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Question

A circuit is made up of a resistance 1 Ω and inductance 0.01 H. An alternating emf of 200 V at 50 Hz is connected, then the phase difference between the current and the emf in the circuit is

​​​​​​​[1 Mark]

A
tan1(π)
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B
tan1(π2)
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C
tan1(π4)
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D
tan1(π3)
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Solution

The correct option is A tan1(π)
We know that,
XL=ωL=(2πfL)=(2π)(50)(0.01)=π Ω
Also, R=1 Ω
Phase difference is given by formula
tanϕ=(XLR)
ϕ=tan1(π)

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