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Question

# In series LCR circuit voltage drop across resistance is 8 V, across inductor is 6 V and across capacitor is 12 V. Then

A
Voltage of the source will be leading current in the circuit
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B
Voltage drop across each element will be less than the applied voltage
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C
Power factor of circuit will be 4/5
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D
None of these
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Solution

## The correct option is C Power factor of circuit will be 4/5correct answer : option D taking I0 as reference ,voltage across R=I0R=8Vvoltage across L=jI0XL=j6Vvoltage across C=−jI0XC=j12Vtotal voltage across LCR=I0R+jI0XL−jI0xC=I0(R+j(XL−XC))=8+j6−j12=8−j6=10∠(−tan−1(6/8))V we see that voltage across LCR lags behind the current.we see that voltage across each element is not less than the applied voltage ( VC=12V).power factor=RZ=R√R2+(XL−XC)2=I0RI0√R2+(XL−XC)2=I0R√I20R2+I20(XL−XC)2=I0R√I20R2+(I0XL−I0XC)2=VR√V2R+(VL−VC)2=8√82+(6−12)2810=45

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