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Question

In series LR circuit, XL=R. Now a capacitor with XC=R added in series. New power factor:

A
Same as initial
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B
12 times the initial
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C
12 times the initial
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D
2 times the initial
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Solution

The correct option is D 2 times the initial

Given that,

XL=R

XC=R

We know that, the impedance is

Z=R2+(XLXC)2

When, XL=R

Then,

Z=R2

Now, the power factor is

P1=RZ

P1=RR2

P1=12

Now, Xc=R

So, Z=R

Now, the new power factor is

P2=RZ

P2=RR

P2=1

Now, the new power factor is

P1=12

P12=1

P2=P12

Hence, the new power factor is 2 initial power factor


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