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. 13.5 V
• D. 10.4 V

Answer 1

Answer: (D)

10.4 V

Answer 2

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 $V _{2}=\frac{12}{ R _{ b }+6} \times R _{ b }=\frac{120}{240+6} \times 240=117 V$ After the heater is connected, the voltage is drop is $V _{1}=\frac{120}{ R +6} \times R =\frac{120}{48+6} \times 48=106.66 V$ So, $V _{2}- V _{1}=10.04 V$