Electricity: Heating Effect of Electric Current

Numerical Problems

These numerical problems are based on the chapter Electricity class 10 science of NCERT books and CBSE syllabus.

Example 1: If an electric heater consumes electricity at the rate of 500W and the potential difference between the two terminals of electric circuit is 250V, calculate the electric current and resistance through the circuit.

Solution: Given, power input (P) = 500 W

Potential difference (V) = 250 V

Electric current (I) =?

Resistance (R) through the circuit =?

We know that power `(P) = VI`

Or, `500 W = 250 V xx I`

Or, `I = 500 W ÷ 250 V = 2 A`

We know, resistance `R = V/I`

Or, `R = 250 V ÷ 2 A = 125 Ω`

Example 2: An electric geyser consumes electricity at the rate of 1000W. If the potential difference through the electric circuit is 250 V, find the resistance offered by geyser and electric current through the circuit.

Solution: Given, power input (P) = 1000 W

Potential difference (V) = 250 V

Electric current (I) =?

Resistance (R) through the circuit =?

We know that power `(P) = VI`

Or, `1000 W = 250 V xx I`

Or, `I = 1000 V ÷ 250 V = 4 A`

We know, resistance `R = V/I`

Or, `R = 250 V ÷ 4 A = 62.5 Ω`

Example 3: An electric heater having resistance equal to 5Ω is connected to electric source. If it produces 180 J of heat in one second, find the potential difference across the electric heater.

Solution: Given, Resistance (R) = 5 Ω, Heat (H) produced per second by heater = 1800 J, time t = 1 s

Potential difference (V) =?

To calculate the potential difference, we need to calculate electric current (I) first.

We know that `H = I^2Rt`

Or, `180 J = I^2 xx 5 Ω xx 1 s`

Or, `I^2 = 180 ÷ 5 = 36`

Or, `I = 6 A`

Now, potential difference `V = IR`

Or, `V = 6 A xx 5 Ω = 30 V`