Class 10 Science Electricity Exam Notes

Electric Current(Change In Motion)

This Class 10 Science exam notes page on Electricity covers electric current and its SI unit, the direction of current, the flow of charge in metals, electric symbols, Ohm's law, resistance and resistivity, series and parallel combinations, solved examples, electric energy and power, and the exercises with answer keys for CBSE and similar boards.

ELECTRIC CURRENT (CHARGE IN MOTION)

Definition: The quantity of electric charge flowing through the cross section of a given conductor in one second is called current.

Thus, if Q is the charge which flows through a conductor in time t, then the current (I) is given by:

Current (I) = Charge (Q) / Time (t)

Electric current is a scalar quantity.

Unit of current: The SI unit of charge (Q) is coulomb (C) and that of time (t) is second (s). So,
SI unit of current = 1 coulomb / 1 second = 1 C s⁻¹ = 1 ampere.

The unit coulomb per second (C s⁻¹) is called ampere (A).

Direction of electric current: The direction of flow of positive charge is taken as the conventional direction of electric current. In an ordinary conductor such as a copper wire, the direction of current is taken as opposite to the direction of flow of electrons.

FLOW OF CURRENT IN A METAL

Metals show a different kind of bonding called metallic bonding. In this bonding, the outermost electrons are not bound to any particular atom and move freely inside the metal. These electrons are called free electrons. They move randomly in all directions with different speeds, so there is no net movement of electrons in any particular direction, and therefore no net flow of current.

[Figure: Metallic wire with electrons moving randomly, so no current flows, when no potential is applied across its ends.]

When the two ends of a wire are connected to the two terminals of a battery, the electrons move from the low potential end to the high potential end, and a current flows.

[Figure: Metal wire connected to a battery, with terminals marked high potential and low potential, showing electrons moving from low potential to high potential.]

ELECTRIC SYMBOLS

Many different kinds of equipment or components are used in setting up electrical circuits. To draw the diagrams of electrical circuits on paper, these components are shown by their symbols.

OHM'S LAW

Definition: According to Ohm's law, at constant temperature the current flowing through a conductor is directly proportional to the potential difference across the conductor.

Thus, if I is the current flowing through a conductor and V is the potential difference (or voltage) across it, then according to Ohm's law:
V ∝ I (when T is constant), or I = V / R ...(i)
where R is a constant called the resistance of the conductor.

Equation (i) may be written as V = I × R ...(ii)

Unit of resistance: The SI unit of resistance (R) is ohm, denoted by the Greek letter Ω (omega). From Ohm's law, R = V / I.

If V = 1 volt and I = 1 ampere, then R = 1 volt / 1 ampere.

Thus, 1 ohm is defined as the resistance of a conductor which allows a current of 1 ampere to flow through it when a potential difference of 1 volt is maintained across it.

Results of Ohm's law:
1. Current flowing through a conductor is directly proportional to the potential difference across the conductor. The graph of V against I is a straight line, and its slope equals the resistance R.
2. When the potential difference in a circuit is kept constant, the current is inversely proportional to the resistance of the conductor, I ∝ 1/R.
3. The ratio of potential difference to current is constant, and this constant equals the resistance of the conductor, so V / I = R.

RESISTANCE OF CONDUCTOR

The movement of electrons gives rise to the flow of current through metals. The moving electrons collide with each other and with the positive ions present in the metal. These collisions slow down the electrons and oppose the flow of electric current.

The property of a conductor by virtue of which it opposes the flow of electric current through it is called its resistance.

Resistance is denoted by the letter R. Its SI unit is ohm (Ω). Resistance is a scalar quantity.

Factors on which the resistance of a conductor depends:

1. Effect of length: The resistance of a conductor is directly proportional to its length, that is, R ∝ l.

2. Effect of area of cross section: The resistance of a conductor is inversely proportional to its area of cross section, that is, R ∝ 1/a.
If the area of cross section of the conductor is doubled, its resistance is halved.

3. Effect of temperature: The resistance of all pure metals increases with a rise in temperature. The resistance of alloys increases very slightly with a rise in temperature. For semiconductors, resistance decreases when temperature increases.

4. Effect of nature of material: Some materials have low resistance, whereas others have much higher resistance. In general, an alloy has higher resistance than the pure metals from which it is made.
Copper, silver and aluminium have very low resistance.
Nichrome and constantan have higher resistance. Nichrome is used for making heating elements of heaters, toasters and electric irons.

RESISTIVITY

Combining the factors above, R ∝ l and R ∝ 1/a, so R ∝ l/a, or R = ρ × l / a ...(i)
where ρ (rho) is called the resistivity of the material of the conductor.

