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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 power dissipated in the circuit |
Answer» The power dissipated in the circuit is `P = I^(2)R` but `I=(i_(m))/(sqrt(2))` and `i_(m)=(upsilon_(m))/(Z)` hence `I=(i_(m))/(sqrt(2))=(upsilon_(m))/(sqrt(2)Z)` given `upsilon m = 283 V` `I = (1)/(sqrt(2))(283)/(5)=40A` Therefore, `P=(40A)^(2)xx3Omega =4800W` `= 4.8 kW`. |
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