \(R_3\).

Answer: \(V_1=18-5-3=10\) V. Supply current \(I=\dfrac{18}{36}=0.5\) A.
\(R_1=\dfrac{10}{0.5}=20\ \Omega\), \(R_2=\dfrac5{0.5}=10\ \Omega\), \(R_3=\dfrac3{0.5}=6\ \Omega\). (Check: \(20+10+6=36\ \Omega\).)

When the switch in the circuit in the figure below is closed, the reading on voltmeter 1 is 30 V and that on voltmeter 2 is 10 V. Determine the reading on the ammeter and the value of resistor \(R_x\).

Answer: Voltmeter 1 measures the supply voltage, 30 V, and voltmeter 2 measures the p.d. across \(R_x\), 10 V. The p.d. across the 5 \(\Omega\) resistor \(=30-10=20\) V, so the current (ammeter reading) is \(I=\dfrac{20}{5}=4\) A. Then \(R_x=\dfrac{10}{4}=2.5\ \Omega\).

A potential difference of 6 V is applied to two resistors of 3 \(\Omega\) and 6 \(\Omega\) connected in parallel. Calculate: (a) the combined resistance (b) the current flowing in the main circuit (c) the current flowing in the 3 \(\Omega\) resistor.

Answer: (a) \(R=\dfrac{3\times6}{3+6}=2\ \Omega\).
(b) \(I=\dfrac62=3\) A.
(c) \(I_3=\dfrac63=2\) A.

Three resistors are connected as shown in the diagram: a 5 \(\Omega\) resistor between A and B, followed by 10 \(\Omega\) and 15 \(\Omega\) resistors in parallel between B and C. Through the 5 \(\Omega\) resistor, a current of 1 A is flowing. (i) What is the current through the other two resistors? (ii) What is the p.d. across AB and across AC? (iii) What is the total resistance?

Answer: The parallel pair has \(R_{BC}=\dfrac{10\times15}{10+15}=6\ \Omega\). The total current 1 A flows into this pair, so \(V_{BC}=1\times6=6\) V.
(i) \(I_{10}=\dfrac6{10}=0.6\) A and \(I_{15}=\dfrac6{15}=0.4\) A.
(ii) \(V_{AB}=1\times5=5\) V; \(V_{AC}=5+6=11\) V.
(iii) Total resistance \(=5+6=11\ \Omega\).

For the circuit shown in the diagram below, what is the value of: (i) the current through the 6 \(\Omega\) resistor? (ii) the p.d. across the 12 \(\Omega\) resistor?

Answer: Each branch is across the 4 V battery.
(i) Upper branch: \(6+3=9\ \Omega\), so the current through the 6 \(\Omega\) resistor is \(\dfrac49=0.44\) A.
(ii) Lower branch: \(12+3=15\ \Omega\), so \(I=\dfrac4{15}=0.267\) A and the p.d. across the 12 \(\Omega\) resistor \(=0.267\times12=3.2\) V.

Calculate the total resistance of the circuit below, as seen by the voltage source.

Answer: \(R_4\) and \(R_5\) in parallel: \(3\) k\(\Omega\). The branch \(R_3+3+R_6=1+3+4=8\) k\(\Omega\). \(R_1\) and \(R_2\) in parallel with this branch: \(\dfrac1R=\dfrac1{16}+\dfrac1{16}+\dfrac18=\dfrac14\) (in k\(\Omega\)\(^{-1}\)). Total resistance \(=4\) k\(\Omega\).

What is the resistance between A and B in the given figure below?

Answer: The branch \(R_3+R_4+R_5=3+4+3=10\) k\(\Omega\). In parallel with 5 k\(\Omega\) and 10 k\(\Omega\): \(\dfrac1R=\dfrac15+\dfrac1{10}+\dfrac1{10}=\dfrac4{10}\Rightarrow R=2.5\) k\(\Omega\).

What is the resistance between A and B in the given figure below?

Answer: \(R_1\parallel R_2=\dfrac{1\times10}{11}=0.909\) k\(\Omega\); \(R_3\parallel R_4\parallel R_5=\dfrac{10}{3}=3.333\) k\(\Omega\); \(R_6\parallel R_7=0.5\) k\(\Omega\); \(R_8=0.091\) k\(\Omega\). These groups are in series: \(R_{AB}=0.909+3.333+0.5+0.091=4.83\) k\(\Omega\).

Resistances of 4 \(\Omega\) and 12 \(\Omega\) are connected in parallel across a 9 V battery. Determine (a) the equivalent circuit resistance, (b) the supply current, and (c) the current in each resistor.

Answer: (a) \(R=\dfrac{4\times12}{4+12}=3\ \Omega\).
(b) \(I=\dfrac93=3\) A.
(c) \(I_4=\dfrac94=2.25\) A and \(I_{12}=\dfrac9{12}=0.75\) A.

Three identical lamps A, B and C are connected in series across a 150 V supply. State (a) the voltage across each lamp, and (b) the effect of lamp C failing.

Answer: (a) The lamps are identical, so the 150 V is shared equally: each lamp has \(\dfrac{150}{3}=50\) V.
(b) If lamp C fails (its filament breaks), the series circuit is broken, so no current flows and lamps A and B also go out.

The p.d.'s measured across three resistors connected in series are 5 V, 7 V and 10 V, and the supply current is 2 A. Determine (a) the supply voltage, (b) the total circuit resistance and (c) the values of the three resistors.

