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The r.m.s. value of potential difference V shown in the figure is
A
$\frac{V_0}{2}$
B
$\frac{V_0}{\sqrt{3}}$
C
$V_0$
D
$\frac{V_0}{\sqrt{2}}$
Detailed Solution
From the graph, in one period T the voltage is $V_0$ from t = 0 to t = T/2 and zero from t = T/2 to t = T.
$V_{rms} = \sqrt{\frac{1}{T}\int_0^T V^2dt}$
$\int_0^T V^2dt = V_0^2\times\frac{T}{2} + 0\times\frac{T}{2} = \frac{V_0^2T}{2}$
$V_{rms} = \sqrt{\frac{V_0^2T/2}{T}} = \sqrt{\frac{V_0^2}{2}}$
$V_{rms} = \frac{V_0}{\sqrt{2}}$
$V_{rms} = \sqrt{\frac{1}{T}\int_0^T V^2dt}$
$\int_0^T V^2dt = V_0^2\times\frac{T}{2} + 0\times\frac{T}{2} = \frac{V_0^2T}{2}$
$V_{rms} = \sqrt{\frac{V_0^2T/2}{T}} = \sqrt{\frac{V_0^2}{2}}$
$V_{rms} = \frac{V_0}{\sqrt{2}}$