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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 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. |
Answer» 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. |
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