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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 10V. Box P contains a capacitance of 1 μF in series with a resistance of 32Ω. Coil Q has a self inductance of 4.9 mH and a resistance of 68Ω in series. The frequency is adjusted so that maximum current flows in P and Q. The voltage across P is

A
12V
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B
7.7V
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C
10V
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D
24V
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Solution

## The correct option is B 7.7VVoltage across P can be given as VP=Irms×ZP .....(i) For Box P, Impedence is given by ZP=√R21+(XC)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 From here we can find the value of XC XC=1ωC=1×7105×10−6=70 Ω Putting this and value of R1 in eq(ii) ZP=√322+702=77 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=0.1×77=7.7 V

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