A periodic voltage V varies with time t as sh | vedclass

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Question

Root mean square value

$ < {\rm{V}} >  = \sqrt {\frac{{\int_0^{{\rm{T}}/4} {{\rm{V}}_0^2} {\rm{dt}} + \int_{{\rm{T}}/4}^{\rm{T}} {(0)} {

Vedclass · Accepted Answer

$\frac{{{V_0}}}{2}$

Vedclass · Answer

Root mean square value

$ < {\rm{V}} >  = \sqrt {\frac{{\int_0^{{\rm{T}}/4} {{\rm{V}}_0^2} {\rm{dt}} + \int_{{\rm{T}}/4}^{\rm{T}} {(0)} {\rm{dt}}}}{{\int_0^{\rm{T}} {dt} }}} $

$=\sqrt{\frac{V_{0}^{2}\left(\frac{T}{4}\right)}{T}}=\sqrt{\frac{V_{0}^{2}}{4}}=\frac{V_{0}}{2}$

A periodic voltage $V$ varies with time $t$ as shown in the figure. $T$ is the time period. The $r.m.s$. value of the voltage is :-

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