Answer: (a) \(V=5+7+10=22\) V.
(b) \(R=\dfrac{22}{2}=11\ \Omega\).
(c) \(R_1=\dfrac52=2.5\ \Omega\), \(R_2=\dfrac72=3.5\ \Omega\), \(R_3=\dfrac{10}{2}=5\ \Omega\).

If three identical lamps are connected in parallel and the combined resistance is 150 \(\Omega\), find the resistance of one lamp.

Answer: \(R_p=\dfrac R3\Rightarrow R=3\times150=450\ \Omega\).

Heating effect of electric current

If the electric circuit is purely resistive, that is, a configuration of resistors only connected to a battery, the source energy continually gets dissipated entirely in the form of heat. This is known as the heating effect of electric current. This effect is utilised in devices such as the electric heater, electric iron, etc.

Work done \(W=Q\times V\). Since \(I=\dfrac Qt\), \(Q=It\); and since \(R=\dfrac VI\), \(V=IR\). So

\[W=(It)(IR)=I^2Rt\]

Heat produced \(H=I^2Rt\) joules. This is known as Joule's law of heating.

The law implies that the heat produced in a resistor is (i) directly proportional to the square of the current for a given resistance (\(I^2\)), (ii) directly proportional to the resistance for a given current (\(R\)), and (iii) directly proportional to the time for which the current flows through the resistor (\(t\)).

Intext questions (page 218)

Why does the cord of an electric heater not glow while the heating element does?

Answer: The heating element of an electric heater is a resistor. The amount of heat produced by it is proportional to its resistance. The resistance of the element of an electric heater is very high. As current flows through the heating element, it becomes too hot and glows red. On the other hand, the resistance of the cord is low. It does not become red when current flows through it.

Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Answer: The heat produced is \(H=VIt\), where \(V=50\) V and \(t=1\text{ h}=3600\) s.
Current \(I=\dfrac{Q}{t}=\dfrac{96000}{3600}=\dfrac{80}{3}\text{ A}\).
\(H=50\times\dfrac{80}{3}\times3600=4.8\times10^{6}\text{ J}\)

An electric iron of resistance 20 \(\Omega\) takes a current of 5 A. Calculate the heat developed in 30 s.

Answer: \(V=IR=5\times20=100\) V. \(H=VIt=100\times5\times30=1.5\times10^{4}\text{ J}\) (or \(H=I^2Rt=25\times20\times30=15000\) J)

Practical applications of the heating effect of electric current

The flowing of current through a conductor produces heat. We use this heat in our daily life in the electric iron, electric bulb, electric fuse, electric heater and more.

1. Electric iron: in an iron the upper part is grooved and a coil is placed in this groove. Mica, an insulator, is placed between the metal part and the coil, so that there is no electrical connection between them. Mica is a bad conductor of electricity but a good conductor of heat. Due to the flow of current through the coil, it becomes heated and the heat is transferred to the metal part through the mica. Finally the metal part becomes heated.
2. Electric bulb: we see a thick metallic wire in the bulb. It is made of the metal tungsten. The tungsten wire is kept sealed in a glass bulb, which is filled with neutral gas or vacuum (neutral gases are commonly used now). The tungsten wire is known as the filament. When current flows through the filament it becomes heated and emits light. Here the heating effect of an electric current is used as a source of light.
3. Electric heater: in an electric heater a coil of a high-resistance material like nichrome is used. The coil is wound in grooves on ceramic or china clay. When electric current flows through the coil it becomes heated, and due to its high resistance the coil glows red.
4. Electric fuse: it protects circuits and appliances by stopping the flow of any unduly high electric current. The fuse is placed in series with the device. It consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminium, copper, iron, lead, etc. If a current larger than the specified value flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit. The fuse wire is usually encased in a cartridge of porcelain or similar material with metal ends. The fuses used for domestic purposes are rated as 1 A, 2 A, 3 A, 5 A, 10 A, etc. For an electric iron which consumes 1 kW electric power when operated at 220 V, a current of \(\dfrac{1000}{220}\) A, that is, 4.54 A, will flow in the circuit. In this case, a 5 A fuse must be used.

Electric power

It is the electrical work done per unit time.

\[P=\frac{W}{t}\]

where \(W\) = work done and \(t\) = time. The S.I. unit of power is the watt (W). If \(W=1\) joule and \(t=1\) second, then power \(=\dfrac{1\text{ joule}}{1\text{ second}}=1\) watt.

One watt: the rate of working of 1 joule per second is the power of 1 watt.

Electric energy

It is the total work done by a current in an electric circuit. Electric power \(=\dfrac{\text{work done by the electric circuit}}{\text{time taken}}\), so

\[\text{work done}=\text{electric power}\times\text{time taken},\qquad E=Pt\]

The electrical energy consumed by an electrical appliance is given by the product of its power rating and the time for which it is used. The unit of electrical energy is the watt-hour (Wh). If \(P=1\) watt and \(t=1\) hour, then \(E=1\) watt \(\times\) 1 hour = 1 watt-hour.

One watt-hour: it is the amount of electrical energy consumed when an electrical appliance of 1 watt power is used for an hour.

The commercial unit of electrical energy is the kilowatt-hour (kWh).

