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Question

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

A
76 Ω, 98 Ω
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B
45 Ω, 72 Ω
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C
85 Ω, 20 Ω
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D
99 Ω, 10 Ω
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Solution

## The correct option is A 76 Ω, 98 ΩThe current is maximum at resonance, when |XL−XC|=0 ⇒(ωL−1ωC)=0 ⇒ωL=1ωC ⇒ω2=1LC ⇒ω=1√(4.9×10−3)(10−6) ∴ω=1057 rad/s The impedance of P is given as ZP=√(R21)+X2C =√R21+(1ωC)2=√(32)2+(7105×110−6)2 =√(32)2+(70)2 =√1024+4900=√5924 ⇒ZP≈76 Ω The impedance of Q is given as, ZQ=√R22+ω2L2 =√(68)2+(1057×4.9×10−3)2 =√(68)2+(70)2 =√9524≈98 Ω Hence, option (A) is correct.

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