Question

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

• A. Zero
• B. 2.9 V
• C. 13.5 V
• D. 10.4 V

Answer 1

Answer: (D)

Resistance of the bulb is say $R_b$ Using, $p=\frac{V^2}{R}$
or $R=\frac{V^2}{P}$
We have,
$R_b=\frac{120^2}{60}=240\Omega$
Similarly for the heater,
$R_n=\frac{120^2}{240}=60\Omega$
Now, the equivalent resistance of the bulb and heater together is
$R=\frac{R_b{R}_n}{R_b+R_n}=\frac{240\times 60}{240+60}=48\Omega$
Before the heater was connected, the voltage drop across the bulbs
$V_2=\frac{12}{R_b+6}\times R_b=\frac{120}{240+6}\times 240=117V$
After the heater is connected, the voltage is drop is
$V_1=\frac{120}{R+6}\times R=\frac{120}{48+6}\times 48=106.66V$
So, $V_2-V_1=10.04V$