One kilowatt-hour: it is the amount of electrical energy consumed when an electrical appliance having a power rating of 1 kilowatt is used for 1 hour.

\[1\text{ kilowatt-hour}=36{,}00{,}000\text{ joules}=3.6\times10^{6}\text{ J}\]

Intext questions (page 220)

What determines the rate at which energy is delivered by a current?

Answer: The rate of consumption of electric energy in an electric appliance is called electric power. Hence, the rate at which energy is delivered by a current is the power of the appliance.

An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Answer: Power \(P=VI=220\times5=1100\text{ W}\).
Time \(t=2\text{ h}=2\times60\times60=7200\) s.
Energy consumed \(=Pt=1100\times7200=7.92\times10^{6}\text{ J}\).
Therefore the power of the motor is 1100 W and the energy consumed is \(7.92\times10^{6}\) J.

Numerical problems

What will be the current drawn by an electric bulb of 40 W when it is connected to a source of 220 V?

Answer: \(I=\dfrac PV=\dfrac{40}{220}=0.1818\text{ A}\)

A bulb is rated as 250 V; 0.4 A. Find its power and resistance.

Answer: \(P=VI=250\times0.4=100\text{ W}\). \(R=\dfrac VI=\dfrac{250}{0.4}=625\ \Omega\)

An electric bulb is connected to a 220 V power supply line. If the bulb draws a current of 0.5 A, calculate the power of the bulb.

Answer: \(P=VI=220\times0.5=110\text{ W}\)

An electric bulb is connected to a 250 V generator. The current is 0.50 A. What is the power of the bulb?

Answer: \(P=VI=250\times0.5=125\text{ W}\)

What current will be taken by a 920 W appliance if the supply voltage is 230 V?

Answer: \(I=\dfrac PV=\dfrac{920}{230}=4\text{ A}\)

When an electric lamp is connected to a 12 V battery, it draws a current of 0.5 A. Find the power of the lamp.

Answer: \(P=VI=12\times0.5=6\text{ W}\)

Calculate the power used in the 2 \(\Omega\) resistor in each of the following: (i) a 6 V battery in series with 1 \(\Omega\) and 2 \(\Omega\) resistors (ii) a 4 V battery in parallel with 12 \(\Omega\) and 2 \(\Omega\) resistors.

Answer: (i) In series \(R=1+2=3\ \Omega\), \(I=\dfrac63=2\) A through the 2 \(\Omega\) resistor. \(P=I^2R=2^2\times2=8\) W.
(ii) In parallel each resistor has the full 4 V. \(P=\dfrac{V^2}{R}=\dfrac{4^2}{2}=8\) W.

A 100 W electric light bulb is connected to a 250 V supply. Determine (a) the current flowing in the bulb, and (b) the resistance of the bulb.

Answer: (a) \(I=\dfrac PV=\dfrac{100}{250}=0.4\text{ A}\).
(b) \(R=\dfrac VI=\dfrac{250}{0.4}=625\ \Omega\)

Calculate the power dissipated when a current of 4 mA flows through a resistance of 5 k\(\Omega\).

Answer: \(P=I^2R=(4\times10^{-3})^2\times5000=0.08\text{ W}=80\text{ mW}\)

An electric kettle has a resistance of 30 \(\Omega\). What current will flow when it is connected to a 240 V supply? Find also the power rating of the kettle.

Answer: \(I=\dfrac VR=\dfrac{240}{30}=8\text{ A}\). \(P=VI=240\times8=1920\text{ W}\)

A current of 5 A flows in the winding of an electric motor, the resistance of the winding being 100 \(\Omega\). Determine (a) the p.d. across the winding, and (b) the power dissipated by the coil.

Answer: (a) \(V=IR=5\times100=500\text{ V}\).
(b) \(P=I^2R=25\times100=2500\text{ W}\)

The current/voltage relationship for two resistors A and B is as shown in the figure below. Determine the value of the resistance of each resistor and also find the power dissipated through each resistor.

Answer: From the graph, resistor A carries 20 mA at 20 V: \(R_A=\dfrac{20}{20\times10^{-3}}=1000\ \Omega=1\) k\(\Omega\). Resistor B carries 5 mA at 16 V: \(R_B=\dfrac{16}{5\times10^{-3}}=3200\ \Omega=3.2\) k\(\Omega\).
Power dissipated at 20 V: \(P_A=VI=20\times0.02=0.4\) W and \(P_B=\dfrac{V^2}{R_B}=\dfrac{20^2}{3200}=0.125\) W.

The hot resistance of a 240 V filament lamp is 960 \(\Omega\). Find the current taken by the lamp and its power rating.

Answer: \(I=\dfrac VR=\dfrac{240}{960}=0.25\text{ A}\). \(P=VI=240\times0.25=60\text{ W}\)

A 12 V battery is connected across a load having a resistance of 40 \(\Omega\). Determine the current flowing in the load, the power consumed and the energy dissipated in 2 minutes.

Answer: \(I=\dfrac{12}{40}=0.3\text{ A}\); \(P=VI=12\times0.3=3.6\text{ W}\); energy in 2 min \(=Pt=3.6\times120=432\text{ J}\)

A source of e.m.f. 15 V supplies a current of 2 A for six minutes. How much energy is provided in this time?

