The supply voltage to a room is 120 V. The re | vedclass

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(A) Power of bulb $P_b = 60\, W$. Resistance of bulb $R_b = \frac{V^2}{P_b} = \frac{120^2}{60} = 240\,\Omega$.Resistance of lead wires $R_L = 6\,\Omeg

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$10.4$

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(A) Power of bulb $P_b = 60\, W$. Resistance of bulb $R_b = \frac{V^2}{P_b} = \frac{120^2}{60} = 240\,\Omega$.Resistance of lead wires $R_L = 6\,\Omega$.Voltage across bulb before heater is switched on: $V_1 = \frac{R_b}{R_b + R_L} 	imes 120 = \frac{240}{240 + 6} 	imes 120 = \frac{240}{246} 	imes 120 \approx 117.07\, V$.Power of heater $P_h = 240\, W$. Resistance of heater $R_h = \frac{V^2}{P_h} = \frac{120^2}{240} = 60\,\Omega$.When heater is in parallel with bulb,equivalent resistance $R_p = \frac{R_b 	imes R_h}{R_b + R_h} = \frac{240 	imes 60}{240 + 60} = \frac{14400}{300} = 48\,\Omega$.Voltage across bulb after heater is switched on: $V_2 = \frac{R_p}{R_p + R_L} 	imes 120 = \frac{48}{48 + 6} 	imes 120 = \frac{48}{54} 	imes 120 \approx 106.67\, V$.Decrease in voltage $= V_1 - V_2 = 117.07 - 106.67 = 10.4\, V$.

The supply voltage to a room is $120\ V$. The resistance of the lead wires is $6\,\Omega$. $A$ $60\ W$ bulb is already switched on. What is the decrease in voltage across the bulb when a $240\ W$ heater is switched on in parallel to the bulb? ............. $V$

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