NCERT Solutions for CBSE Class 10 Science Chapter 12 – Electricity

Electricity is one of the highest-scoring chapters in CBSE Class 10 Physics — but it’s also formula-heavy, so clear step-by-step solutions matter a lot here. Below are complete, freshly-worked NCERT solutions covering every important question from the chapter.

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Exercise Questions (Intro Section)

Q1. What does an electric circuit mean?

Answer: An electric circuit is a continuous, closed path made of conducting wires and components through which electric current can flow. A basic circuit needs a source (cell/battery), conductors (wires), a switch to control the flow, and a load (like a bulb or resistor) that consumes the electrical energy.

Q2. Define the unit of electric current.

Answer: The SI unit of electric current is the ampere (A). A current of 1 ampere means 1 coulomb of charge is flowing past a given point in the circuit every second.

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

Answer: Charge on one electron, e = 1.6 × 10⁻¹⁹ C
Using Q = n × e:
n = Q / e = 1 / (1.6 × 10⁻¹⁹) = 6.25 × 10¹⁸ electrons

Page 202 Questions

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

Answer: A battery (or electric cell) maintains the potential difference needed to keep current flowing through a conductor.

Q2. What does it mean to say the potential difference between two points is 1 V?

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

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

Answer: Using V = W/Q, so W = V × Q = 6 V × 1 C = 6 J of energy per coulomb.

Page 209 Questions

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

Answer:

• Length – resistance increases with length
• Cross-sectional area – resistance decreases as area increases
• Nature of material – different materials have different resistivity
• Temperature – resistance generally increases as temperature rises

Q2. Does current flow more easily through a thick or thin wire of the same material? Why?

Answer: A thick wire allows current to flow more easily. Since R = ρl/A, resistance is inversely proportional to the cross-sectional area — a larger area gives more room for charge carriers to move, resulting in lower resistance.

Q3. If the potential difference across a component is halved while resistance stays constant, what happens to the current?

Answer: By Ohm’s Law, I = V/R. If V becomes V/2 and R is unchanged, the new current I’ = (V/2)/R = I/2. The current is also reduced to half.

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

Answer: Alloys (like nichrome) have much higher resistivity than pure metals, so they generate more heat for the same current. They also have a higher melting point and resist oxidation at high temperatures, making them durable for repeated heating.

Q5. (a) Which is a better conductor — iron or mercury? (b) Which material is the best conductor?

Answer:
(a) Iron is a better conductor than mercury, since mercury has a higher resistivity.
(b) Silver is the best conductor among common materials, as it has the lowest resistivity (1.60 × 10⁻⁸ Ωm).

Page 213 Questions

Q1. Draw a circuit with three 2V cells, resistors of 5Ω, 8Ω, 12Ω, and a plug key, all in series.

Answer: Three 2V cells in series give a total EMF of 6V. The circuit diagram shows the 6V battery connected in series with the 5Ω, 8Ω, and 12Ω resistors, along with a plug key to open/close the circuit.

Q2. Add an ammeter and a voltmeter (across the 12Ω resistor) to the above circuit. Find their readings.

Answer: The ammeter is connected in series (to measure current through the circuit), and the voltmeter is connected in parallel across the 12Ω resistor.

Total resistance = 5 + 8 + 12 = 25 Ω
Current, I = V/R = 6/25 = 0.24 A (ammeter reading)
Voltage across 12Ω resistor = I × R = 0.24 × 12 = 2.88 V (voltmeter reading)

Page 216 Questions

Q1. Find the equivalent resistance for: (a) 1Ω and 10⁶Ω in parallel, (b) 1Ω, 10³Ω, and 10⁶Ω in parallel.

Answer:
(a) 1/R = 1/1 + 1/10⁶ ≈ 1.000001 → R ≈ 1 Ω (the large resistor barely affects the result)
(b) 1/R = 1 + 1/1000 + 1/10⁶ ≈ 1.001 → R ≈ 0.999 Ω

Q2. A 100Ω lamp, 50Ω toaster, and 500Ω water filter are in parallel on a 220V line. What resistance and current would an electric iron need to draw the same total current?

Answer:
1/R = 1/100 + 1/50 + 1/500 = 0.01 + 0.02 + 0.002 = 0.032
R = 31.25 Ω
Current drawn = V/R = 220/31.25 = 7.04 A

Q3. What are the advantages of connecting devices in parallel rather than in series?