Answer: \(E=VIt=15\times2\times(6\times60)=10800\text{ J}\)

Electrical equipment in an office takes a current of 13 A from a 240 V supply. Estimate the cost per week of electricity if the equipment is used for 30 hours each week and 1 kWh of energy costs 7p.

Answer: \(P=VI=240\times13=3120\) W \(=3.12\) kW. Energy per week \(=3.12\times30=93.6\) kWh. Cost \(=93.6\times7=655.2\) p (about £6.55).

An electric heater consumes 3.6 MJ when connected to a 250 V supply for 40 minutes. Find the power rating of the heater and the current taken from the supply.

Answer: \(P=\dfrac Et=\dfrac{3.6\times10^6}{40\times60}=1500\text{ W}\). \(I=\dfrac PV=\dfrac{1500}{250}=6\text{ A}\)

Determine the power dissipated by the element of an electric fire of resistance 20 \(\Omega\) when a current of 10 A flows through it. If the fire is on for 6 hours, determine the energy used and the cost if 1 unit of electricity costs 7p.

Answer: \(P=I^2R=10^2\times20=2000\) W \(=2\) kW. Energy used \(=2\times6=12\) kWh (12 units). Cost \(=12\times7=84\) p.

A business uses two 3 kW fires for an average of 20 hours each per week, and six 150 W lights for 30 hours each per week. If the cost of electricity is 7p per unit, determine the weekly cost of electricity to the business.

Answer: Fires: \(2\times3\times20=120\) kWh. Lights: \(6\times0.15\times30=27\) kWh. Total \(=147\) kWh. Cost \(=147\times7=1029\) p (about £10.29).

If 5 A, 10 A and 13 A fuses are available, state which is most appropriate for the following appliances, which are both connected to a 240 V supply: (a) an electric toaster having a power rating of 1 kW (b) an electric fire having a power rating of 3 kW.

Answer: (a) \(I=\dfrac{1000}{240}=4.167\text{ A}\), so a 5 A fuse.
(b) \(I=\dfrac{3000}{240}=12.5\text{ A}\), so a 13 A fuse.

The hot resistance of a 250 V filament lamp is 625 \(\Omega\). Determine the current taken by the lamp and its power rating.

Answer: \(I=\dfrac{250}{625}=0.4\text{ A}\). \(P=VI=250\times0.4=100\text{ W}\)

Determine the resistance of a coil connected to a 150 V supply when a current of (a) 75 mA (b) 300 \(\mu\)A flows through it. Determine the power dissipated in each case.

Answer: (a) \(R=\dfrac{150}{75\times10^{-3}}=2000\ \Omega=2\text{ k}\Omega\); \(P=VI=150\times0.075=11.25\text{ W}\).
(b) \(R=\dfrac{150}{300\times10^{-6}}=5\times10^{5}\ \Omega=500\text{ k}\Omega\); \(P=VI=150\times300\times10^{-6}=0.045\text{ W}=45\text{ mW}\).

Determine the resistance of an electric fire which takes a current of 12 A from a 240 V supply. Find also the power rating of the fire and the energy used in 20 h.

Answer: \(R=\dfrac{240}{12}=20\ \Omega\). \(P=VI=240\times12=2880\) W \(=2.88\) kW. Energy in 20 h \(=2.88\times20=57.6\) kWh.

Determine the power dissipated when a current of 10 mA flows through an appliance having a resistance of 8 k\(\Omega\).

Answer: \(P=I^2R=(10\times10^{-3})^2\times8000=0.8\text{ W}\)

85.5 J of energy are converted into heat in nine seconds. What power is dissipated?

Answer: \(P=\dfrac Et=\dfrac{85.5}{9}=9.5\text{ W}\)

A current of 4 A flows through a conductor and 10 W is dissipated. What p.d. exists across the ends of the conductor?

Answer: \(V=\dfrac PI=\dfrac{10}{4}=2.5\text{ V}\)

Find the power dissipated when: (a) a current of 5 mA flows through a resistance of 20 k\(\Omega\), (b) a voltage of 400 V is applied across a 120 k\(\Omega\) resistor, (c) the voltage applied to a resistor is 10 kV and the current flow is 4 mA.

Answer: (a) \(P=I^2R=(5\times10^{-3})^2\times20000=0.5\text{ W}\).
(b) \(P=\dfrac{V^2}{R}=\dfrac{400^2}{120000}=1.333\text{ W}\).
(c) \(P=VI=10000\times4\times10^{-3}=40\text{ W}\).

A battery of e.m.f. 15 V supplies a current of 2 A for 5 min. How much energy is supplied in this time?

Answer: \(E=VIt=15\times2\times300=9000\text{ J}\)

In a household during a particular week three 2 kW fires are used on average 25 h each and eight 100 W light bulbs are used on average 35 h each. Determine the cost of electricity for the week if 1 unit of electricity costs 7p.

Answer: Fires: \(3\times2\times25=150\) kWh. Bulbs: \(8\times0.1\times35=28\) kWh. Total \(=178\) kWh. Cost \(=178\times7=1246\) p (about £12.46).

Calculate the power dissipated by the element of an electric fire of resistance 30 \(\Omega\) when a current of 10 A flows in it. If the fire is on for 30 hours in a week, determine the energy used. Determine also the weekly cost of energy if electricity costs 7.2p per unit.

Answer: \(P=I^2R=100\times30=3000\) W \(=3\) kW. Energy used \(=3\times30=90\) kWh. Cost \(=90\times7.2=648\) p (about £6.48).

