Question

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

• A. no change
• B. 10 V
• C. 20 V
• D. more than 10 V

Answer 1

The correct option is D more than 10 V

Using $P=V^2/R$ the resistance of bulb is $R_{60}=\frac{{120}^2}{60}=240\Omega$.
Current , $I=\frac{120}{240+6}=\frac{120}{246}A$
Thus Voltage across bulb is $V_{60}=R_{60}I=240\times \frac{120}{246}=117.07V$

When heater is switch on, the bulb and heater are in parallel combination.
Resistance of heater is $R_{240}=\frac{{120}^2}{240}=60\Omega$
Resistance of parallel combination $R_{||}=\frac{240\times 60}{240+60}=48\Omega$
Total resistance $R_t=6+48=54\Omega$
Now the current , $I^{\prime }=\frac{120}{54}=2.22A$
Voltage across parallel combination $V^{\prime }=I^{\prime }{R}_{||}=2.22\times 48=106.67V$. This is also voltage across heater.

Charge in voltage $=V_{60}-V^{\prime }=117.07-106.67=10.4V$ which is more than 10.