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Question

# A 12 Ω resistance and an inductance of 0.05π H with negligible resistance are connected in series. Across the ends of this circuit a 130 V alternating voltage of frequency 50 Hz is connected. Calculate the alternating current in the circuit and the potential difference across the resistance.

A
10 A, 50 V
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B
10 A, 120 V
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C
5 A, 60 V
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D
5 A, 120 V
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Solution

## The correct option is B 10 A, 120 VThe impedance of the circuit is given by Z=√(R2+X2L)=√(R2+ω2L2) =√[R2+(2πfL)2]=√(12)2+{2×3.14×50×0.053.14}2 =√(144+25)=13 Ω Current in the circuit i=VZ=13013=10 A Potential difference across resistance VR=iR=10×12=120 V Hence, option (B) is correct.

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