If l = 1 m and a = 1 m², then R = ρ ...(ii)
Thus, the resistivity of a substance is the resistance of a 1 metre long piece of the substance having a cross-sectional area of 1 square metre.

Resistivity can also be defined as follows: the resistance offered by a cube of a substance having a side of 1 metre, when current flows perpendicular to its opposite faces, is called its resistivity.

Unit of resistivity: From equation (i), ρ = R × a / l. So, the SI unit of resistivity is ohm × m² / m = ohm metre (Ω m).

Classification of materials on the basis of resistivity:
1. Very low resistivity: Substances with very low resistivities allow the flow of electric current through them. These are called conductors, for example copper, gold, silver, aluminium and electrolytic solutions.
2. Moderate resistivity: Substances with moderate resistivity offer appreciable resistance to the flow of current. These are called resistors, for example alloys such as nichrome, manganin and constantan, and carbon.
3. Very high resistivity: Substances with very high resistivities do not allow electricity to pass through them. These are called insulators, for example rubber, plastics and dry wood.

COMBINATION OF RESISTANCES

Series combination: When two or more resistances are joined end to end so that the same current flows through each of them, they are said to be connected in series.

[Figure: Resistors R1, R2 and R3 joined end to end in a line, with a key and a battery in the same circuit, and the same current I marked through each.]

Law of combination of resistances in series: When a number of resistances are connected in series, their equivalent resistance is equal to the sum of the individual resistances.
R = R1 + R2 + R3 + ... ...(i)

Derivation:
1. Let R1, R2 and R3 be connected in series, I be the current through each, and V1, V2 and V3 be the potential differences across them.
2. From Ohm's law, V1 = IR1, V2 = IR2 and V3 = IR3 ...(ii)
3. The total potential difference is V = V1 + V2 + V3 ...(iii)
4. If R is the equivalent resistance, then V = IR ...(iv)
5. Using equations (i) to (iv): IR = IR1 + IR2 + IR3, so R = R1 + R2 + R3.

Therefore, when resistances are combined in series, the equivalent resistance is greater than each individual resistance.

Results about series combination:
1. The total resistance of resistors in series is the sum of all the individual resistances.
2. The same current flows through each resistor in series.
3. The voltage of the battery equals the sum of the voltage drops across the individual resistors.

Parallel combination: When two or more resistances are connected between two common points so that the same potential difference is applied across each of them, they are said to be connected in parallel.

Derivation of the parallel formula:
1. Let V be the potential difference across the two common points A and B.
2. Current through R1, I1 = V / R1 ...(i)
3. Current through R2, I2 = V / R2 ...(ii)
4. Current through R3, I3 = V / R3 ...(iii)
5. If R is the equivalent resistance, then the total current is I = V / R ...(iv)
6. Also, I = I1 + I2 + I3 ...(v)
7. Substituting, V / R = V / R1 + V / R2 + V / R3 ...(vi)
8. Cancelling V gives 1 / R = 1 / R1 + 1 / R2 + 1 / R3.

The equivalent resistance of a parallel combination is less than each of the individual resistances.

Important results about parallel combination:
1. The total current through the circuit is equal to the sum of the currents through the branches.
2. The potential difference across each resistor is the same and equal to the applied voltage, that is, V1 = V2 = V3 = V.
3. The current through each resistor is inversely proportional to its resistance, so a higher resistance carries a lower current.

SOLVED EXAMPLES

Ex.1 A TV set shoots out a beam of electrons. The beam current is 10 µA.
(a) How many electrons strike the TV screen in each second?
(b) How much charge strikes the screen in a minute?

Answer:
1. Beam current I = 10 µA = 10 × 10⁻⁶ A, and time t = 1 s.
2. (a) Charge flowing per second, Q = I × t = 10 × 10⁻⁶ A × 1 s = 10 × 10⁻⁶ C.
3. Charge on an electron = 1.6 × 10⁻¹⁹ C.
4. Number of electrons per second = (10 × 10⁻⁶) / (1.6 × 10⁻¹⁹) = 6.25 × 10¹³.
5. (b) Charge striking the screen per minute = 10 × 10⁻⁶ C/s × 60 s = 6.0 × 10⁻⁴ C.
6. Hence, (a) 6.25 × 10¹³ electrons and (b) 6.0 × 10⁻⁴ C.

Teacher's Note:
a) The printed solution gives 6.25 × 10¹⁴ and 6.0 × 10⁻³ C, which are wrong for 10 µA; the correct values are 6.25 × 10¹³ and 6.0 × 10⁻⁴ C.
b) The question's printed "10A" is taken as 10 µA, as the solution itself uses.
c) Convert µA to A before using the charge formula.

