CBSE Class 8 Science Current Electricity and its Effects Chapter Notes

Class 8 Science Study Guide: CBSE Class 8 Science Current Electricity and its Effects Chapter Notes

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Class 8 Science: Current Electricity and its Effects

Ohm's Law

Ohm's law gives a relationship between current and potential difference. It states that at constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends. Thus, according to Ohm's law,

class_8_science_useful_01

\(I \propto V\)   or   \(I = \dfrac{V}{R}\)   or   \(\dfrac{V}{I} = R\)

  1. \(V\) = voltage applied
  2. \(I\) = current flowing in the circuit
  3. \(R\) is a constant of proportionality called the resistance of the conductor. The SI unit of resistance is ohm (\(\Omega\)).

The graph between \(V\) and \(I\) is a straight line passing through the origin, as shown in the figure.

VIOVoltage vs. current for an ohmic material

Resistance of a Conductor

The property of a conductor due to which it opposes the flow of current through it is called resistance. The resistance of a conductor is numerically equal to the ratio of the potential difference across its ends to the current flowing through it, i.e.,

class_8_science_useful_02

\(\text{Resistance} = \dfrac{\text{potential difference}}{\text{current}}\)   or   \(R = \dfrac{V}{I}\)

Now, when \(V = 1\) volt and \(I = 1\) ampere, then

\(R(1\,\Omega) = \dfrac{1\,\text{V}}{1\,\text{A}} = 1\,\text{VA}^{-1}\)   or   \(1\,\Omega = 1\,\text{VA}^{-1}\)

Thus, \(1\,\Omega\) is the resistance of a conductor if the potential difference across its ends is 1 V and the current through it is 1 A.

Factors on which Resistance of a Conductor Depends

The resistance of a conductor depends on the following factors:

  1. Length of the conductor
  2. Area of cross-section of the conductor (or thickness of the conductor)
  3. Nature of the material of the conductor
  4. Temperature of the conductor.

\(R \propto \dfrac{l}{A}\)

where \(l\) = length of the conductor and \(A\) = cross-sectional area of the conductor.

\(R = \rho\,\dfrac{l}{A}\)    ....(iii)

Here \(\rho\) (rho) is a constant known as the resistivity (or specific resistance) of the material of the conductor. The SI unit of resistivity is ohm-metre (\(\Omega\) m).

Now in relation (iii), if we put \(A = 1\) and \(l = 1\), we get \(\rho = R\).

Thus, the resistivity of the material of a conductor is defined as the resistance of the conductor of unit length and unit area of cross-section.

Note: The resistivity of a substance does not depend on its length or thickness. It depends only on the nature and temperature of the substance.

Electrical resistivities of some substances at 20°C are given in the following table:

class_8_science_useful_03

TypeMaterialResistivity (\(\Omega\) m)
ConductorsSilver\(1.60\times10^{-8}\)
Copper\(1.62\times10^{-8}\)
Aluminium\(2.63\times10^{-8}\)
Tungsten\(5.20\times10^{-8}\)
AlloysConstantan (alloy of Cu and Ni)\(49\times10^{-6}\)
Manganin (alloy of Cu, Mn and Ni)\(44\times10^{-6}\)
Nichrome (alloy of Ni, Cr, Mn and Fe)\(100\times10^{-6}\)
InsulatorsGlass\(10^{10} - 10^{14}\)
Hard rubber\(10^{13} - 10^{16}\)
Ebonite\(10^{15} - 10^{17}\)
Diamond\(10^{12} - 10^{13}\)
Paper (dry)\(10^{12}\)

From the above table we find that

Resistivity of metals < Resistivity of alloys < Resistivity of insulators

Thus, the metals and alloys have very low resistivity, in the range of \(10^{-8}\) to \(10^{-6}\,\Omega\) m. They are good conductors of electricity.

For example, copper and aluminium metals are generally used for electrical transmission lines. Tungsten metal is used almost exclusively for filaments of electric bulbs. This is because tungsten has a high melting point and does not oxidise (or burn) easily at high temperatures.

The resistivity of alloys is generally higher than that of its constituent metals. Also, alloys do not oxidise readily at high temperature.

