In this short piece article, we will discuss the Average power associated with a resistor, so let’s get started…

Average power associated with a resistor derivation

In the case of a pure resistor, the voltage and current are always in the same phase. So we can write the instantaneous values of voltage and current as follows :
$$V=V_0 \sin \omega t\: \text{and} \:I=I_0 \sin \omega t$$
Work done in small time $d t$ will be

\begin{aligned}d W &=P d t=V I d t=V_0 I_0 \sin ^2 \omega t d t\\
&=\frac{V_0 I_0}{2}(1-\cos 2 \omega t) d t\end{aligned}

The average power dissipated per cycle in the resistor will be

\begin{aligned}
P_{a v} &=\frac{W}{T}=\frac{1}{T} \int_0^T d W \\
&=\frac{V_0 I_0}{2 T} \int_0^T(1-\cos 2 \omega t) d t=\frac{V_0 I_0}{2 T}\left[t-\frac{\sin 2 \omega t}{2 \omega}\right]_0^T \\ &=\frac{V_0 I_0}{2 T}[(T-0)-0]=\frac{V_0 I_0}{2}=\frac{V_0^2}{2 R} \\ \text { or } P_{a v} &=\frac{V_0 I_0}{\sqrt{2} \sqrt{2}}=V_{r m s} I_{r m s}=\frac{V_{r m s s}^2}{R} \quad\left[\because \frac{V_0}{\sqrt{2}}=V_{r m s}\right]
\end{aligned}

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