Ex.2 A current of 10 A exists in a conductor. Assuming that this current is entirely due to the flow of electrons (a) find the number of electrons crossing the area of cross section per second, (b) if such a current is maintained for one hour, find the net flow of charge.

Answer:
1. Current I = 10 A, so the charge flowing in one second Q = 10 C.
2. (a) Number of electrons per second = 10 C / (1.6 × 10⁻¹⁹ C) = 6.25 × 10¹⁹.
3. (b) Charge in one hour = I × t = 10 A × (1 × 60 × 60) s = 36000 C.
4. Hence, (a) 6.25 × 10¹⁹ electrons and (b) 36000 C.

Teacher's Note:
a) One ampere means one coulomb passes per second.
b) Always convert hours to seconds before multiplying by current.

Ex.3 A current of 5.0 A flows through a circuit for 15 min. Calculate the amount of electric charge that flows through the circuit during this time.

Answer:
1. Given I = 5.0 A and t = 15 min = 15 × 60 s = 900 s.
2. Charge Q = I × t = 5.0 A × 900 s = 4500 C.
3. Hence, the charge that flows is 4500 C.

Teacher's Note:
a) Minutes must be changed to seconds first.
b) One ampere second is the same as one coulomb.

Ex.4 A piece of wire is redrawn by pulling it until its length is doubled. Compare the new resistance with the original value.

Answer:
1. The volume of the wire remains the same, so when the length is doubled, the area of cross section is halved.
2. Original resistance R = ρ × l / a.
3. New resistance R' = ρ × 2l / (a/2) = ρ × 4l / a = 4R.
4. Hence, the new resistance is four times the original resistance.

Teacher's Note:
a) Keeping the volume constant is the key assumption.
b) Resistance depends on the square of the length change here, since both length and area change.

Ex.5 Calculate the resistance of 100 m long copper wire. The diameter of the wire is 1 mm.

Answer:
1. Resistivity of copper ρ = 1.6 × 10⁻⁸ Ω m (1.6 × 10⁻⁶ Ω cm).
2. Radius r = 1/2 mm = 0.5 × 10⁻³ m.
3. Area of cross section a = π r² = 3.14 × (0.5 × 10⁻³ m)² = 7.85 × 10⁻⁷ m².
4. R = ρ l / a = (1.6 × 10⁻⁸ × 100) / (7.85 × 10⁻⁷) = 2.04 Ω.
5. Hence, the resistance of the wire is 2.04 Ω.

Teacher's Note:
a) The diameter must be halved to get the radius before finding the area.
b) Keep all lengths in metres so that the units of resistivity match.
c) The printed value of ρ in the solution is in ohm cm, so convert it before using metres.

Ex.6 If four resistances each of values 1 ohm are connected in series. Calculate equivalent resistance.

Answer:
1. In series, R = R1 + R2 + R3 + R4.
2. R = 1 + 1 + 1 + 1 = 4 Ω.
3. Hence, the equivalent resistance is 4 Ω.

Teacher's Note:
a) In series, resistances simply add.
b) Equal resistors in series give n times one resistance.

Ex.7 Suppose a 6-volt battery is connected across a lamp whose resistance is 20 ohm, the current in the circuit is 0.25 A. Calculate the value of the resistance of the resistor which must be used.

Answer:
1. Total circuit resistance R_s = V / I = 6 / 0.25 = 24 Ω.
2. The extra resistor R' is in series with the lamp, so R_s = R + R'.
3. R' = R_s - R = 24 - 20 = 4 Ω.
4. Hence, the resistor must have a resistance of 4 Ω.

Teacher's Note:
a) Find the total resistance first using Ohm's law.
b) Subtracting the lamp's resistance gives the series resistor needed.

Ex.8 A resistance of 6 ohms is connected in series with another resistance of 4 ohms. A potential difference of 20 volts is applied across the combination. Calculate the current through the circuit and potential difference across the 6 ohm resistance.

[Figure: Circuit with a 6 Ω resistor and a 4 Ω resistor in series, connected across a 20 V battery, with the current I shown flowing from the positive to the negative terminal.]

Answer:
1. Total resistance R_s = 6 Ω + 4 Ω = 10 Ω.
2. Circuit current I = V / R_s = 20 / 10 = 2 A.
3. Potential difference across 6 Ω, V1 = I × R1 = 2 × 6 = 12 V.
4. Hence, the current is 2 A and the potential difference across 6 Ω is 12 V.

Teacher's Note:
a) Series resistors share the battery voltage in proportion to their resistances.
b) A proper circuit diagram helps in placing the current arrows correctly.

Ex.9 Two resistances are connected in series as shown in the diagram.
(i) What is the current through the 5 ohm resistance?
(ii) What is the current through R?
(iii) What is the value of R?
(iv) What is the value of V?