A television set having a power rating of 120 W and an electric lawnmower of power rating 1 kW are both connected to a 240 V supply. If 3 A, 5 A and 10 A fuses are available, state which is the most appropriate for each appliance.

Answer: Television: \(I=\dfrac{120}{240}=0.5\text{ A}\), so a 3 A fuse.
Lawnmower: \(I=\dfrac{1000}{240}=4.167\text{ A}\), so a 5 A fuse.

For a heater rated at 4 kW and 220 V, calculate: (a) the current (b) the resistance of the heater (c) the energy consumed in 2 hours and (d) the cost if 1 kWh is priced at Rs. 4.60.

Answer: (a) \(I=\dfrac PV=\dfrac{4000}{220}=18.18\text{ A}\).
(b) \(R=\dfrac{V^2}{P}=\dfrac{220^2}{4000}=12.1\ \Omega\).
(c) \(E=Pt=4\times2=8\) kWh.
(d) Cost \(=8\times4.60=\) Rs. 36.80.

A radio set of 60 W runs for 50 hrs. How much electrical energy is consumed?

Answer: \(E=Pt=60\times50=3000\) Wh \(=3\) kWh (3 units).

A current of 4 A flows through a 12 V car headlight bulb for 10 min. How much energy transfer occurs during this time?

Answer: \(E=VIt=12\times4\times600=28800\text{ J}\)

Calculate the energy transferred by a 5 A current flowing through a resistor of 2 \(\Omega\) for 30 min.

Answer: \(E=I^2Rt=5^2\times2\times1800=90000\text{ J}\)

A bulb is rated at 200 V–100 W. What is its resistance? Five such bulbs burn for 4 hrs. What is the electrical energy consumed? Calculate the cost if the rate is Rs. 4.60 per unit.

Answer: \(R=\dfrac{V^2}{P}=\dfrac{200^2}{100}=400\ \Omega\).
Energy \(=5\times100\text{ W}\times4\text{ h}=2000\text{ Wh}=2\) kWh. Cost \(=2\times4.60=\) Rs. 9.20.

A refrigerator having a power rating of 350 W operates for 10 hours a day. Calculate the cost of electrical energy to operate it for a month of 30 days. The rate of electrical energy is Rs. 3.40 per kWh.

Answer: Energy \(=0.35\times10\times30=105\) kWh. Cost \(=105\times3.40=\) Rs. 357.

What will be the current drawn by an electric bulb of 40 W when it is connected to a source of 220 V?

Answer: \(I=\dfrac PV=\dfrac{40}{220}=0.1818\text{ A}\)

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, find the power consumed.

Answer: \(R=\dfrac{V^2}{P}=\dfrac{220^2}{100}=484\ \Omega\) (taken as constant). \(P^{\prime}=\dfrac{110^2}{484}=25\text{ W}\)

An electric heater draws a current of 10 A from a 220 V supply. What is the cost of using the heater for 5 hrs every day for 30 days if the cost of 1 unit is Rs. 5.20?

Answer: \(P=VI=220\times10=2200\) W \(=2.2\) kW. Energy \(=2.2\times5\times30=330\) kWh. Cost \(=330\times5.20=\) Rs. 1716.

In a house two 60 W electric bulbs are lighted for 4 hrs and three 100 W bulbs for 5 hrs every day. Calculate the electrical energy consumed in 30 days.

Answer: Per day: \(2\times60\times4+3\times100\times5=480+1500=1980\) Wh \(=1.98\) kWh. In 30 days: \(1.98\times30=59.4\) kWh.

An electric motor takes a current of 5 A from a 220 V supply line. Calculate the power of the motor and the electrical energy consumed by it in 2 hrs.

Answer: \(P=VI=220\times5=1100\) W. Energy \(=1.1\times2=2.2\) kWh (\(=7.92\times10^6\) J).

An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220 V. What are the current and the resistance in each case?

Answer: Maximum heating: \(I=\dfrac{840}{220}=3.818\text{ A}\), \(R=\dfrac{V^2}{P}=\dfrac{220^2}{840}=57.62\ \Omega\).
Minimum heating: \(I=\dfrac{360}{220}=1.636\text{ A}\), \(R=\dfrac{220^2}{360}=134.4\ \Omega\).

An electric refrigerator rated 400 W operates 8 hours/day. What is the cost of the energy to operate it for 30 days at Rs. 3.00 per kWh?

Answer: Energy \(=0.4\times8\times30=96\) kWh. Cost \(=96\times3.00=\) Rs. 288.

An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Answer: \(P=VI=220\times5=1100\) W. Energy \(=1100\times7200=7.92\times10^{6}\) J (\(=2.2\) kWh).

Two lamps, one rated 100 W at 220 V and the other 60 W at 220 V, are connected in parallel to the electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Answer: In parallel each lamp has 220 V. \(I_1=\dfrac{100}{220}\) A and \(I_2=\dfrac{60}{220}\) A. Current from the line \(=\dfrac{100+60}{220}=0.7273\text{ A}\)

Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Answer: TV: \(250\times1=250\) Wh. Toaster: \(1200\times\dfrac{10}{60}=200\) Wh. The TV set uses more energy.

Two bulbs A and B are rated 100 W–120 V and 10 W–120 V respectively. They are connected across a 120 V source in series. Which will consume more energy?

