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A capacitor of capacity `C` is connected in `A.C` circuit. If the applied emf is `V = V_(0) sin omega t`, then the current isA. `I = (V_(0))/(L omega) sin omega t`B. `I = (V_(0))/(omega C) sin (omega t + (pi)/(2))`C. `I = V_(0) C omega sin omega t`D. `t = V_(0) C omega sin (omega t + (pi)/(2))` |
Answer» Correct Answer - 4 | |