The r.m.s. value of potential difference $V$ shown in the figure is\

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Professor PC Thomas Classes · Answer

(c) :\ $V=V_{0}$ for $0 \leq t \leq \frac{T}{2} \Rightarrow V=0$ for $\frac{T}{2} \leq t \leq T$\ $V_{\mathrm{rms}}=\left[\frac{\int_{0}^{T} V^{2} d t}{\int_{0}^{T} d t}ight]^{1 / 2}=\left[\frac{\int_{0}^{T / 2} V_{0}^{2} d t+\int_{T / 2}^{T}(0) d t}{\int_{0}^{T} d t}ight]^{1 / 2}$\ $=\left[\frac{V_{0}^{2}}{T}[t]_{0}^{T / 2}ight]^{1 / 2}=\left[\frac{V_{0}^{2}}{T}\left(\frac{T}{2}ight)ight]^{1 / 2}=\left[\frac{V_{0}^{2}}{2}ight]^{1 / 2}$\ $	herefore V_{\mathrm{rms}}=\frac{V_{0}}{\sqrt{2}}$