Answer: \(R_A=\dfrac{120^2}{100}=144\ \Omega\), \(R_B=\dfrac{120^2}{10}=1440\ \Omega\). In series the same current flows through both, and \(P=I^2R\), so the bulb with the larger resistance uses more power. Bulb B (10 W–120 V) consumes more energy.

Two bulbs A and B are rated 100 W–120 V and 10 W–120 V respectively. They are connected across a 120 V source in series. Find the current in each bulb. Which will consume more energy?

Answer: \(R_A=144\ \Omega\), \(R_B=1440\ \Omega\). Total \(R=1584\ \Omega\). The same current flows in both: \(I=\dfrac{120}{1584}=0.07576\text{ A}\).
\(P_A=I^2R_A=0.8264\text{ W}\), \(P_B=I^2R_B=8.264\text{ W}\). Bulb B consumes more energy.

An electric kettle is rated at 230 V, 1000 W. What is the resistance of its element? What maximum current can pass through its element?

Answer: \(R=\dfrac{V^2}{P}=\dfrac{230^2}{1000}=52.9\ \Omega\). Maximum current \(I=\dfrac PV=\dfrac{1000}{230}=4.348\text{ A}\)

An electric geyser has the rating 1000 W, 220 V marked on it. What should be the minimum rating, in whole number, of a fuse wire that may be required for safe use with this geyser?

Answer: \(I=\dfrac PV=\dfrac{1000}{220}=4.545\text{ A}\). The minimum rating of the fuse in whole number is 5 A.

The mains power supply of a house is through a 5 A fuse. How many 100 W, 220 V bulbs can be used in this house at the correct voltage?

Answer: Maximum power \(=VI=220\times5=1100\) W. Number of bulbs \(=\dfrac{1100}{100}=11\).

An electrician puts a fuse of rating 5 A in that part of a domestic electrical circuit in which an electrical heater of rating 1.5 kW, 220 V is operating. What is likely to happen in this case and why? What change, if any, needs to be made?

Answer: Current drawn by the heater \(I=\dfrac{1500}{220}=6.818\text{ A}\), which is more than 5 A. So the fuse wire will get too hot and melt, breaking the circuit, and the heater will not work. A fuse of higher rating, such as 10 A, should be used.

Two bulbs of ratings 40 W–220 V and 60 W–220 V are connected in series and this combination is connected to a supply of 220 V. Calculate the current from the supply line.

Answer: \(R_1=\dfrac{220^2}{40}=1210\ \Omega\), \(R_2=\dfrac{220^2}{60}=806.7\ \Omega\). Total \(R=2017\ \Omega\). \(I=\dfrac{220}{R}=0.1091\text{ A}\)

Two bulbs have the ratings 40 W–200 V and 20 W–110 V. What is the ratio of their resistances?

Answer: \(R_1=\dfrac{200^2}{40}=1000\ \Omega\), \(R_2=\dfrac{110^2}{20}=605\ \Omega\). \(\dfrac{R_1}{R_2}=\dfrac{1000}{605}=\dfrac{200}{121}\), i.e. \(R_1:R_2=200:121\approx1.65:1\)

I can spend Rs. 9 per month (30 days) on electric light. If power costs 30 paise per kWh and I use 5 identical bulbs for 5 hours a day, what should be the power of each bulb?

Answer: Energy that can be bought \(=\dfrac{9}{0.30}=30\) kWh \(=30000\) Wh. Total bulb-hours \(=5\times5\times30=750\) h. Power of each bulb \(=\dfrac{30000}{750}=40\) W.

Compute the number of electrons passing per minute through an electric bulb of 60 W, 220 V.

Answer: \(I=\dfrac PV=\dfrac{60}{220}=0.2727\text{ A}\). Charge per minute \(Q=It=0.2727\times60=16.36\text{ C}\). Number of electrons \(n=\dfrac Qe=\dfrac{16.36}{1.6\times10^{-19}}=1.02\times10^{20}\)

If electrical energy costs Rs. 3 per unit, what is the total cost of leaving 4 light bulbs rated at 100 W each switched on for 8 hours?

Answer: Energy \(=4\times0.1\times8=3.2\) kWh. Cost \(=3.2\times3=\) Rs. 9.60.

An electric heater of resistance 8 \(\Omega\) draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.

Answer: Rate of heat production \(P=I^2R=15^2\times8=1800\text{ W}\) (J/s)

100 J of heat are produced each second in a 4 \(\Omega\) resistance. Find the potential difference across the resistor.

Answer: \(P=I^2R\Rightarrow I=\sqrt{\dfrac PR}=\sqrt{\dfrac{100}{4}}=5\text{ A}\). \(V=IR=5\times4=20\text{ V}\)

Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Answer: \(H=VIt=VQ=50\times96000=4.8\times10^{6}\text{ J}\)

An electric iron of resistance 20 \(\Omega\) takes a current of 5 A. Calculate the heat developed in 30 s.

Answer: \(H=I^2Rt=5^2\times20\times30=1.5\times10^{4}\text{ J}\)

A p.d. of 250 V is applied across a resistance of 500 \(\Omega\) in an electric iron. Calculate (i) the current (ii) the heat energy produced in joules in 10 s.