[Figure: Two resistors in series. The first resistor is 5 Ω, with a potential difference of 10 V across it. The second resistor is R, with a potential difference of 6 V across it. A battery of V volts is connected across both.]

Answer:
1. (i) Current through 5 Ω, I = V1 / R1 = 10 / 5 = 2 A.
2. (ii) Resistors in series carry the same current, so the current through R is 2 A.
3. (iii) R = V2 / I = 6 / 2 = 3 Ω.
4. (iv) V = V1 + V2 = 10 + 6 = 16 V.
5. Hence, (i) 2 A, (ii) 2 A, (iii) 3 Ω, (iv) 16 V.

Teacher's Note:
a) Both parts of a series circuit carry the same current.
b) The battery voltage is the sum of the drops across each resistor.
c) Check the answer: 3 Ω with 2 A gives 6 V, as given.

Ex.10 Resistors R1, R2 and R3 having values 5 Ω, 10 Ω and 30 Ω respectively are connected in parallel across a battery of 12 volt. Calculate (a) the current through each resistor, (b) the total current in the circuit and (c) the total circuit resistance.

Answer:
1. (a) I1 = V / R1 = 12 / 5 = 2.4 A.
2. I2 = 12 / 10 = 1.2 A.
3. I3 = 12 / 30 = 0.4 A.
4. (b) Total current I = I1 + I2 + I3 = 2.4 + 1.2 + 0.4 = 4 A.
5. (c) 1 / Rp = 1/5 + 1/10 + 1/30 = 10/30 = 1/3, so Rp = 3 Ω.
6. Check: V / Rp = 12 / 3 = 4 A, which matches the total current.
7. Hence, (a) 2.4 A, 1.2 A and 0.4 A, (b) 4 A and (c) 3 Ω.

Teacher's Note:
a) In parallel, each resistor has the full battery voltage across it.
b) The equivalent resistance must be less than the smallest branch resistance.
c) Use the reciprocal method carefully with a common denominator.

Ex.11 Resistors R1 = 10 Ω, R2 = 40 Ω, R3 = 30 Ω, R4 = 20 Ω and R5 = 60 Ω and a 12 volt battery are connected as shown. Calculate (a) the total resistance and (b) the total current flowing in the circuit.

[Figure: R1 and R2 connected in parallel, R3, R4 and R5 connected in parallel, and the two parallel groups connected in series across a 12 V battery.]

Answer:
1. For R1 and R2 in parallel: 1/RP1 = 1/10 + 1/40 = 5/40 = 1/8, so RP1 = 8 Ω.
2. For R3, R4 and R5 in parallel: 1/RP2 = 1/30 + 1/20 + 1/60 = 6/60 = 1/10, so RP2 = 10 Ω.
3. (a) RP1 and RP2 are in series, so R = 8 + 10 = 18 Ω.
4. (b) I = V / R = 12 / 18 = 2/3 A, which is about 0.67 A.
5. Hence, (a) 18 Ω and (b) 0.67 A.

Teacher's Note:
a) Find each parallel group first, then add the groups in series.
b) Keep the fraction 2/3 before rounding to 0.67 A.

Ex.12 In the circuit diagram given below. find
(i) total resistance of the circuit
(ii) total current flowing in the circuit
(iii) potential difference across R1

[Figure: Resistors of 8 Ω and 12 Ω connected in parallel, and this combination connected in series with a 7.2 Ω resistor (R1) across a 6 V battery.]

Answer:
1. (i) 8 Ω and 12 Ω in parallel: (8 × 12)/(8 + 12) = 4.8 Ω.
2. This comes in series with 7.2 Ω, so total resistance = 7.2 + 4.8 = 12 Ω.
3. (ii) Total current I = V / R = 6 / 12 = 0.5 A.
4. (iii) Potential difference across R1 = I × R1 = 0.5 × 7.2 = 3.6 V.
5. Hence, (i) 12 Ω, (ii) 0.5 A and (iii) 3.6 V.

Teacher's Note:
a) Use the product-over-sum rule for two resistors in parallel.
b) The potential difference across R1 is found after the current is known.

Ex.13 Three resistances are connected as shown in diagram through the resistance 5 ohms, a current of 1 ampere is flowing:
(i) What is the current through the other two resistors?
(ii) What is the potential difference (p.d.) across AB and across AC?
(iii) What is the total resistance.

[Figure: A 5 Ω resistor between points A and B. Between points B and C, a 10 Ω resistor and a 15 Ω resistor are connected in parallel.]

