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 of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?

We have a mains (supply) voltage of $$120\ \text{V}$$ feeding the room through lead-in wires whose total resistance is given as $$R_L = 6\ \Omega$$. Any current that flows must first pass through these leads, so whatever appliances are inside the room receive a slightly smaller voltage than the full 120 V because part of the supply voltage is dropped across $$R_L$$.

A 60 W incandescent bulb is already glowing. Because every manufacturer rates a lamp at the voltage that should appear directly across it, we can find its resistance from the power formula

$$P = \frac{V^2}{R}\; ,$$ so $$R = \frac{V^2}{P}.$$

Substituting the rating figures of the bulb,

$$R_B = \frac{(120\ \text{V})^2}{60\ \text{W}} = \frac{14400}{60} = 240\ \Omega.$$

First we calculate the situation before the heater is switched on. The circuit is simply the lead resistance $$R_L$$ in series with the bulb resistance $$R_B$$, and the current through the series combination is

$$I_1 = \frac{V_{\text{supply}}}{R_L + R_B} = \frac{120}{6 + 240} = \frac{120}{246} \approx 0.4878\ \text{A}.$$

The voltage that actually appears across the bulb is then

$$V_{B1} = I_1 \, R_B = 0.4878 \times 240 \approx 117.1\ \text{V}.$$

Thus the initial drop across the lead wires is $$120 - 117.1 \approx 2.9\ \text{V}$$, but we keep the exact figure for later comparison.

Now a 240 W heater is added in parallel with the bulb. We first find the heater’s resistance from the same power formula:

$$R_H = \frac{(120\ \text{V})^2}{240\ \text{W}} = \frac{14400}{240} = 60\ \Omega.$$

The bulb and heater form a parallel combination. The equivalent resistance of two resistors in parallel is given by

$$\frac{1}{R_{\text{eq}}} = \frac{1}{R_B} + \frac{1}{R_H},$$

so

$$R_{\text{eq}} = \frac{R_B\,R_H}{R_B + R_H} = \frac{240 \times 60}{240 + 60} = \frac{14400}{300} = 48\ \Omega.$$

Now the supply sees the series combination of the lead resistance $$R_L = 6\ \Omega$$ and the equivalent load resistance $$R_{\text{eq}} = 48\ \Omega$$. Hence the total current drawn from the mains after the heater is switched on is

$$I_2 = \frac{V_{\text{supply}}}{R_L + R_{\text{eq}}} = \frac{120}{6 + 48} = \frac{120}{54} \approx 2.222\ \text{A}.$$

The voltage that now reaches the parallel junction (and therefore appears across both the bulb and the heater) is

$$V_{B2} = I_2 \, R_{\text{eq}} = 2.222 \times 48 \approx 106.7\ \text{V}.$$

Finally, the decrease in the bulb’s voltage is

$$\Delta V = V_{B1} - V_{B2} \approx 117.1\ \text{V} - 106.7\ \text{V} \approx 10.4\ \text{V}.$$

Hence, the correct answer is Option B.