Question

The supply voltage to a room is 120 $V$. The resistance of the lead wires is 6 $\Omega$. A $60W,120V$ bulb is already switched on. What is the decrease of voltage (in Volts ) across the bulb, when a $240W,120V$ heater is switched on in parallel to the bulb?

• A. 2.9
• B. 13.3
• C. 10.4
• D. 0

Answer 1

Answer: (C)

10.4

Resistance of bulb $=\frac{120\times 120}{60}=240\Omega$
Resistance of Heater $=\frac{120\times 120}{240}=60\Omega$
Voltage across bulb before heater is switched on, $V_1=\frac{120}{246}\times 240$

Voltage across bulb after heater is switched on, $V_2=\frac{120}{54}\times 48$

Decrease in the voltage is $V_1-V_2=10.4$

Note : Here supply voltage is taken as rated voltage