A) 100 watt bulb
B) 25 watt bulb
C) None of them
D) Both of them
Correct Answer: B
Solution :
Resistance of 25 W bulb \[=\,\,\frac{220\times 220}{25}=1936\,\Omega \] Its safe current \[\frac{220}{1936}=0.11\,amp\] Resistance of 100 W bulb \[=\,\,\,\frac{220\times 220}{100}=484\,\,\Omega \] Its safe current \[\frac{220}{484}=0.48\,\,amp\] When connected in series to 440 V supply, then the \[I=\frac{440}{(1936+484)}=0.18\,\,amp\]You need to login to perform this action.
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