# Power in AC Circuit: The Power Factor

So we have seen that DC and AC are two different entities. The power delivered to the circuit by a DC is a result of the flow of charges or in other words the electric potential. Here we will see what the power factor in an AC circuit is. We use AC in our household circuits and the concept of the power consumed in an AC circuit is very interesting. Let us see!

## Mathematical Analysis

Suppose a Voltage V is applied to a LCR circuit, where V is give by:

$$V=V_{m}\times \sin \omega t$$

The current in this case is written by:

$$I=I_{m}\times \sin \left ( \omega t+\phi \right )$$

Where, V= Voltage Amplitude, Im = Current Amplitude, ω = Angular Frequency, ø = Phase Constant

(Source: Wikipedia)

Now, Current Amplitude is related to Voltage Amplitude as:

$$I_{m}=\frac{V_{m}}{Z}$$

Where, Z = Impedance of circuit and is given by:

$$Z=\sqrt{\left ( X_{L}-X_{C} \right )^{2}+R^{2}}$$

$$\tan \phi =\frac{X_{L}-X_{C}}{R}$$

$$\phi =\arctan \frac{X_{L}-X_{C}}{R}$$

## Power Consumption

Now we will use all the above formulae to derive Power in AC circuit. We know that P=VI. So we can write power consumed in an AC circuit is:

$$P=\left (V_{m}\times \sin \omega t \right )\times \left (I_{m}\times \sin \left ( \omega t+\phi \right ) \right )$$

Note: Since P= V x I. So, If V=0 or I=0 then P=0. Also If V & I both are positive then P will be positive and If anyone of either V & I is negative then power will be negative and If V&I both are negative then P will be positive.

### Now Finding the Average Value of Power

$$P_{avg} = \left (V_{m}\times \sin \omega t \right )\times \left (I_{m}\times \sin \left ( \omega t+\phi \right ) \right )$$

Using trigonometric identity: $$2\sin A\sin B= \cos \left ( A-B \right ) – \cos \left ( A+B \right )$$

We get:

$$P_{avg}=Average \ of \ \frac{V_{m}I_{m}}{2}\left \lfloor \cos \phi – \cos \left ( 2\omega t+\phi \right ) \right \rfloor$$

$$P_{avg}=Average \ of \ \frac{V_{m}I_{m}}{2}\cos\phi-Average \ of \ \frac{V_{m}I_{m}}{2}\cos \left ( 2\omega t+\phi \right )$$

$$P_{avg}= \frac{V_{m}I_{m}}{2}\cos\phi$$

$$Average \ of \ \frac{V_{m}I_{m}}{2}\cos \left ( 2\omega t+\phi \right )=0$$

$$P_{avg}=\frac{V_{m}I_{m}}{2}\cos \phi \\ P_{avg}=\frac{V_{m}}{\sqrt{2}}\times \frac{I_{m}}{\sqrt{2}}\times \cos \phi \\ P_{avg}=V_{rms}\times R_{rms}\times \cos \phi$$

### Resistive Circuit

Now for a Resistive Circuit we know that, Φ = 0 which implies cos Φ = 1

$$P_{avg}=V_{rms}\times R_{rms}$$

### Inductive Circuit

But for a Inductive Circuit we know that, Φ = 90º which implies cos Φ = 0. This is as we know that voltage across the inductor leads the current by 90 degrees.

$$P_{avg}=0$$

### Capacitive Circuit

But for a Capacitive Circuit we know that, Φ = -90º which implies cos Φ = 0. This is as we know that voltage across the inductor lags the current by 90 degrees.

$$P_{avg}=0$$

## Power Factor

The power factor of an AC electrical power system is defined as the ratio of the real power flowing to the load to the apparent power in the circuit. It is dimensionless quantity and in the closed interval of  -1 to 1.

$$Power \ Factor=\frac{True \ Power}{Apparent \ Power}$$

Suppose a graph in which Voltage leads the current in a LCR circuit by .We can resolve this voltage into two components that will be:

$$V I\cos\phi- resistive power$$

$$V I\sin\phi – reactive power$$

Reactive power is the power loss in the circuit which takes place due tpo reactive components. So, $$V I\sin \phi$$ is referred to as wattles component of the voltage because it does not produce any power loss.

## Solved Example for You

Question: Two loads of 10KW each, are operating at a power factor 0.8 lagging (each).What is their combined power factor?

Solution: In the above question, it is not specified whether the loads are connected in series or parallel, our answer does not depend whether it is connected in series or parallel because in both the case the answer will be same.

$$Power \ Factor=\frac{True \ Power}{Apparent \ Power}$$

$$Apparent \ Power=\frac{True \ Power}{Power \ Factor}$$

$$Apparent \ Power=\frac{10KW}{0.8}$$

$$Apparent \ Power=12.5 KVA$$

$$Apparent \ Power=12.5 KVA$$

Combined Power Factor:

$$Combined \ Power \ Factor=\frac{Total \ True \ Power}{Total \ Apparent \ Power}\\ Combined \ Power \ Factor=\frac{10+10}{12.5+12.5} \\ Combined \ Power \ Factor=0.8 lag$$

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