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Question

An RLC series circuit has L=10 mH, R=3 Ω and C=1 μF connected in series to a source of E=15cosωt V. If the frequency of the voltage is 10% less than the resonant frequency, then

A
Peak value of current is 0.704 A
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B
Impedance of the circuit is 21.32 Ω
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C
Power factor of the circuit is 0.141
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D
Resonant frequency of the circuit is 9×103 rad/s
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Solution

The correct option is C Power factor of the circuit is 0.141
Given:-

L=10 mH; C=1 μF

R=3 Ω; E=15 cos ωt

Resonant frequency of the circuit is given as

ωR=1LC

ωR=110×103×106

ωR=104 rad/s

Now, 10% less frequency than the resonant frequency is given as

ω=10410100×104=9×103 rad/s

At this frequency

XL=ωL=9×103×(10×103)

XL=90 Ω

and XC=1ωC=19×103×106

XC=111.1 Ω

Impedance of the circuit,

Z=R2+(XLXC)2

Z=R2+(90111.1)2

Z=32+(21.1)2

Z=21.32 Ω

Peak value of current is given as

I0=E0Z=1521.32

I0=0.704 A

Power factor of the circuit is given as

cosϕ=RZ=321.32

cosϕ=0.141

Hence, options (A),(B) and (C) are the correct answers.

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