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# 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 0.141Given:- L=10 mH; C=1 μF R=3 Ω; E=15 cos ωt Resonant frequency of the circuit is given as ωR=1√LC ωR=1√10×10−3×10−6 ωR=104 rad/s Now, 10% less frequency than the resonant frequency is given as ω=104−10100×104=9×103 rad/s At this frequency XL=ωL=9×103×(10×10−3) XL=90 Ω and XC=1ωC=19×103×10−6 XC=111.1 Ω Impedance of the circuit, Z=√R2+(XL−XC)2 Z=√R2+(90−111.1)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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