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 - Physics MCQ JEE Main 2026 | ExamTest.live

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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?

Solution & Step-by-step Explanation

1. Resistance of bulb R_{b} = \frac{V ^{2}}{P} = \frac{12 0 ^{2}}{60} = 240 Ω .2. Resistance of heater R_{h} = \frac{12 0 ^{2}}{240} = 60 Ω .3. Before heater: Total resistance R_{t o t a l} = 6 + 240 = 246 Ω .Current I_{1} = \frac{120}{246} A .Voltage across bulb V_{1} = I_{1} \times 240 = \frac{120 \times 240}{246} \approx 117.07 V .4. After heater: R_{b} and R_{h} are in parallel. R_{p} = \frac{240 \times 60}{240 + 60} = 48 Ω .Total resistance R_{t o t a l}^{'} = 6 + 48 = 54 Ω .Current from supply I_{2} = \frac{120}{54} = \frac{20}{9} A .Voltage across parallel part V_{2} = I_{2} \times 48 = \frac{20 \times 48}{9} \approx 106.67 V .5. Decrease in voltage = V_{1} - V_{2} = 117.07 - 106.67 = 10.4 V .

Try it yourself before checking the explanation above.

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 V

13.3 V

10.04 V

zero volt

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