Question

The supply voltage to 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 volt
• B. $2.9$ volt
• C. $13.3$ volt
• D. $10.4$ volt

Answer 1

The correct option is D $10.4$ volt

Resistance of bulb $=\frac{120\times 120}{60}=240\Omega$

Resistance of heater $=\frac{120\times 120}{240}=60\Omega$

Voltage across bulb before heater switched, on

$V_1=\frac{120}{24}\times 240=117.07$

Voltage across bulb after heater switched on $V_2$

$V_2=\frac{120}{54}\times 48=100.60$

Decrease of water $=V_1-V_2=10.04$