Answer: (i) \(I=\dfrac{250}{500}=0.5\text{ A}\).
(ii) \(H=I^2Rt=0.5^2\times500\times10=1250\text{ J}\)

Calculate the heat produced when 96000 C of charge is transferred in 1 hour through a p.d. of 50 V.

Answer: \(H=VQ=50\times96000=4.8\times10^{6}\text{ J}\)

A resistance of 40 \(\Omega\) and one of 60 \(\Omega\) are arranged in series across a 220 V supply. Find the heat in joules produced by this combination of resistances in half a minute.

Answer: \(R=40+60=100\ \Omega\), \(I=\dfrac{220}{100}=2.2\) A. \(H=I^2Rt=2.2^2\times100\times30=14520\text{ J}\)

When a current of 4 A passes through a certain resistor for 10 min, \(2.88\times10^4\) J of heat is produced. Calculate (a) the power of the resistor (b) the voltage across the resistor.

Answer: (a) \(P=\dfrac Ht=\dfrac{2.88\times10^4}{600}=48\text{ W}\).
(b) \(V=\dfrac PI=\dfrac{48}{4}=12\text{ V}\)

A heating coil has a resistance of 200 \(\Omega\). At what rate will heat be produced in it when a current of 2.5 A flows through it?

Answer: \(P=I^2R=2.5^2\times200=1250\text{ W}\) (J/s)

An electric heater of resistance 8 \(\Omega\) takes a current of 15 A from the mains supply line. Calculate the rate at which heat is developed in the heater.

Answer: \(P=I^2R=15^2\times8=1800\text{ W}\)

A resistance of 25 \(\Omega\) is connected to a 12 V battery. Calculate the heat energy in joules generated per minute.

Answer: \(H=\dfrac{V^2}{R}t=\dfrac{12^2}{25}\times60=345.6\text{ J}\)

How much heat will an instrument of 12 W produce in one minute if it is connected to a battery of 12 V?

Answer: \(H=Pt=12\times60=720\text{ J}\) (the voltage is not needed)

Exercise questions (page 221)

A piece of wire of resistance \(R\) is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is \(R'\), then the ratio \(R/R'\) is

1. \(\dfrac1{25}\)
2. \(\dfrac15\)
3. 5
4. 25

Answer: (d) 25. The resistance of a wire is proportional to its length, so each part has resistance \(\dfrac R5\). For five such parts in parallel \(\dfrac1{R^{\prime}}=5\times\dfrac5R=\dfrac{25}{R}\), so \(R^{\prime}=\dfrac R{25}\) and \(\dfrac R{R^{\prime}}=25\).

Which of the following terms does not represent electrical power in a circuit?

1. \(I^2R\)
2. \(IR^2\)
3. \(VI\)
4. \(V^2/R\)

Answer: (b) \(IR^2\). Power \(P=VI\). Using \(V=IR\), \(P=I^2R\); using \(I=\dfrac VR\), \(P=\dfrac{V^2}{R}\). Power cannot be expressed as \(IR^2\).

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be

1. 100 W
2. 75 W
3. 50 W
4. 25 W

Answer: (d) 25 W. \(R=\dfrac{V^2}{P}=\dfrac{220^2}{100}=484\ \Omega\) (taken as constant). At 110 V: \(P^{\prime}=\dfrac{110^2}{484}=25\) W.

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be

1. 1 : 2
2. 2 : 1
3. 1 : 4
4. 4 : 1

Answer: (c) 1 : 4. Let each wire have resistance \(R\). Series: \(R_s=2R\), so \(H_s=\dfrac{V^2t}{2R}\). Parallel: \(R_p=\dfrac R2\), so \(H_p=\dfrac{V^2t}{R/2}=\dfrac{2V^2t}{R}\). Ratio \(\dfrac{H_s}{H_p}=\dfrac{1/2}{2}=\dfrac14\).

How is a voltmeter connected in the circuit to measure the potential difference between two points?

Answer: To measure the potential difference between two points, a voltmeter should be connected in parallel across the two points.

A copper wire has diameter 0.5 mm and resistivity of \(1.6\times10^{-8}\ \Omega\) m. What will be the length of this wire to make its resistance 10 \(\Omega\)? How much does the resistance change if the diameter is doubled?

Answer: \(R=\rho\dfrac lA\) with \(A=\pi\left(\dfrac{0.0005}{2}\right)^2\). \(l=\dfrac{RA}{\rho}=\dfrac{10\times3.14\times(0.00025)^2}{1.6\times10^{-8}}=122.7\) m.
If the diameter is doubled to 1 mm, the area becomes 4 times, so \(R^{\prime}=\dfrac{1.6\times10^{-8}\times122.7}{3.14\times(0.0005)^2}=2.5\ \Omega\).
Therefore the length of the wire is 122.7 m and the new resistance is 2.5 \(\Omega\).

The values of current \(I\) flowing in a given resistor for the corresponding values of potential difference \(V\) across the resistor are given below. Plot a graph between \(V\) and \(I\) and calculate the resistance of that resistor.

Answer: The plot of voltage against current is the \(I\)–\(V\) characteristic. Voltage is plotted on the x-axis and current on the y-axis; the points lie on a straight line through the origin. The slope of the line is \(\dfrac1R\). Taking points A and B on the line, \(\text{slope}=\dfrac1R=\dfrac{BC}{AC}=\dfrac{2}{6.8}\Rightarrow R=\dfrac{6.8}{2}=3.4\ \Omega\). Therefore the resistance of the resistor is 3.4 \(\Omega\).

