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Question

A coil has an inductance of 0.7H and is joined in series with a resistance of 220 Ω. When an alternating e.m.f. of 220V at 50 c.p.s. is applied to it, then the watt less component of the current in the circuit is


A

5 ampere

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B

0.5 ampere

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C

0.7 ampere

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D

7 ampere

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Solution

The correct option is B

0.5 ampere


The correct option is (B).

Step 1: Given data:

Inductance = 0.7H

Resistance = 220 Ω

Alternating e.m.f. = 220V

f=50cps

Step 2: Calculation:

The impedance of the circuit:

Now, Z=R2+(2πfL)2

Z=2202+(2π×50×0.7)2

Z2202π

The power factor:

cosϕ=RZ=π4

The total current in the circuit:

I=VZI=2202202I=12A

Watt less component of current in the circuit:

Ia=Isin(ϕ)Ia=12×12Ia=0.5A


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