question_answer

The supply voltage to 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?       [JEE MAIN 2013]

A) zero Volt

B) 2.9 Volt

C) 13.3 Volt

D) None of these

Correct Answer: D

Solution :

(no option matches)

\[{{R}_{b}}=\frac{{{(120)}^{2}}}{60}=240\Omega \],\[{{R}_{H}}=\frac{240}{4}=60\Omega \]

\[{{V}_{1}}=120\frac{240}{243}=120,\frac{40}{41}V\]

\[{{V}_{2}}=120\frac{({{R}_{b}}||{{R}_{H}})}{({{R}_{b}}||{{R}_{H}})+6}=120\frac{48}{54}=120\frac{8}{9}V\]

Loss in potential \[={{V}_{1}}-{{V}_{2}}\]

\[=120\left( \frac{40}{41}-\frac{8}{9} \right)\]

\[=10.40\text{ }V.\]