For example, manganin (alloy of copper, manganese and nickel) and constantan (alloy of copper and nickel) are used to make resistors (resistance wires) used in electrical appliances.

Nichrome (an alloy of nickel, chromium, iron and manganese) is used for making the heating elements of electrical appliances like electric iron, room heaters, toaster, immersion rod, etc.

Note: The resistivity of semiconductors like silicon and germanium is in between those of conductors and insulators and decreases on increasing the temperature.

Combination of Resistances (or Resistors)

There are two methods of joining resistors together: (i) in series and (ii) in parallel.

Resistors in Series

When two (or more) resistances are connected end to end consecutively, they are said to be connected in series. Let three resistances \(R_1\), \(R_2\) and \(R_3\) be connected in series, as shown in the figure.

 CBSE Class 8 Science Current Electricity and its Effects Chapter Notes

AR₁R₂R₃B

Their equivalent or effective resistance is given by

 CBSE Class 8 Science Current Electricity and its Effects Chapter Notes

CBSE Class 8 Science Current Electricity and its Effects Chapter Notes

\(R_s = R_1 + R_2 + R_3\)

Thus, we conclude that when several resistors are joined in series, the resistance of the combination \(R_s\) equals the sum of their individual resistances.

Resistors in Parallel

When two (or more) resistances are connected between the same two points, they are said to be connected in parallel. Let us consider the arrangement of three resistors joined in parallel, as shown in the figure.

 CBSE Class 8 Science Current Electricity and its Effects Chapter Notes

ABR₁R₂R₃

\(\dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}\)

Thus we conclude that the reciprocal of the combined resistance of a group of resistances connected in parallel is equal to the sum of the reciprocals of all the individual resistances.

Note: When a number of resistances are connected in parallel, their combined or equivalent resistance is always less than the smallest individual resistance.

Advantages of Connecting Electrical Appliances in Parallel

  1. In a series combination, when one component fails, the circuit is broken and none of the components in the circuit works. But in a parallel combination, if one component fails, the working of the other components is not affected.
  2. In a series circuit the current is the same throughout the circuit. So it is impracticable to connect an electric bulb and an electric heater in series, because they need currents of different values to operate properly. On the other hand, a parallel circuit divides the current through the electrical gadgets, and this is helpful when each gadget has a different resistance and requires a different current to operate properly.

Electric Power

The rate at which electric energy is dissipated (or consumed) is called electric power. The electric power \(P\) is given by

\(P = \dfrac{\text{Work done}}{\text{Time taken}}\)   or   \(P = \dfrac{W}{t}\)

The SI unit of electric power is the watt (W).

\(1\,\text{watt} = \dfrac{1\,\text{joule}}{1\,\text{second}} = 1\,\text{Js}^{-1}\)

Thus, the power is said to be 1 watt when an electrical appliance consumes electrical energy at the rate of 1 joule per second. Other units of electric power used for commercial purposes are the kilowatt and the megawatt.

1 kilowatt (1 kW) = \(10^3\) W     1 megawatt (1 MW) = \(10^6\) W

 CBSE Class 8 Science Current Electricity and its Effects Chapter Notes

Formula for Calculating Electric Power

As we know,

\(P = \dfrac{\text{Work done}}{\text{Time taken}} = \dfrac{W}{t}\)    ...(i)

Also, the amount of work done when a charge \(Q\) moves against a potential difference \(V\) is \(W = Q \times V\), or

\(W = I \times t \times V\)    ...(ii)    [\(\because\ I = Q/t\)]

From (i) and (ii), we have

\(P = \dfrac{I \times t \times V}{t}\)   or   \(P = V \times I\)    ...(iii)

or Electric power = Potential difference \(\times\) Current. Also, we can write

\(P = VI\)   or   \(P = I^2R\) [\(\because\ V = RI\)]   or   \(P = \dfrac{V^2}{R}\) [\(\because\ I = V/R\)]

Now, if we put \(V = 1\) volt and \(I = 1\) ampere in equation (iii), we get an electric power of 1 watt, i.e., Power = 1 watt = 1 volt \(\times\) 1 ampere, or 1 watt = 1 VA.

