Q. The main supply voltage to a room is 120 V. The resistance of the lead wires is 6$\Omega$. A 60 W bulb is already giving light. What is the decrease in voltage across the bulb when a 240 W heater is switched on?
A) No change
B) 10 V
C) 20 V
D) More than 10 V
Explanation
(d) $R_{60} = \frac{120 \times 120}{60} = 240\Omega$
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Current $= \frac{120}{240 + 6}A = \frac{120}{246}A$
Voltage across bulb
$= \frac{120}{246} \times 240\ volt = 117.1\ volt$
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$R_{240} = \frac{120 \times 120}{240}60\Omega$
Resistance of parallel combination
$\ = \frac{60 \times 240}{60 + 240} = 48\Omega.$
Total resistance $= (48 + 6)\Omega = 54\Omega.$
Current $I = \frac{120}{54}A$
Voltage across parallel combination
$\ = \frac{120}{54} \times 48\ volt = 106.7\ volt$
Change in voltage $= (117.1 - 106.7) = 10.4\ $V.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.