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  • A coil of inductance 0.12 H and resistance   is connected to an AC–source of 13 V, <img alt="\fr

A coil of inductance 0.12 H and resistance 5\Omega  is connected to an AC–source of 13 V, \frac{50}{\pi}Hz. The power factor of this circuit is

Option: 1

\mathrm{\frac{12}{5}}


Option: 2

\mathrm{\frac{12}{13}}


Option: 3

5


Option: 4

\frac{5}{13}


Answers (1)

best_answer

The power factor is given by,

\mathrm{\begin{aligned} & \cos \phi=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+(\omega \mathrm{L})^2}} \\ & =\frac{5}{\sqrt{5^2+(12)^2}}=\frac{5}{13} . \end{aligned}}

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chirag

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