SSC JE Electrical Engineering 2017 Set 2 Paper

Question : 129 of 200

Marks: +1, -0

In series LCR circuit, at resonance____ :

Solution:

In series LCR circuit at resonance

X_L = X_C

Z=√{R^2+( X_L- X _ C)^2}

Z = R (minimum)

I_ {\max} =V _S/Z

So current is maximum and power factor is unity because

\cos ϕ=R/√{R^2+( X_L- X _ C)^2}

X_L = X_C

\cos ϕ =R/R=1

, unity power factor

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