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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 impedance of P at this frequency is

A
77 Ω
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B
36 Ω
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C
40 Ω
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D
125 Ω
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Solution

The correct option is A 77 Ω
For Box P, Impedence is given by
Z=R21+(XC)2 .......(i)
Now if the maximum current flows through P and Q, then circuit is at resonance. Also at resonance ω=1LC
ω=14.9×103×106ω=1057
From here we can find the value of XC
XC=1ωC=1×7105×106=70 Ω
Putting this and value of R1 in eq(i)
Z=322+702=77 Ω

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