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A bulb with rating 60 W,120 V, with the connection of wire 6Ω is switched on in a room. Then the decrease in the voltage across the bulb, when a 240W,120 V heater is switched on in parallel to the bulb can be calculated as: (Given that, the supply voltage is 120 V.)

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13.3 Volt

10.04 Volt

zero Volt

2.9 Volt

answer is B.

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Detailed Solution

Rating of a bulb 60 W, 120 V

$P = \frac{v^{2}}{R}$

Resistance of Bulb

$(R) = \frac{120 \times 120}{60} = 240 Ω$

$\begin{array}{l} R_{eq} = 240 + 6 = 246 Ω \\ v_{1} = \frac{240}{246} \times 120 = 117.073 volt \end{array}$

$Resistance of the heater R = \frac{120 \times 120}{240} = 60 Ω equivalent resistance = \frac{60 \times 240}{300} = 48 Ω V_{2} = \frac{48}{54} \times 120 = 106.66 volt$

$v_{1} - v_{2} = 10.04 Volt$

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