A) Nil
B) 23 V
C) 32 V
D) 12 V
Correct Answer: B
Solution :
\[{{R}_{Bulb}}=\frac{{{220}^{2}}}{100}=484\,\,\Omega \], \[{{R}_{Geyser}}=\frac{{{220}^{2}}}{1000}=48.4\,\,\Omega \] (i) When only bulb is ON, \[{{V}_{Bulb}}=\frac{220\times 484}{490}=217.4\,V\] (ii) When geyser is also switched ON, equivalent resistance of bulb and geyser is \[R=\frac{484\times 48.4}{484+48.4}=44\,\Omega \] Voltage across the bulb \[{{V}_{Bulb}}=\frac{220\times 44}{50}=193.6\,V\] Hence the potential drop is \[217.4-193.6=23.8\,\,V\]You need to login to perform this action.
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