Question

An electric bulb is rated 240V-60W and is working at 100% efficiency.
(i) Calculate the resistance of bulb.
(ii) (a) Draw the circuit diagram.
(b) What is the rate of conversion of energy in each bulb?
(c) Total power used by the bulbs.

Answer 1

(i) From P = `"V"^2/"R"`

Resistance of bulb (R) = `"V"^2/"P" = (240 xx 240)/60 = 960 Omega`

(ii) (a)

(b) Resistance of bulbs in series = (960 + 960) Ω = 1920 Ω

∴ Current in series circuit I = `"V"/"R" = 240/1920 = 0.125` A

∴ Rate of conversion of energy in any one bulb in 1 second = I²Rt

= (0.125)² × 960 × 1 = 15.5

(c) Power of 1 bulb in series circuit = 15 W

So poewr of 2 bulb in series circuit = 2 × 15 W = 30 W.