Question
The supply voltage to a room is 120 V. The resistance of the lead wires is 6 Ω. 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?
-
zero
-
2.9V
-
13.3 V
-
10.04 V
Solution
D.
10.04 V
Resistance of bulb =(120x 120)/60 = 240Ω
Resistance of Heater =(120x 120)/240 = 60Ω
Voltage across bulb before heater is switched on,
Voltage across bulb after heater is switched on,
A decrease in the voltage is V1 − V2 = 10.04 (approximately)
