Chapter 11 · CBSE Class 10 Physics (NCERT)

Electricity

Learn Ohm's law, how to combine resistors in series and parallel, and how to calculate electrical power and energy.

⚡ Chapter 11

11.1

Ohm's Law

At constant temperature, the current through a conductor is directly proportional to the voltage across it:

V = I × R
1

A resistor has 12 V across it and 2 A flowing through it. Find its resistance.

R = V/I = 12/2 = 6 Ω

11.2

Series and Parallel Resistance

Series: R = R₁ + R₂ + ...
Parallel: 1/R = 1/R₁ + 1/R₂ + ...
1

Find the total resistance of two 4 Ω and 6 Ω resistors connected in series.

R = 4 + 6 = 10 Ω

2

Find the total resistance of a 6 Ω and a 3 Ω resistor connected in parallel.

1/R = 1/6 + 1/3 = 1/2, so R = 2 Ω

Why it matters — Household appliances are wired in parallel (not series) specifically so each device gets the full supply voltage and keeps working even if another appliance is switched off.
11.3

Electrical Power

P = V × I
1

A device draws 2 A at 10 V. Find its power.

P = V × I = 10 × 2 = 20 W

2

A 100 W bulb runs for 5 hours. How much energy does it use, in kWh (units)?

Energy = Power × time = (100 × 5) / 1000 = 0.5 kWh

Practice

Exercises

1. Using V=IR, find R if V=20V and I=4A.

2. Find the series combination of 2 Ω, 3 Ω, and 5 Ω resistors.

3. Find the parallel combination of two equal 10 Ω resistors.

4. Using P=VI, find the power of a device with V=5V and I=3A.

5. Is total resistance in a parallel combination always less than the smallest individual resistor?