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Question

# A series LCR circuit containing a resistor of 120 Ω has an angular resonant frequency 4×105 rad s−1. At resonance the voltages across resistor and inductor are 60 V and 40 V respectively. At what frequency, the current in the circuit lags the voltage by 45∘?

A
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Solution

## The correct option is C 8×105 rad s−1 After resolving this phasor we get From the above phasor, we can say tan45∘=XL−XCR⇒1=ωL−1ωCR .....(i) Now given that R=120 Ω and Now we know that when the circuit was given at resonance, voltage across the inductor VL=40 V So XL=XC and also VL=VC and the circuit will behave as purely resistive. ⇒Z=R ∴Irms=VrmsZ=VrmsR⇒Irms=60120=12 A ⇒VL=Irms×ω×L ⇒40=0.5×4×105×L⇒L=0.2 mH VC=irms×1ωC⇒40=12×1ωC⇒C=132 μF Putting these values in eq(i) 120=ω×2×10−4−1ω132×10−6 On solving the quadratic equation we get ω=8×105 rad/s

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