Answer:

• Each device gets the full supply voltage, ensuring consistent performance
• Devices work independently — if one fails or is switched off, the others keep running
• Adding more devices reduces total resistance, allowing appropriate current to flow to each one

Q4. Connect three resistors — 2Ω, 3Ω, 6Ω — to get a total resistance of (a) 4Ω, (b) 1Ω.

Answer:
(a) Connect 3Ω and 6Ω in parallel: (3×6)/(3+6) = 2Ω. Add this in series with the 2Ω resistor: 2 + 2 = 4Ω
(b) Connect all three in parallel: 1/R = 1/2 + 1/3 + 1/6 = 1 → R = 1Ω

Q5. What is the highest and lowest total resistance from four coils of 4Ω, 8Ω, 12Ω, 24Ω?

Answer:
Highest (series): 4 + 8 + 12 + 24 = 48 Ω
Lowest (parallel): 1/R = 1/4 + 1/8 + 1/12 + 1/24 = 1/2 → R = 2 Ω

Page 218 Questions

Q1. Why doesn’t the cord of an electric heater glow, while the heating element does?

Answer: The heating element is made of a high-resistance alloy, so it heats up intensely (and glows) when current flows. The cord, made of low-resistance copper/aluminium, generates very little heat and stays cool — this also keeps the cord safe to touch.

Q2. Calculate the heat generated when 96,000 C of charge flows in 1 hour through a 50V potential difference.

Answer: Using H = V × Q (since Q = It, this simplifies from H = VIt):
H = 50 × 96,000 = 4.8 × 10⁶ J

Q3. An electric iron of 20Ω resistance draws 5A current. Find the heat developed in 30 seconds.

Answer: H = I²Rt = (5)² × 20 × 30 = 25 × 20 × 30 = 1.5 × 10⁴ J

Page 220 Questions

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

Answer: The rate of energy delivery is called electric power, given by P = VI — it depends on both the current flowing and the potential difference across the device.

Q2. An electric motor draws 5A from a 220V line. Find its power and energy consumed in 2 hours.

Answer:
P = VI = 220 × 5 = 1100 W
E = P × t = 1100 × (2 × 3600) = 7.92 × 10⁶ J

Page 221 – MCQs and Numericals

Q1. A wire of resistance R is cut into 5 equal parts and connected in parallel. Find R/R′.

Answer: Each part has resistance R/5. In parallel: R′ = (R/5)/5 = R/25
So R/R′ = 25 (option d)

Q2. Which does NOT represent electrical power?
(a) I²R (b) IR² (c) VI (d) V²/R

Answer: (b) IR² — Power is correctly given by VI, I²R, or V²/R (all derived from Ohm’s Law), but not IR².

Q3. A bulb rated 220V, 100W is run on 110V. Find the power consumed.

Answer: R = V²/P = (220)²/100 = 484 Ω (fixed value)
At 110V: P = V²/R = (110)²/484 = 25 W (option d)

Q4. Two identical wires connected first in series, then in parallel, across the same voltage. Find the ratio of heat produced (series : parallel).

Answer: Let each wire have resistance r.
Series: Rs = 2r, Power = V²/2r
Parallel: Rp = r/2, Power = 2V²/r
Ratio = (V²/2r) : (2V²/r) = 1 : 4 (option c)

Q5. How is a voltmeter connected to measure potential difference?

Answer: A voltmeter is always connected in parallel across the two points where the potential difference is to be measured, so it doesn’t disturb the current flow in the main circuit.

Q6. A copper wire has diameter 0.5 mm and resistivity 1.6 × 10⁻⁸ Ωm. Find its length for 10Ω resistance. How does resistance change if the diameter is doubled?

Answer:
Radius = 0.25 mm = 2.5 × 10⁻⁴ m
Area, A = πr² = π × (2.5×10⁻⁴)² ≈ 1.9635 × 10⁻⁷ m²
Using R = ρl/A → l = RA/ρ = (10 × 1.9635×10⁻⁷)/(1.6×10⁻⁸) ≈ 122.7 m

If the diameter is doubled, the area becomes 4× larger, so resistance becomes ¼ of original:
New R = 10/4 = 2.5 Ω

Q7. Plot V–I data and find resistance:

I (A)	0.5	1.0	2.0	3.0	4.0
V (V)	1.6	3.4	6.7	10.2	13.2

Answer: The V–I graph is a straight line through the origin (confirming Ohm’s Law). Taking two points, say (1.0, 3.4) and (3.0, 10.2):
Slope = ΔV/ΔI = (10.2 − 3.4)/(3.0 − 1.0) = 6.8/2 = 3.4
Since V = IR, the slope equals resistance: R = 3.4 Ω

Q8. A 12V battery gives a current of 2.5 mA through an unknown resistor. Find its resistance.