Answer:
1. (i) Currents in parallel branches are inversely proportional to resistances, so I1 : I2 = 15 : 10 = 3 : 2.
2. Also I1 + I2 = 1 A, so I1 = 0.6 A (through 10 Ω) and I2 = 0.4 A (through 15 Ω).
3. (ii) P.D. across AB = I × R = 1 × 5 = 5 V.
4. P.D. across BC = I1 × 10 = 0.6 × 10 = 6 V.
5. P.D. across AC = 5 + 6 = 11 V.
6. (iii) Parallel resistance of 10 Ω and 15 Ω: Rp = (10 × 15)/(10 + 15) = 6 Ω.
7. Total resistance = 5 + 6 = 11 Ω.
8. Hence, (i) 0.6 A and 0.4 A, (ii) 5 V and 11 V, (iii) 11 Ω.

Teacher's Note:
a) The two parallel branches have the same potential difference, which links them.
b) The potential difference across AC is the sum of AB and BC.
c) Check: 11 V across 11 Ω gives the 1 A current.

Ex.14 In the diagram shown below (Fig.), the cells and the ammeter both have negligible resistance. The resistors are identical. With the switch K open, the ammeter reads 0.6 A. What will be the ammeter reading when the switch is closed?

[Figure: Identical resistors R connected in parallel across a cell with a switch K and an ammeter in the circuit. With the switch closed, a third identical resistor is added in parallel.]

Answer:
1. Let the cell have potential difference V and each resistor have resistance R.
2. With K open, two resistors are in parallel: RP1 = R/2. The current is I1 = 0.6 A, so V = I1 × RP1 = 0.6 × R/2 = 0.3 R.
3. With K closed, three resistors are in parallel: RP2 = R/3.
4. Current with K closed: I2 = V / RP2 = 0.3 R / (R/3) = 0.9 A.
5. Hence, the ammeter reads 0.9 A.

Teacher's Note:
a) Adding a branch in parallel lowers the total resistance and raises the current.
b) The potential difference V stays the same in both cases.

Ex.15 In the circuit diagram.
Find (i) total resistance
(ii) current shown by the ammeter A.

[Figure: A 3 Ω and 2 Ω resistor in series, in parallel with a 5 Ω resistor, across a 4 V battery, with ammeter A in the main circuit.]

Answer:
1. 3 Ω and 2 Ω in series give 5 Ω.
2. This 5 Ω is in parallel with the other 5 Ω: Rp = (5 × 5)/(5 + 5) = 2.5 Ω.
3. (i) Total resistance = 2.5 Ω.
4. (ii) Circuit current I = V / Rp = 4 / 2.5 = 1.6 A.
5. Hence, (i) 2.5 Ω and (ii) the ammeter reads 1.6 A.

Teacher's Note:
a) Combine the series resistors first, then the parallel pair.
b) The ammeter in the main line measures the total circuit current.

Ex.16 For the circuit shown in the following diagram, what is the value of (i) current through the 6 Ω resistor and (ii) potential difference across the 12 Ω resistor?

[Figure: A 4 V battery connected across two parallel branches. The first branch has a 6 Ω and a 3 Ω resistor in series. The second branch has a 12 Ω and a 3 Ω resistor in series.]

Answer:
1. (i) Branch 1 has 6 Ω + 3 Ω = 9 Ω. Current in this branch I = 4 / 9 = 0.44 A.
2. This is also the current through the 6 Ω resistor, since they are in series.
3. (ii) Branch 2 has 12 Ω + 3 Ω = 15 Ω. Current in this branch I = 4 / 15 A.
4. P.D. across 12 Ω = (4 / 15) × 12 = 3.2 V.
5. Hence, (i) 0.44 A and (ii) 3.2 V.

Teacher's Note:
a) Each parallel branch gets the full 4 V.
b) Series resistors in one branch share the same current.
c) The 3 Ω resistor in each branch is not part of the 12 Ω or 6 Ω reading.

ELECTRIC ENERGY

When a potential difference is applied across a wire, current starts flowing in it. The free electrons collide with the positive ions of the metal and lose energy. The energy taken from the battery is dissipated and is given to the metal ions during collisions, so the temperature of the wire rises. The energy taken from the battery is transferred into heat. This energy is called electrical energy, and this effect is called the heating effect of current.

If R is the resistance of the wire, I the current in it and V the potential difference across it, then the flow of charge in time t is It. Energy dissipated W = Vq = VIt.
Using V = IR, W = I²Rt = (V²/R) t.

This energy is equal to the work done by the battery, or the heat produced in the wire.

ELECTRICAL POWER

The rate of loss of energy in an electrical circuit is called electrical power. It is denoted by P.
P = W / t = VI = I²R = V²/R.

Units of power are joule per second, watt and horse power. 1 watt = 1 joule per second, and 1 HP = 746 watt.
Units of electrical energy are watt second and kilowatt hour. 1 kilowatt hour (kWh) = 3.6 × 10⁶ joule.

