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Chapter 11

Electricity

NCERT solutions and explanations for Class 10 Science Chapter 11 Electricity — covering electric current and circuit, potential difference, Ohm's law, resistance and resistivity, series and parallel resistor combinations, heating effect of electric current (Joule's law), electric power, and the commercial unit of energy (kWh).

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Questions

10
Q1

Define electric current and an electric circuit. How many electrons make up one coulomb of charge?

Electric current is the rate of flow of electric charge through a conductor. An electric circuit is a continuous and closed conducting path along which an electric current flows. Nearly 6×10186 \times 10^{18} electrons constitute one coulomb of charge.
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Q2

Define potential difference and explain how a voltmeter is connected in a circuit. How much energy is given to each coulomb of charge passing through a 6 V battery?

Potential difference between two points is the work done in moving a unit charge from one point to the other. A voltmeter is always connected in parallel across the points between which the potential difference is to be measured. Each coulomb of charge passing through a 6 V battery receives 6 J of energy.
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Q3

State Ohm's law. What does a V-I graph for a metallic conductor look like, and what does it indicate?

Ohm's law states that the potential difference VV across the ends of a given metallic conductor in an electric circuit is directly proportional to the current II flowing through it, provided its temperature remains constant (V∝IV \propto I, or V=IRV = IR). The V-I graph is a straight line passing through the origin, indicating that V/IV/I (or RR) is constant.
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Q4

What is a rheostat? Explain how it can be used to regulate the current in a circuit without changing the voltage source.

A rheostat is a variable resistance device. By changing the effective length of the resistance wire introduced into the circuit (through a sliding contact), the total resistance of the circuit changes. Since I=V/RI = V / R from Ohm's law, increasing resistance reduces current and decreasing resistance increases current, all while keeping the voltage source unchanged.
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Q5

On what factors does the resistance of a conductor depend? Write the formula relating resistance and resistivity, and define resistivity.

Resistance of a conductor depends on (i) its length ll (R∝lR \propto l), (ii) its area of cross-section AA (R∝1/AR \propto 1/A), and (iii) the nature of the material (resistivity ρ\rho). The formula is R=ρl/AR = \rho l / A. Resistivity is the resistance offered by a conductor of unit length and unit cross-sectional area at a given temperature.
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Q6

Derive the equivalent resistance of three resistors R1, R2, R3 connected in series. How does current and potential difference behave in a series combination?

For resistors in series, the same current II flows through each resistor. The total potential difference VV across the combination equals the sum of individual potential differences: V=V1+V2+V3=IR1+IR2+IR3=I(R1+R2+R3)V = V_1 + V_2 + V_3 = I R_1 + I R_2 + I R_3 = I (R_1 + R_2 + R_3). Comparing with V=IRsV = I R_s, the equivalent series resistance is Rs=R1+R2+R3R_s = R_1 + R_2 + R_3.
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Q7

Derive the equivalent resistance of three resistors R1, R2, R3 connected in parallel. How does current and potential difference behave in a parallel combination?

For resistors in parallel, the potential difference VV across each resistor is the same. The total current II drawn from the source equals the sum of individual currents through each resistor: I=I1+I2+I3I = I_1 + I_2 + I_3. Using Ohm's law, I1=V/R1I_1 = V/R_1, I2=V/R2I_2 = V/R_2, I3=V/R3I_3 = V/R_3. Therefore, I=V(1/R1+1/R2+1/R3)I = V(1/R_1 + 1/R_2 + 1/R_3). Comparing with I=V/RpI = V/R_p, we get 1/Rp=1/R1+1/R2+1/R31/R_p = 1/R_1 + 1/R_2 + 1/R_3.
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Q8

Why are household electric circuits arranged in parallel rather than in series? Explain the advantages of a parallel arrangement.

Household circuits use parallel arrangement because: (i) each appliance gets the full supply voltage so it operates at its rated power, (ii) each appliance has its own switch and can be operated independently, (iii) if one appliance fails or is switched off, other appliances continue to work, and (iv) the overall resistance of a parallel circuit is lower, allowing each appliance to draw its required current without affecting others.
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Q9

State Joule's law of heating. Explain two practical applications of the heating effect of electric current.

Joule's law of heating states that the heat HH produced in a resistor is (i) directly proportional to the square of the current (H∝I2H \propto I^2), (ii) directly proportional to the resistance (H∝RH \propto R), and (iii) directly proportional to the time for which the current flows (H∝tH \propto t). Therefore, H=I2RtH = I^2 R t. Applications include: (1) Electric fuse — a thin wire of low melting point melts and breaks the circuit when current exceeds a safe limit, protecting appliances. (2) Heating elements in devices like electric iron and geyser — made of alloys like nichrome which have high resistivity and high melting point, so they produce large amounts of heat without melting.
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Q10

Define electric power. Derive the different forms of the power formula, and explain the commercial unit of electrical energy (kWh).

Electric power is the rate at which electric work is done or electrical energy is consumed. Power P=W/t=VIP = W/t = VI. Using Ohm's law (V=IRV = IR), this can also be written as P=I2RP = I^2 R and P=V2/RP = V^2 / R. The commercial unit of electrical energy is the kilowatt-hour (kWh), where 1 kWh=3.6×106 J1\ \text{kWh} = 3.6 \times 10^6\ \text{J} — it is the energy consumed when a device of 1 kW power operates for 1 hour.
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