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A circuit has a resistance of 12 ohm and an impedance of 15 ohm. The power factor of the circuit will be –


  • Option A) 1.25
  • Option B) 0.125
  • Option C) 0.8
  • Option D) 0.4

As we all know that,

R.L Circuit Voltage –

V_{R}= IR
V_{L}= {I}X_{L}

– wherein

Power factor –

\cos \phi = \frac{R}{\sqrt{R^{2}+X_{c}^{2}}}

Solution is correct

Capacitive current (C) Current –

{i}'= {i}'_{0}\sin \left ( \omega t+\frac{\pi }{2} \right )

– wherein

Inductive circuit (L) Current –

{i}'= {i}'_{0}\sin \left ( \omega t-\frac{\pi }{2} \right )

– wherein

Power\; factor\; cos\phi =\frac{R}{Z}=\frac{12}{15}=0.8

So the correct answer is (C)

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