Electricity | Class 10 Science (NCERT Ch 11)

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Electricity

NCERT Chapter 11 • Current, Resistance, and Power

NCERT 2025–26 Physics Numericals

Illustration of glowing filament and abstract electric current.

Electricity is a controllable and convenient form of energy. In this chapter, we explore how current flows, what controls it (resistance), and how we use it for heating and lighting.

1. Electric Current and Circuit

Electric Current is expressed by the amount of charge flowing through a particular area in unit time. It is the rate of flow of electric charges.

Conventionally, the direction of electric current is opposite to the flow of electrons.
SI Unit: Ampere (A). One ampere is the flow of one coulomb of charge per second ( $1 A = 1 C/s$ ).

I = \dfrac{Q}{t}

Where $I$ is current, $Q$ is net charge, and $t$ is time.

Schematic diagram of a simple electric circuit. — A simple electric circuit containing a cell, bulb, and ammeter.

Example 11.1 Question: 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.
Solution:
Given: $I = 0.5 \text{ A}, t = 10 \text{ min} = 600 \text{ s}$ .
From $Q = It$ :
$Q = 0.5 \times 600 = 300 \text{ C}$ .

Q1. What does an electric circuit mean?

A continuous and closed path of an electric current is called an electric circuit.

Q2. Define the unit of current.

The unit of current is Ampere (A). One ampere is constituted by the flow of one coulomb of charge per second.

Q3. Calculate the number of electrons constituting one coulomb of charge.

Charge on 1 electron $e = 1.6 \times 10^{-19} \text{ C}$ .
Total charge $Q = n \times e$ .
For $Q = 1 \text{ C}$ , $n = \dfrac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18}$ electrons (approx $6 \times 10^{18}$ ).

2. Electric Potential and Potential Difference

Electrons move only if there is a difference of electric pressure, called Potential Difference. This is often produced by a battery.

Electric potential difference ( $V$ ) between two points is the work done ( $W$ ) to move a unit charge ( $Q$ ) from one point to the other.

V = \dfrac{W}{Q}

SI Unit: Volt (V). $1 \text{ Volt} = 1 \text{ Joule} / 1 \text{ Coulomb}$ .

Example 11.2 Question: How much work is done in moving a charge of 2 C across two points having a potential difference 12 V?
Solution:
$W = VQ = 12 \text{ V} \times 2 \text{ C} = 24 \text{ J}$ .

Q1. Name a device that helps to maintain a potential difference across a conductor.

A battery or an electric cell.

Q2. What is meant by saying that the potential difference between two points is 1 V?

It means that 1 joule of work is done to move a charge of 1 coulomb from one point to the other.

Q3. How much energy is given to each coulomb of charge passing through a 6 V battery?

Energy = Work Done = $V \times Q$ .
For each coulomb ( $Q=1$ ), Energy = $6 \text{ V} \times 1 \text{ C} = 6 \text{ Joules}$ .

3. Ohm’s Law

Ohm’s Law states that the potential difference ( $V$ ) across the ends of a given metallic wire is directly proportional to the current flowing through it ( $I$ ), provided the temperature remains the same.

V = IR

Here, $R$ is the Resistance, a property of the conductor to resist the flow of charges. Its SI unit is Ohm ( $\Omega$ ).

4. Factors on which Resistance Depends

The resistance of a conductor depends on:

Length ( $l$ ): Directly proportional ( $R \propto l$ ).
Area of Cross-section ( $A$ ): Inversely proportional ( $R \propto 1/A$ ).
Nature of Material: Represented by Resistivity ( $\rho$ ).

R = \rho \dfrac{l}{A}

$\rho$ (rho) is the Electrical Resistivity. Metals have very low resistivity ( $10^{-8} \Omega m$ ), while insulators have high resistivity.

Example 11.6 Question: A wire of given material having length $l$ and area $A$ has a resistance of $4 \Omega$ . What would be the resistance of another wire of the same material having length $l/2$ and area $2A$ ?
Solution:
$R_1 = \rho \dfrac{l}{A} = 4 \Omega$ .
$R_2 = \rho \dfrac{l/2}{2A} = \dfrac{1}{4} \rho \dfrac{l}{A}$ .
$R_2 = \dfrac{1}{4} R_1 = \dfrac{1}{4} \times 4 = 1 \Omega$ .

