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Question

A series LCR circuit with 100 Ω resistance is connected to an AC source of 200 V and angular frequency 300 rad/s. When only the capacitance is removed, the current lags behind the voltage by 60. When only the inductance is removed, the current leads the voltage by 60. The average power dissipated in the LCR circuit is -

A
100 W
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B
200 W
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C
300 W
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D
400 W
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Solution

The correct option is D 400 W
When capacitance is removed, then

tanϕ=XLR

tan60=XLR

XL=3R ...(1)

When inductance is removed, then

tanϕ=XCR

tan60=XCR

XC=3R ...(2)

From equations (1) and (2),

XL=XC

LCR circuit is in resonance.

So, Z=R

Therefore,

irms=VrmsZ=200100=2 A

Further, power dissipated in the LCR circuit,

Pavg=Vrmsirmscosϕ

Pavg=200×2×cos0 [At resonance, ϕ=0]

Pavg=400 W

Hence, option (D) is the correct answer.

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