The supply voltage to a room is 120 V. The resistance of the lead wires is 6Ω. 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 ?

Engineering

Physics

Heating Effects of Current and Electric Power

Question

2.9 Volt

13.3 Volt

zero Volt

10.04 Volt

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$R_{B} = \frac{V^{2}}{P} = \frac{120 \times 120}{60} = 240 Ω$

$R_{H} = \frac{V^{2}}{P} = \frac{120 \times 120}{240} = 60 Ω$

Initially $V_{B} = \frac{120}{246} \times 240 = 117 V$

Later on $V_{B} = \frac{120}{54} \times 48 = 106 .6 V$