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A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which `R = 3 Omega. L = 25.48mH`. And `C = 796mu F`. The phase difference between the voltage across the source and the current |
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Answer» Phase difference, `phi=tan^(-1)(X_(C )-X_(L))/(R )` `= tan^(-1)((4-8)/(3))=-53.1^(@)` Since `phi` is negative, the current in the circuit lags the voltage across the source. |
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