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?     AIEEE Solevd Paper-2013

A) zero Volt             

B)        2.9 Volt                

C)        13.3 Volt             

D)        10.04 Volt

Correct Answer: D

Solution :

\[{{R}_{b}}=\frac{{{(120)}^{2}}}{60}=240\Omega \],\[{{R}_{H}}=\frac{240}{4}=60\Omega \] \[{{V}_{1}}=120\frac{240}{243}=120,\frac{40}{41}V\] \[{{V}_{2}}=120\frac{({{R}_{b}}||{{R}_{H}})}{({{R}_{b}}||{{R}_{H}})+6}=120\frac{48}{54}=120\frac{8}{9}V\] Loss in potential               \[={{V}_{1}}-{{V}_{2}}\] \[=120\left( \frac{40}{41}-\frac{8}{9} \right)\] \[=10.40\text{ }V.\]