POINTS TO BE REMEMBERED

Current: The rate of flow of charge (Q) through a conductor is called current, I = Q / t. The SI unit of current is ampere (A), where 1 A = 1 C/s. An ammeter measures the current and is connected in series with the conductor. The direction of current is taken as the direction of flow of positive charge.

Ohm's law: At any constant temperature, the current (I) flowing through a conductor is directly proportional to the potential difference (V) applied across it. Mathematically, I = V/R or V = IR.

Resistance: Resistance is the property of a conductor by virtue of which it opposes the flow of electricity through it. It is measured in ohms and is a scalar quantity.

Resistivity: The resistance offered by a cube of a substance having a side of 1 metre, when current flows perpendicular to its opposite faces, is called its resistivity (ρ). Its SI unit is ohm metre.

Equivalent resistance: A single resistance which can replace a combination of resistances so that the current through the circuit remains the same is called equivalent resistance.

Series combination: The equivalent resistance is R = R1 + R2 + R3 + ... and it is greater than each individual resistance.

Parallel combination: The reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances, 1/R = 1/R1 + 1/R2 + 1/R3 + ... The equivalent resistance is less than each individual resistance.

EXERCISE - 1

A. VERY SHORT ANSWER TYPE QUESTIONS

Q.1 Define current.
Q.2 Define one ampere.
Q.3 What is the conventional direction of electric current? How does it differ from the direction of flow of electrons?
Q.4 What do you mean by elementary charge?
Q.5 State Ohm's law.
Q.6 Define one ohm.
Q.7 Write the formula for resistance of a wire of length l and cross-section A.
Q.8 Define specific resistance.
Q.9 Write the unit of specific resistance.
Q.10 Distinguish between resistance and resistivity.
Q.11 Two resistors R1 and R2 are joined in series. Find the equivalent resistance.
Q.12 Two resistors R1 and R2 are joined in parallel. Find the equivalent resistance.

B. SHORT ANSWER TYPE QUESTIONS

Q.13 Define current. Is it a scalar quantity or a vector quantity? What is meant by the conventional direction of current?
Q.14 Define resistance.
Q.15 On what factors does the resistance of a conductor depend?
Q.16 Define resistivity. Write the formula for resistivity.
Q.17 What is the formula for the combination of resistances when they are combined in: (i) series and (ii) parallel?
Q.18 Why is the series arrangement not used for domestic circuits?
Q.19 How does the resistance of a wire vary with its cross-sectional area?
Q.20 A piece of wire is redrawn by pulling it until its length is doubled. Compare the new resistance with the original value.

C. LONG ANSWER TYPE QUESTIONS

Q.21 Define charge. What do you understand by positive and negative charge? Write down the expression for force between two charges.
Q.22 State Ohm's law. How it can be verified experimentally?
Q.23 Describe the conditions for constituting an electric current. Explain the mechanism of flow of electrons in a conductor.
Q.24 Derive the expression for the equivalent resistance when two resistors are joined in series.
Q.25 Derive the expression for the equivalent resistance when two resistors are joined in parallel.

D. NUMERICAL PROBLEMS

Q.26 If the charge on an electron be 1.6 × 10⁻¹⁹ C, how many electrons should pass through a conductor in 1 second to constitute 1 ampere current?
Q.27 How many electrons pass through a lamp in one minute if the current be 200 mA? (Charge on an electron, e = 1.6 × 10⁻¹⁹ C.)
Q.28 A conductor carries a current of 0.2 A. Find the amount of charge that will pass through the cross-section of the conductor in 30 s. How many electrons will flow in this time-interval? (Charge on an electron, e = 1.6 × 10⁻¹⁹ C.)
Q.29 The potential difference between the two points of a wire carrying 2 amperes current is 0.1 volt. Calculate the resistance between these points.
Q.30 A resistance of 12 ohm is connected in parallel with another resistor X. The resultant resistance of the combination is 4.8 ohms. What is the resistance X?
Q.31 Three resistances 12 ohms each are connected in parallel. Three such combinations are connected in series. What is the total resistance?
Q.32 How will you connect three resistors of 3 Ω, 4 Ω and 7 Ω respectively so as to obtain a resultant resistance of 3.5 Ω?
Q.33 Find the current through the circuit shown in figure. Also find the potential difference across the 20-ohm resistor.
[Figure: Circuit with a 20 Ω resistor and a 15 Ω resistor in series with a 7 V cell.]
Q.34 Find (a) the equivalent resistance, (b) the current passing through the cell, and (c) the current passing through the 30-ohm resistor in the circuit shown in figure.
[Figure: Circuit with a 15 Ω resistor and a 30 Ω resistor connected across a 6 V cell.]
Q.35 Find the current supplied by the cell in the circuit shown in figure.
[Figure: Circuit with 5 Ω, 10 Ω and 10 Ω resistors and a 5 V cell.]
Q.36 Figure shows a part of an electric circuit. The reading of the ammeter is 3.0 A. Find the currents through the 10-ohm and 20-ohm resistors.
[Figure: A 10 Ω and a 20 Ω resistor in parallel, in series with an ammeter A.]
Q.37 Three resistors of resistances 10 Ω, 20 Ω and 30 Ω are connected in parallel with a 6-V cell. Find (a) the current through each resistor, (b) the current supplied by the cell, and (c) the equivalent resistance of the circuit.
Q.38 When two resistors are joined in series, the equivalent resistance is 90 Ω. When the same resistors are joined in parallel, the equivalent resistance is 20 Ω. Calculate the resistances of the two resistors.
Q.39 Consider the circuit shown in figure. Calculate the current through the 3-ohm resistor.
[Figure: A 4 Ω resistor in series with a parallel combination of a 3 Ω resistor and a 6 Ω resistor, connected across a 12 V cell. Currents i, i1 and i2 are marked.]
Q.40 (a) How will you join three resistors of resistances 4 Ω, 6 Ω and 12 Ω to get an equivalent resistance of 8 Ω? (b) What would be the highest and the lowest equivalent resistances possible by joining these resistors?