Answer: R = V/I = 12/(2.5×10⁻³) = 4800 Ω (4.8 kΩ)

Q9. A 9V battery is in series with resistors 0.2Ω, 0.3Ω, 0.4Ω, 0.5Ω, and 12Ω. Find the current through the 12Ω resistor.

Answer: In series, current is the same everywhere.
Total R = 0.2+0.3+0.4+0.5+12 = 13.4 Ω
I = V/R = 9/13.4 ≈ 0.671 A

Q10. How many 176Ω resistors in parallel are needed to draw 5A from a 220V line?

Answer: Required total resistance = V/I = 220/5 = 44Ω
Number of resistors, n: 176/n = 44 → n = 4 resistors

Q11. Connect three 6Ω resistors to get (i) 9Ω, (ii) 4Ω.

Answer:
(i) Two resistors in parallel: (6×6)/(6+6) = 3Ω. Add the third in series: 3 + 6 = 9Ω
(ii) Two resistors in series: 6 + 6 = 12Ω. Put this in parallel with the third: (12×6)/(12+6) = 4Ω

Q12. 10W bulbs on a 220V line, max allowable current 5A. How many bulbs can be connected in parallel?

Answer: Current per bulb = P/V = 10/220 ≈ 0.0455 A
Number of bulbs = Total current / current per bulb = 5/0.0455 ≈ 110 bulbs

Q13. Two 24Ω coils (A and B) on a 220V line, used separately, in series, and in parallel. Find the current in each case.

Answer:
Separately: I = 220/24 ≈ 9.17 A (each coil)
Series: Total R = 48Ω, I = 220/48 ≈ 4.58 A
Parallel: Total R = 12Ω, I = 220/12 ≈ 18.33 A

Q14. Compare power in the 2Ω resistor: (i) 6V battery with 1Ω & 2Ω in series, (ii) 4V battery with 12Ω & 2Ω in parallel.

Answer:
(i) Total R = 3Ω, I = 6/3 = 2A. Power in 2Ω = I²R = 4×2 = 8W
(ii) Voltage across 2Ω (parallel) = 4V. Power = V²/R = 16/2 = 8W
Both cases give equal power of 8W.

Q15. Two lamps (100W and 60W, both at 220V) are connected in parallel to 220V mains. Find the current drawn from the line.

Answer:
I₁ = P/V = 100/220 ≈ 0.455 A
I₂ = P/V = 60/220 ≈ 0.273 A
Total current = I₁ + I₂ ≈ 0.727 A

Q16. Which uses more energy — a 250W TV for 1 hour, or a 1200W toaster for 10 minutes?

Answer:
TV: E = 250 × 3600 = 9 × 10⁵ J
Toaster: E = 1200 × 600 = 7.2 × 10⁵ J
The TV consumes more total energy, despite the toaster having higher power, because it runs much longer.

Q17. An 8Ω heater draws 15A. Find the rate of heat production.

Answer: P = I²R = (15)² × 8 = 225 × 8 = 1800 W

Q18. Explain:

(a) Why is tungsten used for lamp filaments?

Tungsten has a very high melting point and high resistivity, so it can glow white-hot without melting when current passes through it.

(b) Why are alloys used in heating devices instead of pure metals?

Alloys have higher resistivity, produce more heat for the same current, and resist oxidation and melting at high operating temperatures.

(c) Why isn’t series arrangement used in domestic circuits?

In series, all devices share the same current, and if one device fails or is switched off, the entire circuit breaks. Also, voltage gets divided, so no device gets the full supply voltage.

(d) How does resistance vary with cross-sectional area?

Resistance is inversely proportional to cross-sectional area — thicker wires have less resistance.

(e) Why are copper and aluminium used for electricity transmission?

Both have very low resistivity, making them efficient conductors that minimize power loss as heat over long transmission distances. They’re also cost-effective compared to silver or gold.

Quick Formula Recap

Concept	Formula
Ohm’s Law	V = IR
Series resistance	R = R₁ + R₂ + …
Parallel resistance	1/R = 1/R₁ + 1/R₂ + …
Resistance & dimensions	R = ρl/A
Electric Power	P = VI = I²R = V²/R
Heat (Joule’s Law)	H = I²Rt

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