Question

The supply voltage in 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?

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

Answer 1

The correct option is D 10.4 V


Resistance of the bulb is say $R_b$
Using, $P=\frac{V^2}{R}\Rightarrow R=\frac{V^2}{P}$
We have,
$R_b=\frac{{120}^2}{60}=240\Omega$
Similarly for the heater,
$R_h=\frac{{120}^2}{240}=60\Omega$
Now, the equivalent resistance of the bulb and heater together is,
$R=\frac{R_b{R}_h}{R_b+R_h}=\frac{240\times 60}{240+60}=48\Omega$
Before the heater was connected, the voltage drop across the bulb is,
$V_2=\frac{120}{R_b+6}\times R_b=\frac{120}{240+6}\times 240=117V$
After the heater is connected, the voltage drop is,
$V_1=\frac{120}{R+6}\times R=\frac{120}{48+6}\times 48=106.66V$