Question

The supply voltage to room is 120 V. The resistance of the lead wires is `6Omega`. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?
A. zero
B. 2.9 Volt
C. 13.3 Volt
D. 10.04 Volt

Answer 1

Correct Answer - D
(d)
Power of bulb = 60 W (given)
Resistance of bulb `= (120 xx 120 )/60 = 240 Omega [.: P = V^2/R]`
Power of heater = 240 W (given)
Resistance of heater `= (120 xx 120)/240 = 60 Omega`
Voltage across bulb before heater is switched on,
`V_1 = (240)/(246) xx 120 = 117.73 volt`
Voltage across bulb after heater is switched on,
`V_2 = 48/54 xx 120 = 106.66 volt.
Hence decrease in voltage
`V_1 - V_2 = 117.073 - 106.66 = 10.04 Volt (approximately)`.