# Power in an AC circuit: definition, and formula derivation

In this article, we will discuss the Power of an AC circuit: definition, and formula derivation, so let’s get started…

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## Power in an AC circuit

Power in an AC circuit definition: The rate at which electric energy is consumed in an electric circuit is called its power.

In a DC circuit, power is given by the product of voltage and current. But in an AC circuit, both voltage $\varepsilon$ and current $I$ vary sinusoidally with time and are generally not in phase: So for an AC circuit, we define instantaneous power as the product of the instantaneous voltage and instantaneous current.

Suppose in an a.c. circuit, the voltage, and current at any instant are given by
$$\varepsilon=\varepsilon_0 \sin \omega t \text { and } I=I_0 \sin (\omega t-\phi)$$
where $\phi$ is the phase angle by which the voltage $\varepsilon$ leads the current $I$.

## Power in an AC circuit derivation

The instantaneous power is given by
\begin{aligned} P=& \varepsilon_I=\varepsilon_0 I_0 \sin \omega t \cdot \sin (\omega t-\phi) \\ =& \frac{\varepsilon_0 I_0}{2}[2 \sin \omega t \cdot \sin (\omega t-\phi)] \\ =& \frac{\varepsilon_0 I_0}{2}[\cos \phi-\cos (2 \omega t-\phi)] \\ [\because 2 \sin A \sin B&=\cos (A-B)-\cos (A+B)] \end{aligned}
Average power dissipated per cycle
$$=\text { Average of } \frac{\varepsilon_0 I_0}{2}[\cos \phi-\cos (2 \omega t-\phi)] \text {. }$$
The second cosine term $[\cos (2 \omega t-\phi)]$ is time-dependent. Its average over a cycle is zero.
$$\therefore \quad P_{a v}=\frac{\varepsilon_0 I_0}{2} \cos \phi$$
or$$P_{a v}=\frac{\varepsilon_0}{\sqrt{2}} \cdot \frac{I_0}{\sqrt{2}} \cdot \cos \phi$$or$$P_{a v}=\varepsilon_{r m s} I_{r m s} \cos \phi=\varepsilon_{r m s} I_{r m s} \cdot \frac{R}{Z}$$

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