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

Explanation

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`.