Average Value of Current

Average Value of Current

Average value of any current in a given time is defined as that constant value of current which will send the same amount of charge in a given circuit as sent by actual current in the same time interval.

$${{I}_{avg}}\left( {{t}_{2}}-{{t}_{1}} \right)=\int\limits_{{{t}_{1}}}^{{{t}_{2}}}{I(t)dt}$$.

$$\Rightarrow \,Average\,\,Cureent\left( \,{{I}_{avg}} \right)=\frac{\int\limits_{{{t}_{1}}}^{{{t}_{2}}}{I(t)dt}}{{{t}_{2}}-{{t}_{1}}}$$.

Now, the average value of I = I₀ sin ωt;

1. Over first half cycle: for this t₁ = 0, $${{t}_{2}}=\frac{T}{2}=\frac{\pi }{\omega }$$,

$${{I}_{avg}}\left( \frac{T}{2}-0 \right)=\int\limits_{0}^{T/2}{{{I}_{0}}\sin \omega t\,dt}=\left[ \frac{-{{I}_{0}}\cos \omega t}{\omega } \right]_{0}^{\pi /\omega }$$,

$$\Rightarrow \,\,{{I}_{avg}}\left( \frac{T}{2} \right)=\frac{2{{I}_{0}}}{\omega }=\frac{2{{I}_{0}}T}{2\pi }$$,

$$\Rightarrow \,{{I}_{avg}}=\frac{2{{I}_{0}}}{\pi }=0.637{{I}_{0}}$$.

2. Over full cycle: for this t₁ = 0, $${{t}_{2}}=T=\frac{2\pi }{\omega }$$,

$${{I}_{avg}}\left( T-0 \right)=\int\limits_{0}^{T}{{{I}_{0}}\,\sin \omega t\,dt}=\left[ \frac{-{{I}_{o}}\,\cos \omega t}{\omega } \right]_{0}^{2\pi /\omega }$$,

Iavg = 0

So, average value of current for the full cycle is zero. We can also think in this way: Average value during positive half cycle is 0.637 I₀ and the average value during the negative half cycle is – 0.637 I₀. So, over full cycle, Iavg = 0.637 I₀ – 0.637 I₀ = 0.