A room is supplied with electricity at 120 V. The lead wires have a total resistance of $6\,\Omega$. A 60 W bulb is already running. What is the decrease in voltage across the bulb when a 240 W heater is switched on in parallel with the bulb?
Resistance of the bulb: $R_b = \frac{V^2}{P} = \frac{(120)^2}{60} = 240\,\Omega$
Before heater is switched on: Current $= \frac{120}{240+6} = \frac{120}{246}\,\text{A}$. Voltage across bulb $= \frac{120 \times 240}{246} \approx 116.9\,\text{V}$.
Resistance of heater: $R_h = \frac{(120)^2}{240} = 60\,\Omega$. Parallel combination: $R_p = \frac{240 \times 60}{300} = 48\,\Omega$.
After heater is switched on: Voltage across combination $= \frac{120 \times 48}{48+6} = \frac{5760}{54} \approx 106.67\,\text{V}$.
Decrease $= 116.9 - 106.67 \approx 10.04\,\text{V}$. Answer: D.
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