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JEE Main 2013 Paper

Section: Physics

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Question : 63 of 90

Marks: +1, -0

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?

2.9 Volt
13.3 Volt
10.04 Volt
zero volt

Solution:

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 = 10.04

(approximately)

Note: Here supply voltage is taken as rated voltage.

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