A) 2420 and 311 V
B) 4840 and 311 V
C) 484 Q and 440 V
D) 242 Q and 440 V
Correct Answer: B
Solution :
The resistance of bulb \[R=\frac{{{V}^{2}}}{P}\] \[=\frac{220\times 220}{100}\] \[=484\,\Omega \] The peak voltage of the source \[{{V}_{0}}=220\sqrt{2}=311\,V\]You need to login to perform this action.
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