SSC JE Electrical Engineering 2012 Paper

Question : 117 of 200

Marks: +1, -0

In a series RLC circuit

R = 20 Ω, X_L = 30 Ω

and

X_C = 30 Ω

. If the supply voltage across the combination is v = 100 sin (100 πt + 30°) Volts, the instantaneous current and the power factor of the circuit are respectively

I = 3.536 sin (100 πt + 30°) Amps, p.f. = 0.866
I = 5 sin (100 πt + 30°) Amps, p.f. = unity
I = 3.536 sin (100 πt + 30°) Amps, p.f. = unity
I = 5 sin (100 πt + 30°) Amps, p.f. = 0.866

Validate

Solution:

R = 20 Ω, X_L = 30 Ω

and

X_C = 30 Ω

X_L-X_C=0

Circuit is in Resonance
So

I=V/R=5\sin(100πt+30°)

P.F. = unit y

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