NCERT Solution Class 10 Science Chapter 11: Electricity

NCERT Exercise Solutions (Easy Explanation for Beginners)

Q1.

A wire of resistance R is cut into five equal parts and all parts are connected in parallel. Find the ratio R / R′.

Answer:
Each part has resistance = R/5
In parallel,
1/R′ = 5 × (5/R) = 25/R
⇒ R′ = R/25
✅ Ratio R / R′ = 25 : 1

Q2.

Which of the following does not represent electrical power in a circuit?
(a) IR² (b) VI (c) V²/R (d) V/I

✅ Answer: (a) IR²
Reason: Power = VI = I²R = V²/R, not IR².

Q3.

A bulb is rated 220 V, 100 W. When operated at 110 V, find power consumed.

Answer:
R = V² / P = (220)² / 100 = 484 Ω
New power = (110)² / 484 = 25 W
✅ Power = 25 W

Q4.

Two wires of same material and length are connected in (i) series and (ii) parallel. Find the ratio of heat produced.

Answer:
Heat = V²t / R
In series: R₁ = 2R
In parallel: R₂ = R/2
H₁ / H₂ = R₂ / R₁ = (R/2) / (2R) = 1/4
✅ Ratio = 1 : 4

Q5.

How is a voltmeter connected to measure potential difference?

Answer:
A voltmeter is always connected in parallel with the component across which potential difference is measured.

Q6.

A copper wire has diameter 0.5 mm, resistivity 1.6×10⁻⁸ Ωm.
Find the length if resistance is 10 Ω.
Also find new resistance if diameter is doubled.

Answer:
r = 0.25 mm = 0.25×10⁻³ m
A = πr² = 3.14 × (0.25×10⁻³)² = 1.96×10⁻⁷ m²
R = ρL/A
L = (R×A)/ρ = (10×1.96×10⁻⁷)/(1.6×10⁻⁸) = 122.5 m
If diameter doubles, area increases 4 times ⇒ resistance becomes ¼.
✅ New resistance = 2.5 Ω

Q7.

Given data:

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

Find resistance.

Answer:
R = V/I
= 1.6/0.5 = 3.2 Ω, 3.4/1.0 = 3.4 Ω, 6.7/2.0 = 3.35 Ω, etc.
✅ Average Resistance ≈ 3.4 Ω

Q8.

A 12 V battery gives 2.5 mA current. Find resistance.

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

Q9.

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

Answer:
Total R = 13.4 Ω
I = V/R = 9/13.4 = 0.67 A
✅ Current = 0.67 A

Q10.

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

Answer:
R = V/I = 220/5 = 44 Ω
For n resistors in parallel: 176/n = 44 ⇒ n = 4
✅ 4 resistors required

Q11.

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

Answer:
(i) For 9 Ω: two in parallel (gives 3 Ω) + one in series → 3 + 6 = 9 Ω
(ii) For 4 Ω: two in series (12 Ω) + one in parallel → (12×6)/(12+6) = 4 Ω
✅ (i) 9 Ω → 2 parallel + 1 series
✅ (ii) 4 Ω → 2 series + 1 parallel

Q12.

Each bulb is 10 W, 220 V. How many can be connected in parallel if total current is 5 A?

Answer:
I = P/V = 10/220 = 0.0455 A
No. of bulbs = 5 / 0.0455 = 110
✅ 110 bulbs

Q13.

Two coils A and B, each 24 Ω. Find current if connected
(i) separately, (ii) in series, (iii) in parallel.

Answer:
(i) Single: R = 24 Ω → I = 220/24 = 9.17 A
(ii) Series: R = 48 Ω → I = 220/48 = 4.58 A
(iii) Parallel: R = 12 Ω → I = 220/12 = 18.33 A
✅ Currents = 9.17 A, 4.58 A, 18.33 A

Q14.

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

Answer:
(i) Series: R = 3 Ω, I = 6/3 = 2 A
P = I²R = 4×2 = 8 W
(ii) Parallel: voltage across 2 Ω = 4 V
P = V²/R = 16/2 = 8 W
✅ Same power = 8 W

Q15.

Two lamps 100 W and 60 W at 220 V connected in parallel. Find total current.

Answer:
I₁ = 100/220 = 0.455 A
I₂ = 60/220 = 0.273 A
Total = 0.728 A
✅ Total current = 0.73 A

Q16.

Which uses more energy:
(a) 250 W TV in 1 hour
(b) 1200 W toaster in 10 minutes

Answer:
TV: 250 × 1 = 250 Wh
Toaster: 1200 × 1/6 = 200 Wh
✅ TV uses more energy

Q17.

Electric heater: R = 44 Ω, I = 5 A, t = 2 hours. Find heat produced.

Answer:
H = I²Rt = 5²×44×2×3600 = 7.92×10⁶ J
✅ Heat = 7.92 MJ

Q18.

Explain:

(a) Why tungsten is used in bulbs?
It has a very high melting point (about 3380°C) and gives light when heated.

(b) Why alloys are used in heaters?
They have higher resistivity and do not oxidize easily.

(c) Why series circuits are not used in homes?
Because if one appliance fails, others stop working, and voltage is not distributed equally.

(d) How resistance changes with area?
Resistance decreases when area increases (R ∝ 1/A).

(e) Why copper and aluminium are used for transmission?
Because they are good conductors and have very low resistivity.

🔍 Frequently Asked Questions (FAQs)

Q1. What is the formula for electric current?

Answer: Electric current is the rate of flow of electric charge.
It is given by the formula I = Q / t,
where I is current, Q is charge, and t is time.

Q2. What is Ohm’s Law?

Answer: According to Ohm’s Law,
the current flowing through a conductor is directly proportional to the potential difference across it,
if temperature remains constant.
Mathematically, V = IR

Q3. What is the SI unit of resistance?

Answer: The SI unit of resistance is ohm (Ω).

Q4. What happens to resistance when wire length or thickness changes?

Answer:

Resistance increases when the length of the wire increases.
Resistance decreases when the thickness (area) of the wire increases.

Q5. Why are copper and aluminium used for electrical wires?

Answer: Copper and aluminium are good conductors of electricity and have low resistivity.
They allow electricity to pass easily with very little heat loss.

Q6. What is the relationship between power, voltage, and current?

Answer:
Electrical power (P) is the rate at which electrical energy is used.
Formula: P = VI = I²R = V²/R

Q7. What causes heat in an electrical circuit?

Answer:
When current flows through a resistor, energy is converted into heat.
This is called the Joule heating effect and is given by
H = I² R t

Q8. Why are household appliances connected in parallel, not in series?

Answer:
Because in parallel connection, each appliance gets the same voltage.
If one device fails, others keep working properly.

Q9. What is the SI unit of electric power?

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

Q10. What is the function of a fuse in an electric circuit?

Answer:
A fuse protects the circuit from damage.
If current becomes too high, the fuse wire melts and breaks the circuit to prevent fire or appliance damage.