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Question

# A box P and a coil Q are connected in series with an ac source of variable frequency. The emf of the source is constant at 10 V. Box P contains a capacitance of 1 μF in series with a resistance of 32 Ω. Coil Q has a self inductance of 4.9 mH and a resistance of 68 Ω in series. The frequency is adjusted so that maximum current flows in P and Q. The voltage across Q is

A
20 V
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B
7.7 V
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C
10 V
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D
952410 V
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Solution

## The correct option is D √952410 VVoltage across Q can be given as VQ=Irms×ZQ .....(i) For impedence in coil Q Z=√R22+(XL)2 ....(ii) Now if the maximum current flows through P and Q, then circuit is at resonance. Also at resonance ω=1√LC ω=1√4.9×10−3×10−6ω=1057 So we can find XL as XL=ωL=1057×4.9×10−3=70 Ω Putting values in eq(i) Z=√682+702=√9524 Ω ......(iii) To calulate Irms we need to understand that the circuit is at resonance and it behaves as purely resistive ciruit. So Irms=VrmsR1+R2=1068+32=0.1 A .....(iv) From (iii) and (iv) VP=0.1×√9524=√952410 V

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