Thus, the power consumed by an electrical appliance is said to be 1 watt when 1 ampere of current flows at a potential difference of 1 volt.

Electric Energy

We know that electric power \(= \dfrac{\text{work done by electric current}}{\text{time taken}}\). According to the law of conservation of energy, work done by electric current = electric energy consumed. Therefore, power \(P = \dfrac{W}{t}\), or

\(W = P \times t\)   or   \(E = P \times t\)

Commercial Unit of Electrical Energy

For commercial purposes we use a bigger unit of electrical energy, called the kilowatt-hour (kWh). One kilowatt-hour is the amount of electrical energy consumed when an electrical appliance having a power rating of 1 kilowatt is used for 1 hour.

\(1\ \text{kWh} = 1000\ \text{W} \times 60 \times 60\ \text{s} = 1000\ \text{J/s} \times 3600\ \text{s} = 3600000\ \text{J} = 3.6\times10^6\ \text{J}\)

Heating Effect of Electric Current

Whenever electric current is passed through a metallic conductor, it becomes hot after some time. This indicates that electrical energy is being converted into heat energy. This effect is known as the heating effect of current.

Cause of Heating Effect of Current

When a potential difference is applied across the ends of a conductor, an electric field is set up across its ends. A large number of free electrons present in the conductor get accelerated towards the positive end and acquire kinetic energy, due to which an electric current flows through the conductor. These accelerated electrons on their way suffer frequent collisions with the ions or atoms of the conductor and transfer their gained kinetic energy to them. As a result, the average kinetic energy of vibration of the ions or atoms of the conductor rises and consequently the temperature of the conductor rises. Thus the conductor gets heated due to the flow of electric current through it.

Heat Produced in a Conductor

CBSE Class 8 Science Current Electricity and its Effects Chapter Notes

AB+–VIIV

Consider a conductor AB connected to a cell as shown in the figure. Let \(V\) be the potential difference applied across the ends of AB and \(I\) be the current flowing through AB in time \(t\). Now, by definition of potential difference, the work done in carrying the charge \(Q\) from A to B is

\(W = V \times Q = VIt = I^2Rt\)   [\(\because\ Q = It,\ V = RI\)]

This work done is equal to the heat (\(H\)) produced in the conductor, i.e.,

\(H = I^2Rt\)   or   \(H = \dfrac{V^2}{R}\,t\)

Applications of Heating Effect of Current

  1. The heating effect of current is used in electrical heating appliances like electric iron, room heater, toaster, electric kettle, electric oven, etc. All these heating appliances contain coils of high-resistance wire made of nichrome alloy.
  2. The heating effect of electric current is utilised in electric bulbs for producing light. Here the filament must retain as much of the heat generated as possible, so that it gets very hot and emits light. It must not melt at such high temperature. A strong metal with a high melting point, such as tungsten (about 3380°C), is used for making bulb filaments. The filament should be thermally isolated as much as possible using insulating supports, etc. The bulbs are usually filled with chemically inactive nitrogen and argon gases to prolong the life of the filament.
  3. An electric fuse is a safety device connected in series with the electric circuit. A fuse wire is generally prepared from copper or a tin-lead alloy whose melting point is very low. When the current in a domestic electric circuit rises too much, the fuse wire gets heated, melts and breaks the circuit, thereby preventing fire and damage to various electrical appliances.

Magnetism

Magnetic field: Around every magnet there is a certain zone within which the magnetic force can be experienced. This zone is called the magnetic field.

SNField Lines Around a Bar Magnet

Magnetic field lines: Within the field the magnetic force acts along some lines. These lines are known as magnetic field lines. A current-carrying straight conductor or a circular conductor produces a magnetic field around it.

Magnetic field due to a current through a straight conductor: When a straight conductor carries electric current, a magnetic field is set up around the conductor. The field lines around the conductor are concentric circles whose centres lie on the conductor, as shown in the figure.

AVKXY

The magnetic field produced causes the deflection in the compass needle. It is found that the deflection in the needle increases when the current increases. Again, if the needle is moved away from the conductor, the deflection in the needle decreases, i.e., the magnetic field produced by a given current in the conductor decreases as the distance from it increases.

