Q.
The wiring of a house has resistance 6$\ \Omega$. A 100 W bulb is glowing as shown in figure. If a geyser of 1000 W is switched on, the change in potential drop across the bulb is nearly
A) Nil
B) 12 V
C) 24 V
D) 32 V
Explanation
(c) $R_{bulb} = \frac{220^{2}}{100} = 484\Omega;$
$R_{geyser} = \frac{220^{2}}{1000} = 48.4\Omega$
When only bulb is on,
$V_{bulb} = \frac{220 \times 484}{484 + 6} = 217.4$ volt
When geyser is also switched on, effective resistance of bulb and geyser
$\ = \frac{484 \times 48.4}{484 + 48.4} = 44\ \Omega$
$V_{bulb} = \frac{220 \times 44}{(44 + )} = 193.6$ V
Hence, the potential drop $= 217.4 \times 193.6$
$\ = 23.8$ V$\approx$ 24 V.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.