In the circuit shown, the power consumed through 5 Ω resistor is 10 W. Then the power factor of circuit is

This question was previously asked in

ESE Electronics 2016 Paper 1: Official Paper

Option 2 : 0.6

__Concept:__

**Power factor cos(ϕ)**

It** **is an important part of an AC circuit that can also be expressed in terms of circuit impedance of **circuit power.**

- Power factor is defined as the ratio of real power (P) to apparent power (S) and is generally expressed as either a decimal value, for example, 0.95, or as a percentage: 95%.
- Power factor
**defines**the phase**angle**between the current and voltage waveforms, were I and V are the magnitudes of RMS values of the current and voltage. - Note that it does not matter whether the phase angle is the difference of the current with respect to the voltage, or the voltage with respect to the current.

The mathematical relationship is given as:

**Power factor = Watts / Volt-amperes**

VIcosϕ / VI = cosϕ

Cosϕ = R_{eq} / Z_{eq}

__Calculation:__

Given power consumed by 10 Ω resistor is 10 W.

(I_{5 Ω})^{2} = 10/5 = 2 A

\({I_{5{\rm{\Omega }}}} = \sqrt 2 \;A\)

Given voltage source is 50Cosωt = 50∠ 0°

Power factor will be given by

\(pf = \frac{{{V_{Req}}}}{V}\)

V: RMS value

V_{Req} = I × R_{eq }

\({V_{Req}} = \sqrt 2 \times 15\)

\(pf = \frac{{15\sqrt 2 }}{{\frac{{50}}{{\sqrt 2 }}}}\)

\(pf = \frac{30}{50} = 0.6\)

As the reactance part is **inductive **so we get** lagging** power factor.