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Question

An ideal inductor takes a current of 10 A when connected to a 125V, 50 Hz ac supply. A pure resistor across the same source takes 12.5 A. If the two are connected in series across a 100V, 40 Hz supply, the current through the circuit will be

A
7A
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B
12.5A
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C
20A
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D
25A
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Solution

The correct option is A 7A
X2=VI=12510; LW=12.5; L=12.52πfL=0.125π
R=VI=12512.5=10Ω
Irms=VrmsZ=Vrms(R)2+(LW)2
=100102+(0.125π×2π×40)2=100102
=52=5×1.414

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