Q1. On what factors does the resistance of a conductor depend?

It depends on: (1) Its length, (2) Its area of cross-section, and (3) The nature of its material.

Q2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Current will flow more easily through a thick wire. This is because resistance is inversely proportional to the area of cross-section ( $R \propto 1/A$ ). A thicker wire has less resistance.

Q3. Let the resistance of an electrical component remain constant while the potential difference across it decreases to half. What change will occur in the current?

According to Ohm’s Law, $I = V/R$ . If $V$ becomes $V/2$ and $R$ is constant, the current $I$ will also become half.

Q4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

(1) Alloys have higher resistivity than pure metals, producing more heat.
(2) Alloys do not oxidise (burn) readily at high temperatures.

Q5. Use Table 11.2 (Resistivity) to answer: (a) Which is a better conductor, Iron or Mercury? (b) Which material is the best conductor?

(a) Iron (lower resistivity = $10.0 \times 10^{-8}$ ) is a better conductor than Mercury ( $94.0 \times 10^{-8}$ ).
(b) Silver is the best conductor (lowest resistivity = $1.60 \times 10^{-8}$ ).

5. Resistance of a System of Resistors

Comparison of Series and Parallel combinations of resistors. — Resistors in Series (Left) vs Resistors in Parallel (Right).

A. Resistors in Series

Joined end to end. The current is the same in every part of the circuit.

R_s = R_1 + R_2 + R_3

The total potential difference is the sum of potential differences across individual resistors ( $V = V_1 + V_2 + V_3$ ).

Q1. 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, and a 12 $\Omega$ resistor, and a plug key, all connected in series.

(Imagine a circuit loop: Battery (6V total) -> Key -> 5Ω -> 8Ω -> 12Ω -> back to Battery).

Q2. Redraw the circuit of Q1, 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?

Total Resistance $R_s = 5 + 8 + 12 = 25 \Omega$ .
Total Voltage $V = 6 \text{ V}$ .
Current (Ammeter Reading) $I = V/R = 6/25 = \textbf{0.24 A}$ .
Voltage across 12 $\Omega$ (Voltmeter Reading) $V = I \times R = 0.24 \times 12 = \textbf{2.88 V}$ .

B. Resistors in Parallel

Connected between the same two points. The potential difference is the same across each resistor.

\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}

The total current is the sum of currents through individual branches ( $I = I_1 + I_2 + I_3$ ).

Q1. 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 $10^6 \Omega$ .

In parallel, the equivalent resistance is always less than the smallest individual resistance.
(a) Slightly less than 1 $\Omega$ .
(b) Slightly less than 1 $\Omega$ .

Q2. An electric lamp of 100 $\Omega$ , a toaster of 50 $\Omega$ , and a water filter of 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?

$\dfrac{1}{R_p} = \dfrac{1}{100} + \dfrac{1}{50} + \dfrac{1}{500} = \dfrac{5+10+1}{500} = \dfrac{16}{500}$ .
$R_p = \dfrac{500}{16} = \textbf{31.25 } \Omega$ .
Current $I = V/R = 220 / 31.25 = \textbf{7.04 A}$ .

Q3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

(1) In parallel, each device gets the full voltage.
(2) If one device fails, others continue to work (unlike series where the circuit breaks).
(3) Different devices can draw different currents according to their resistance.

Q4. 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 $\Omega$ ?

(a) 4 $\Omega$ : Connect 3 $\Omega$ and 6 $\Omega$ in parallel (gives 2 $\Omega$ ), then connect that in series with 2 $\Omega$ . ( $2 + 2 = 4$ ).
(b) 1 $\Omega$ : Connect all three in parallel. ( $\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{3+2+1}{6} = 1$ ).

Q5. 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 $\Omega$ ?

(a) Highest (Series): $4+8+12+24 = \textbf{48 } \Omega$ .
(b) Lowest (Parallel): $\dfrac{1}{R} = \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{12} + \dfrac{1}{24} = \dfrac{6+3+2+1}{24} = \dfrac{12}{24} = \dfrac{1}{2}$ . So $R = \textbf{2 } \Omega$ .

6. Heating Effect of Electric Current

If an electric circuit is purely resistive, the source energy is dissipated entirely as heat. This is the heating effect of current.

