CBSE Class 10 Science 2019 · Paper 31/1/1 · Question 20(b)

Electricity 5 marks Medium

An electric lamp of resistance

20 Ω

and a conductor of resistance

4 Ω

are connected to a 6 V battery as shown in the circuit. Calculate :

(a)the total resistance of the circuit,
(b)the current through the circuit,
(c)the potential difference across the (i) electric lamp and (ii) conductor, and (d) power of the lamp.

Answer

Total resistance $24 Ω$ , current $0.25$ A, potential differences $5$ V across the lamp and $1$ V across the conductor, lamp power $1.25$ W.

(a) The lamp and the conductor are in series: total resistance

R = 20 Ω + 4 Ω = 24 Ω

. (b) Current

I = \frac{V}{R} = \frac{6 V}{24 Ω} = 0.25

A. (c) (i) Across the lamp:

V_{L} = I R_{L} = 0.25 \times 20 = 5

V; (ii) Across the conductor:

V_{c} = I R_{c} = 0.25 \times 4 = 1

V. (d) Power of the lamp:

P = V_{L} I = 5 V \times 0.25 A = 1.25

🔒 Step-by-step solution