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The rms value of potential difference $V$ shown in the figure is
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Verified Answer
The correct answer is:
$\frac{V_0}{\sqrt{2}}$
$V_{\mathrm{rms}}=\left[\frac{1}{T} \int_0^{T / 2} V_0^2 d t\right]^{1 / 2}$
$$
\begin{aligned}
& =\left[\frac{V_0^2}{T}[t]_0^{T / 2}\right]^{1 / 2} \\
& =\left[\frac{V_0^2}{T}\left(\frac{T}{2}\right)\right]^{1 / 2} \\
V_{\mathrm{rms}} & =\left[\frac{V_0^2}{2}\right]^{1 / 2} \\
V_{\mathrm{mms}} & =\frac{V_0}{\sqrt{2}}
\end{aligned}
$$
$$
\begin{aligned}
& =\left[\frac{V_0^2}{T}[t]_0^{T / 2}\right]^{1 / 2} \\
& =\left[\frac{V_0^2}{T}\left(\frac{T}{2}\right)\right]^{1 / 2} \\
V_{\mathrm{rms}} & =\left[\frac{V_0^2}{2}\right]^{1 / 2} \\
V_{\mathrm{mms}} & =\frac{V_0}{\sqrt{2}}
\end{aligned}
$$
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