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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. The frequency is adjusted so that the maximum current flows in P and Q. If ZP and ZQ represents the impedance of box P and Q respectively, then


A
ZP=77 Ω
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B
ZP=38 Ω
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C
ZQ=49 Ω
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D
ZQ=98 Ω
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Solution

The correct option is D ZQ=98 Ω
The current flowing is the LCR circuit is given as

i=E [R2+(ωL1ωC)2]

The current in maximum at resonance, when

(ωL1ωC)=0

ω2=1LC

ω=1(4.9×103)(106)

ω=1057rad/s

The impedance of P is given as

ZP=(R21+X2C)= [R21+(1ωC)2]

ZP=[(32)2+(7105×1106)2]12

ZP=77 Ω

The impedance of Q is given as

ZQ=[R22+ω2L2]

ZQ= [(68)2+(1057×4.9×103)2]

ZQ=98 Ω

Hence, options (A) and (D) are correct.

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