When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Answer: \(R=\dfrac VI=\dfrac{12}{2.5\times10^{-3}}=4.8\times10^3\ \Omega=4.8\) k\(\Omega\).

A battery of 9 V is connected in series with resistors of 0.2 \(\Omega\), 0.3 \(\Omega\), 0.4 \(\Omega\), 0.5 \(\Omega\) and 12 \(\Omega\), respectively. How much current would flow through the 12 \(\Omega\) resistor?

Answer: In a series circuit the same current flows through every resistor. \(R=0.2+0.3+0.4+0.5+12=13.4\ \Omega\), so \(I=\dfrac9{13.4}=0.671\) A.

How many 176 \(\Omega\) resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer: For \(x\) resistors of 176 \(\Omega\) in parallel, \(R=\dfrac{176}{x}\). By Ohm’s law \(R=\dfrac VI=\dfrac{220}{5}=44\ \Omega\). So \(x=\dfrac{176}{44}=4\). Four resistors of 176 \(\Omega\) are required.

Show how you would connect three resistors, each of resistance 6 \(\Omega\), so that the combination has a resistance of (i) 9 \(\Omega\), (ii) 4 \(\Omega\).

Answer: Three in series would give 18 \(\Omega\) and three in parallel 2 \(\Omega\), neither of which is wanted, so we use a mixed arrangement.
(i) Two resistors in parallel give \(\dfrac{6\times6}{6+6}=3\ \Omega\); the third 6 \(\Omega\) resistor in series with this gives \(6+3=9\ \Omega\).

Answer: (ii) Two resistors in series give \(12\ \Omega\); the third 6 \(\Omega\) resistor in parallel with this gives \(\dfrac{12\times6}{12+6}=4\ \Omega\).

Several electric bulbs designed to be used on a 220 V electric supply line are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of the 220 V line if the maximum allowable current is 5 A?

Answer: Resistance of each bulb \(R_1=\dfrac{V^2}{P}=\dfrac{220^2}{10}=4840\ \Omega\). Total resistance allowed \(R=\dfrac VI=\dfrac{220}{5}=44\ \Omega\). For \(x\) bulbs in parallel \(R=\dfrac{R_1}{x}\), so \(x=\dfrac{4840}{44}=110\). 110 lamps can be connected.

A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 \(\Omega\) resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer: (i) Used separately: \(I=\dfrac{220}{24}=9.166\) A.
(ii) In series: \(R=24+24=48\ \Omega\), \(I=\dfrac{220}{48}=4.58\) A.
(iii) In parallel: \(R=\dfrac{24}{2}=12\ \Omega\), \(I=\dfrac{220}{12}=18.33\) A.

Compare the power used in the 2 \(\Omega\) resistor in each of the following circuits: (i) a 6 V battery in series with 1 \(\Omega\) and 2 \(\Omega\) resistors, and (ii) a 4 V battery in parallel with 12 \(\Omega\) and 2 \(\Omega\) resistors.

Answer: (i) In series \(R=1+2=3\ \Omega\), \(I=\dfrac63=2\) A flows through the 2 \(\Omega\) resistor, so \(P=I^2R=2^2\times2=8\) W.
(ii) In parallel the voltage across the 2 \(\Omega\) resistor is 4 V, so \(P=\dfrac{V^2}{R}=\dfrac{4^2}{2}=8\) W.
The power used in the 2 \(\Omega\) resistor is 8 W in both cases.

Two lamps, one rated 100 W at 220 V and the other 60 W at 220 V, are connected in parallel to the electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Answer: In parallel each lamp gets 220 V. Current of the 100 W lamp \(=\dfrac{100}{220}\) A; current of the 60 W lamp \(=\dfrac{60}{220}\) A. Current from the line \(=\dfrac{100+60}{220}=0.727\) A.

Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Answer: Energy \(=Pt\). TV set: \(250\times3600=9\times10^5\) J. Toaster: \(1200\times600=7.2\times10^5\) J. The 250 W TV set in 1 h uses more energy.

An electric heater of resistance 8 \(\Omega\) draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.

Answer: Rate of heat production \(P=I^2R=15^2\times8=1800\) J/s.

Explain the following. (a) Why is tungsten used almost exclusively for the filament of electric lamps? (b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal? (c) Why is the series arrangement not used for domestic circuits? (d) How does the resistance of a wire vary with its area of cross-section? (e) Why are copper and aluminium wires usually employed for electricity transmission?

Answer: (a) The melting point and resistivity of tungsten are very high. It does not burn readily at a high temperature. Electric lamps glow at very high temperatures. Hence tungsten is used almost exclusively for the filament of electric lamps.
(b) The resistivity of an alloy is more than that of a pure metal, so it produces a large amount of heat. Also the alloy does not melt readily at high temperatures.
(c) In a series circuit the voltage is divided among the appliances, so each receives only a part of the supply voltage, and the current is the same in all; if one appliance fails the whole circuit breaks. Also, appliances need different currents. Hence the series arrangement is not used for domestic circuits.
(d) The resistance of a wire is inversely proportional to its area of cross-section, \(R\propto\dfrac1A\).
(e) Copper and aluminium have low resistivity. They are good conductors of electricity, so they are usually employed for electricity transmission.

Potential difference, re VA is potential at point A, VB is