A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω . L = 25.48 m H . And C = 796 μ F .Suppose the frequency of the source in the previous example can be varied.
Calculate the impedance, the current, and the power dissipated at the resonant condition.

Explanation

The impedance Z at resonant condition is equal to the resistance
`Z = R = 3Omega`
The rms current at resonance is,
as `V=(upsilon_(m))/(sqrt(2))`
`I=(V)/(Z)=(V)/(R )=((283)/(sqrt(2)))(1)/(3)=66.7A`
The power dissipated at resonance is
`P=I^(2)xxR=(66.7)^(2)xx3=13.35 kW`
You can see that in the present case, power dissipated at resonance is more than the power dissipated.