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Question

Two resistors are connected in series across a 5 V rms source of alternating potential. The potential difference across 6Ω resistor is 3 V. If R is replaced by a pure inductor L of such magnitude that current remains same, then the same potential difference across L is:

A
1 V
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B
2 V
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C
3 V
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D
4 V
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Solution

The correct option is C 4 V
Let the current across the resistors 6Ω and RΩ in series be I.
Then,
I×6=3VI=0.5A
also,
I×(6+R)=5VR=4Ω
When R replaced by L, such that current remains same,
Let impedance of inductor be XL, then as current remains same,
I=0.5A=VR2+X2L=562+X2LXL=8Ω
Therefore,
P.d across L = I×XL=0.5×8=4V

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