The supply voltage to room is 120 V. The resistance of the lead wires is 6 Ω 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?
Text Solution
Verified by ExpertsThe correct answer is:
D
= 240 Ω Ω
R eq . = 240 + 6 = 246 Ω Ω
V 1 = 
= 117.073 volt
V 1 – V 2 = 10.04 Volt
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