Question
Establish the relation between mean value and peak value of AC.
✓
Answer
Let the instantenous value of alternating current be
$i=I_{0} sin \omega t$
$I_{0}=$ peak value of current,
$t=$ time at any instant.
Also, $\omega=\frac{2\pi}{T}$, $T=$ Time period.
$i=\frac{dQ}{dt}\Rightarrow dQ=idt$
$\int_{0}^{Q}dQ=\int_{0}^{T/2}idt$
$Q|_{0}^{Q}=\int_{0}^{T/2}I_{0}sin \omega tdt$
$Q=\frac{-I_{0}}{\omega}|cos\frac{2\pi}{T}.t|_{0}^{T/2}$
$Q=-\frac{I_{0}}{\omega}|cos\frac{2\pi}{T}\times\frac{T}{2}-cos\frac{2\pi}{T}.0|$
$Q=-\frac{I_{0}}{\omega}(cos \pi-cos 0) = -\frac{I_{0}}{\omega}\times-2 = \frac{2I_{0}}{\omega}$
Also, $Q=I_{m}\times\frac{T}{2}$
from (1) and (2)
$I_{m}\times\frac{T}{2}=\frac{2I_{0}}{2\pi} \times T$
$I_{m}\times\frac{1}{2}=\frac{2I_{0}}{2\pi} \times T$
$I_{m}=\frac{2I_{0}}{\pi}= 0.637 I_{0}$
Mean value of current, $I_{m}=0.637$ times peak value of current.
$i=I_{0} sin \omega t$
$I_{0}=$ peak value of current,
$t=$ time at any instant.
Also, $\omega=\frac{2\pi}{T}$, $T=$ Time period.
$i=\frac{dQ}{dt}\Rightarrow dQ=idt$
$\int_{0}^{Q}dQ=\int_{0}^{T/2}idt$
$Q|_{0}^{Q}=\int_{0}^{T/2}I_{0}sin \omega tdt$
$Q=\frac{-I_{0}}{\omega}|cos\frac{2\pi}{T}.t|_{0}^{T/2}$
$Q=-\frac{I_{0}}{\omega}|cos\frac{2\pi}{T}\times\frac{T}{2}-cos\frac{2\pi}{T}.0|$
$Q=-\frac{I_{0}}{\omega}(cos \pi-cos 0) = -\frac{I_{0}}{\omega}\times-2 = \frac{2I_{0}}{\omega}$
Also, $Q=I_{m}\times\frac{T}{2}$
from (1) and (2)
$I_{m}\times\frac{T}{2}=\frac{2I_{0}}{2\pi} \times T$
$I_{m}\times\frac{1}{2}=\frac{2I_{0}}{2\pi} \times T$
$I_{m}=\frac{2I_{0}}{\pi}= 0.637 I_{0}$
Mean value of current, $I_{m}=0.637$ times peak value of current.
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