EXERCISE - 2

Single Correct Answer Type Questions

Q.1 How many electrons in 1 s constitute a current of 1 A?
(A) 6.25 × 10¹⁸ (B) 6.25 × 10¹²
(C) 6.25 × 10¹¹ (D) 6.25

Q.2 1 Coulomb is equal to -
(A) 1 amp × 1 sec
(B) 1 amp / 1 sec
(C) 1 joule × 1 amp
(D) 1 joule / 1 sec

Q.3 When a body is negatively charged by friction, it means-
(A) the body has acquired excess of electrons
(B) the body has acquired excess of protons
(C) the body has lost some electrons
(D) the body has lost some neutrons

Q.4 If a charged body attracts another body, the charge on the other body-
(A) must be negative
(B) must be positive
(C) must be zero
(D) may be negative or positive or zero

Q.5 A suitable unit for expressing the strength of electric field is -
(A) V/C (B) C/m
(C) N/C (D) C/N

Q.6 One ampere equals -
(A) 10⁶ µA (B) 10⁻⁶ µA
(C) 10⁻³ µA (D) 10 mA

Q.7 What constitutes current in a metal wire?
(A) Electrons (B) Protons
(C) Atoms (D) Molecules

Q.8 If I is the current through a wire and 'e' is the charge of electron then the number of electrons in t seconds will be given by -
(A) Ie/t (B) e/It
(C) It/e (D) Ite

Q.9 Conventionally, the direction of the current is taken as-
(A) the direction of flow of negative charges
(B) the direction of flow of atoms
(C) the direction of flow of positive charges
(D) the direction of flow of molecules

Q.10 Figure shows current in a part of electrical circuit, then the value of current is-

(A) 1.7 A (B) 3.7 A
(C) 13 A (D) 1.0 A

Q.11 When the temperature of a metallic conductor is increased its resistance-
(A) always decrease
(B) always increase
(C) may increase or decrease
(D) remain the same

Q.12 Specific resistance of wire depends upon-
(A) its length (B) its cross-section area
(C) its dimensions (D) its material

Q.13 The unit of resistivity is-
(A) ohm (B) ohm metre
(C) ohm metre⁻¹ (D) mho metre⁻¹

Q.14 A wire of resistance R is cut into n equal parts. These parts are then connected in parallel. The equivalent resistance of combination will be -
(A) nR (B) R/n
(C) n/R (D) R/n²

Q.15 A piece of wire of resistance 4 Ω is bent through 180° at its mid point and the two halves are twisted together, then resistance is -
(A) 1 Ω (B) 2 Ω
(C) 5 Ω (D) 8 Ω

Q.16 Three resistances each of 8 Ω are connected to a triangle. The resistance between any two terminals will be:
(A) 12 Ω (B) 2 Ω
(C) 6 Ω (D) 16/3 Ω

Q.17 In how many parts (equal) a wire of 100 Ω be cut so that a resistance of 1 Ω is obtained by connecting them in parallel?
(A) 10 (B) 5
(C) 100 (D) 50

Q.18 The filament of an electric bulb is made of tungsten because-
(A) its resistance is negligible
(B) it is cheaper
(C) its melting point is high
(D) its filament is easily made