Right Hand Thumb Rule: The direction of the magnetic field associated with a current-carrying straight conductor can be obtained by applying the “right hand thumb rule”. According to this rule: hold the current-carrying conductor in your right hand such that the thumb points in the direction of the current; then the direction in which your fingers encircle the conductor gives the direction of the field lines of the magnetic field.

CurrentMagneticfield

Magnetic field due to a current through a circular loop: When a current is passed through a circular coil, a magnetic field is produced around it. The field lines are circular near the wire but they become straight at the centre of the coil.

Earth magnetism: The earth is also a huge magnet. We shall learn more about magnetism in higher classes.

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EXERCISES

Answer the following questions:

What should be the potential difference applied across a conductor of resistance \(5\,\Omega\) to flow a current of 1.5 A through it?

Answer: \(V = IR = 1.5\times5 = 7.5\) volt

Calculate the resistance of 1 km long copper wire of radius 1 mm (resistivity of copper is \(1.72\times10^{-8}\,\Omega\) m).

Answer: \(R = 5.47\,\Omega\)

When a potential difference of 2 volt is applied across the ends of a wire of 5 m length, a current of 1 A is found to flow through it. Calculate (i) the resistance per unit length of the wire, (ii) the resistance of 2 m length of this wire.

Answer: (i) \(0.4\ \Omega\,\text{m}^{-1}\); (ii) \(0.8\,\Omega\)

Three resistors of \(2\,\Omega\), \(3\,\Omega\) and \(4\,\Omega\) are connected in (i) series; (ii) parallel. Find the equivalent resistance in each case.

Answer: (i) \(R = 9\,\Omega\); (ii) \(R = 0.92\,\Omega\)

Three resistors of \(5\,\Omega\), \(3\,\Omega\) and \(2\,\Omega\) are connected together so that the total resistance is greater than \(2\,\Omega\) but less than \(8\,\Omega\). Draw a diagram to show their arrangement and calculate its total resistance.

A5 Ω3 Ω2 ΩB

Answer: Connect the \(3\,\Omega\) and \(2\,\Omega\) resistors in parallel, and join this combination in series with the \(5\,\Omega\) resistor. The parallel part is \(\dfrac{3\times2}{3+2} = 1.2\,\Omega\), so the total resistance is \(5 + 1.2 = 6.2\,\Omega\), which is more than \(2\,\Omega\) and less than \(8\,\Omega\).

In the circuit diagram shown below, calculate the equivalent resistance between the points A and B.

A6 Ω6 Ω6 Ω6 ΩB

Answer: \(6\,\Omega\)

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: Resistance \(R = \dfrac{V^2}{P} = \dfrac{230^2}{1000} = 52.9\,\Omega\); maximum current \(I = \dfrac{P}{V} = \dfrac{1000}{230} = 4.35\) A

A 6 V, 12 W lamp is connected in series with a resistor \(R\) and a source of 12 V supply. Calculate the value of the resistor \(R\) for the proper working of the lamp. What is the current flowing through the circuit?

Answer: \(I = 2\) A, \(R = 3\,\Omega\)

Three resistors of \(2\,\Omega\), \(4\,\Omega\) and \(6\,\Omega\) are connected in series to a battery of emf 12 V and negligible internal resistance. Draw the circuit diagram and calculate (i) the current drawn from the cell, (ii) the potential difference at the ends of each resistor.

2 Ω4 Ω6 Ω12 V

Answer: (i) \(I = 1\) A; (ii) \(2\) V, \(4\) V and \(6\) V

A nichrome wire has a resistance of \(10\,\Omega\). Find the resistance of another nichrome wire whose length is three times and area of cross-section four times that of the first wire.

Answer: \(R = 7.5\,\Omega\)

Thus,weThus,theirequivalentoreffective resistance isgivenby

 

 

The SIunitofelectricpoweriswatt(W)

Formula for Calculating Electric Power

As we known that

1 kWh = 1000 W × 60 × 60 seconds

= 1000 joules/seconds × 3600 seconds

= 3600000 joules

= 3.6 × 106 J

Heating Effect of Electric Current

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