Question

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

• A. Zero
• B. $2.9V$
• C. $13.3V$
• D. $10.4V$

Answer 1

Answer: (D)

Resistance of the bulb $R_{b}=\frac{V^{2}}{P_{b}}$ $\Rightarrow R_{b}=\frac{\left(120\right)^{2}}{60}=240ohm,$ Resistance of the heater $R_{h}=\frac{\left(120\right)^{2}}{240}=60ohm$ .
Diagram1
From the above diagram, According to voltage division law voltage across bulb $V_{1}=120\times \frac{240}{246}=117.07V.$ Now heater is connected in parallel with the bulb according to the diagram below.
Diagram 2
From the above diagram, equivalent resistance of bulb and heater is $R_{eq}=\frac{60 \times 240}{300}=48ohm$ . So voltage across bulb in this condition is $V=120\times \frac{48}{54}=106.66V$ . Hence the decrease in voltage across bulb will be $V_{1}-V_{2}=117.07-106.66\sim eq10.4V$