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A `120 V, 60 Hz` a.c. power is connected `800 Omega` non-inductive resistance and unknown capacitance in series. The voltage drop across the resistance is found to be `102 V`, then voltage drop across capacitor isA. `8 V`B. `102 V`C. `63 V`D. `55 V` |
Answer» Correct Answer - 3 `V^(2) = V_(R )^(2) + V_(C )^(2)` `V_(C )^(2) = V^(2) - V_(R )^(2)` `V_(C )^(2) = (120)^(2) - (102)^(2)` `V_(C ) = 63 V` |
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