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A series LCR circuit with a resistance o...

A series LCR circuit with a resistance of `100 sqrt(5) Omega` is connected to an ac source of 200 V. When the capacitor is removed from the circuit, current lags behind emf by `45^(@)`. When the inductor is removed from the circuit keeping the capacitor and resistor in the circuit, current leads by an angle of `tan^(-1)((1)/(2))`. Calculate the current and power dissipated in LCR circuit.

Text Solution

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When capacitor is removed, `tanphi = X_L/R` (It is L - R circuit)
`tan phi = tan 45^@ = X_L/R rArr X_L = R`
When inductor is removed. `tan phi = X_C/R = 1/2 = X_C/R rArr X_C = R/2`
Impedance, `Z = sqrt(R^2 + (X_L - X_C)^2) = sqrt(R^2 + (R -R/2)^2) = sqrt5/2R`
Now,` l_(rms) = E_(rms)/Z = 200/250 = 0.8 A, cos phi = R/Z = 2/sqrt5`
`P_(av) = V_(rms)l_(rms) cosphi = 64 sqrt5 W`
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