Joule’s Law of Heating:

H = I^2 R t

Heat ( $H$ ) is directly proportional to square of current ( $I$ ), resistance ( $R$ ), and time ( $t$ ).

Structure of an electric heating element. — Electric Iron heating element based on Joule’s Law.

Q1. Why does the cord of an electric heater not glow while the heating element does?

The heating element is made of an alloy with high resistance, so it produces a lot of heat ( $H \propto R$ ) and glows. The cord is made of copper (low resistance), so negligible heat is produced.

Q2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Heat $H = V \times Q$ .
$H = 50 \text{ V} \times 96000 \text{ C} = 4,800,000 \text{ J} = \textbf{4.8 \times 10^6 J}$ .

Q3. An electric iron of resistance 20 $\Omega$ takes a current of 5 A. Calculate the heat developed in 30 s.

$H = I^2 R t = 5^2 \times 20 \times 30 = 25 \times 20 \times 30 = \textbf{15,000 J}$ .

7. Electric Power

The rate at which electric energy is dissipated or consumed. The SI unit is Watt (W).

P = VI = I^2R = \dfrac{V^2}{R}

Commercial Unit: Kilowatt-hour (kWh). $1 \text{ kWh} = 3.6 \times 10^6 \text{ J}$ .

Q1. What determines the rate at which energy is delivered by a current?

Electric Power.

Q2. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Power $P = VI = 220 \times 5 = \textbf{1100 W}$ .
Energy $E = P \times t = 1100 \text{ W} \times 7200 \text{ s} (2 \text{ hr}) = \textbf{7.92 \times 10^6 J}$ .

8. Chapter Exercises

Practice these NCERT exercise questions to master the chapter:

Q1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance is R’, then the ratio R/R’ is:
(a) 1/25 (b) 1/5 (c) 5 (d) 25

(d) 25.
Each part has resistance $R/5$ . Connected in parallel, $\dfrac{1}{R'} = 5 \times \dfrac{1}{R/5} = \dfrac{25}{R}$ . So $R' = R/25$ , thus $R/R' = 25$ .

Q2. Which term does NOT represent electrical power?
(a) $I^2R$ (b) $IR^2$ (c) $VI$ (d) $V^2/R$

(b) $IR^2$ .

Q3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be:
(a) 100 W (b) 75 W (c) 50 W (d) 25 W

(d) 25 W.
$R = V^2/P = 220^2/100 = 484 \Omega$ .
New Power $P' = (110)^2 / 484 = 12100/484 = 25 \text{ W}$ .

Q4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be:
(a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1

(c) 1:4.
Series resistance $R_s = 2R$ . Parallel resistance $R_p = R/2$ .
Heat $\propto 1/Resistance$ (for same V). Ratio $H_s/H_p = R_p/R_s = (R/2) / 2R = 1/4$ .

Q5. How is a voltmeter connected in the circuit to measure the potential difference between two points?

In parallel.

Q6. A copper wire has diameter 0.5 mm and resistivity $1.6 \times 10^{-8} \Omega$ 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?

Area $A = \pi (d/2)^2$ .
Length $l = RA/\rho = 122.7 \text{ m}$ .
If diameter is doubled, Area becomes 4 times. Resistance becomes 1/4th.

Q10. How many 176 $\Omega$ resistors (in parallel) are required to carry 5 A on a 220 V line?

Required Total Resistance $R = V/I = 220/5 = 44 \Omega$ .
For $n$ resistors in parallel: $R_{eq} = 176/n$ .
$44 = 176/n \Rightarrow n = 176/44 = \textbf{4}$ .

Q12. 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 220 V line if the maximum allowable current is 5 A?

Current for one bulb $I = P/V = 10/220 = 1/22 \text{ A}$ .
Total Max Current = 5 A.
Number of bulbs $n = 5 / (1/22) = 5 \times 22 = \textbf{110}$ .

Q18. Explain the following:
(a) Why is tungsten used for filaments?
(b) Why are alloys used for heating devices?
(c) Why is series arrangement not used for domestic circuits?
(d) How does resistance vary with area?
(e) Why are copper and aluminium used for transmission?

(a) High melting point and doesn’t oxidize easily.
(b) Higher resistivity and high melting point.
(c) Voltage divides, total resistance increases, and if one appliance fails, all stop.
(d) Resistance is inversely proportional to area.
(e) Low resistivity (good conductors).

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