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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 ΩThe current flowing is the LCR circuit is given as i=E ⎷[R2+(ωL−1ωC)2] The current in maximum at resonance, when (ωL−1ωC)=0 ⇒ω2=1√LC ⇒ω=1√(4.9×10−3)(10−6) ⇒ω=1057rad/s The impedance of P is given as ZP=√(R21+X2C)= ⎷[R21+(1ωC)2] ZP=[(32)2+(7105×110−6)2]12 ∴ZP=77 Ω The impedance of Q is given as ZQ=√[R22+ω2L2] ZQ= ⎷[(68)2+(1057×4.9×10−3)2] ∴ZQ=98 Ω Hence, options (A) and (D) are correct.

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