Q.19 If a wire of resistance 1 Ω is stretched to double its length, then the resistance will become-
(A) 1/2 Ω (B) 2 Ω
(C) 1/4 Ω (D) 4 Ω

Q.20 In the given circuit, the effective resistance between points A and C will be -
[Figure: Four resistors of R on the sides AD, DC and AB, and BC, with a 2R resistor on the diagonal AC, forming a square with corners A, D, C and B.]
(A) 3/2 R (B) 6R
(C) 2/3 R (D) 3R

Q.21 Rate of heat generated by electric current in a resistive circuit is expressed in-
(A) IR (B) IR²
(C) I²R (D) √(IR)

Q.22 Two heater wires of equal length are first connected in series and then in parallel with a battery. The ratio of heat produced in the two cases is-
(A) 2 : 1 (B) 1 : 2
(C) 4 : 1 (D) 1 : 4

Q.23 How much electrical energy in kilowatt hour is consumed in operating ten, 50 watt bulbs for 10 hours per day in a month of 30 days?
(A) 15 (B) 150
(C) 1500 (D) 15000

Q.24 An electric iron draws a current of 4 A when connected to a 220 V mains. Its resistance must be -
(A) 40 Ω (B) 55 Ω
(C) 100 Ω (D) none of these

Q.25 The resistance of a conductor is reduced to half its initial value. In doing so the heating effects in the conductor will become.
(A) half (B) one-fourth
(C) four times (D) double

Q.26 Laws of heating are given by-
(A) Faraday (B) Joule
(C) Ohm (D) Maxwell

Q.27 An electric iron is based upon the principle of-
(A) magnetic effect of current
(B) heating effect of current
(C) chemical effect of current
(D) none of these

Q.28 A fuse wire is always connected to the-
(A) neutral wire (B) earth wire
(C) live wire (D) none of these

Q.29 Heating effect of a current carrying conductor is due to-
(A) loss of kinetic energy of moving atoms
(B) loss of kinetic energy of moving electrons
(C) attraction between electrons and atoms
(D) repulsion between electrons and atoms

Q.30 The correct relation between heat produced (H) and electric current I flowing is-
(A) H ∝ I (B) H ∝ 1/I
(C) H ∝ I² (D) H ∝ 1/I²

Q.31 The wire having a red plastic covering is a:
(A) live wire (B) neutral wire
(C) earth wire (D) none of these

Q.32 A switch is always connected to the-
(A) earth wire (B) neutral wire
(C) live wire (D) none of these

Q.33 The wire having a black plastic covering is a-
(A) live wire (B) neutral wire
(C) earth wire (D) none of these

Q.34 The wire having a green plastic covering is a-
(A) live wire (B) neutral wire
(C) earth wire (D) none of these

Q.35 In three pin socket (shoe) the bigger hole is connected to-
(A) any wire (B) live wire
(C) neutral wire (D) earth wire

Q.36 Coming of live wire and neutral wire in direct contact causes:
(A) short-circuiting (B) over loading
(C) no damage (D) unknown effect

Q.37 In electric fittings in a house:
(A) the live wire goes through the switch
(B) the neutral wire goes through the switch
(C) the earth wire goes through the switch
(D) no wire goes through the switch

Q.38 High power electrical appliances are earthed to -
(A) avoid shock
(B) avoid wastage
(C) make the appliance look beautiful
(D) reduce the bill

ANSWER KEY

EXERCISE - 1, Numerical Problems
Q.26: 6.25 × 10¹⁸
Q.27: 7.5 × 10¹⁹
Q.28: 6 C and 3.75 × 10¹⁹ electrons
Q.29: 0.05 Ω
Q.30: 8 Ω
Q.31: 12 Ω
Q.32: Connect 3 Ω and 4 Ω in series, and connect this combination in parallel with 7 Ω.
Q.33: 0.2 A and 4 V
Q.34: (a) 10 Ω (b) 0.6 A (c) 0.2 A
Q.35: 0.5 A
Q.36: 2 A through the 10 Ω resistor and 1 A through the 20 Ω resistor
Q.37: (a) 0.6 A, 0.3 A and 0.2 A (b) 1.1 A (c) ≈ 5.5 Ω
Q.38: 60 Ω, 30 Ω
Q.39: 1.33 A
Q.40: (b) Highest 22 Ω and lowest 2 Ω

EXERCISE - 2, Answer Key
Q.1 to Q.15: A, A, A, D, C, A, A, C, C, A, B, D, B, D, A
Q.16 to Q.30: D, A, C, D, C, C, D, B, B, D, B, B, C, B, C
Q.31 to Q.38: A, C, B, C, D, A, A, A

RESISTANCE OF CONDUCTOR

♦ Classification of Material on Basis of Resistivity