Find the RMS value for the saw-tooth voltage | vedclass

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(D) The voltage waveform is periodic with a period of $T$. Let us consider one cycle from $t = 0$ to $t = T$.In the interval $t = 0$ to $t = T$,the vo

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$\frac{V_0}{\sqrt{3}}$

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(D) The voltage waveform is periodic with a period of $T$. Let us consider one cycle from $t = 0$ to $t = T$.In the interval $t = 0$ to $t = T$,the voltage $V(t)$ varies linearly from $-V_0$ to $+V_0$.The equation of the line passing through $(0, -V_0)$ and $(T, V_0)$ is given by:$V(t) = \frac{V_0 - (-V_0)}{T - 0} t - V_0 = \frac{2V_0}{T} t - V_0$The $RMS$ value $V_{rms}$ is defined as:$V_{rms} = \sqrt{\frac{1}{T} \int_0^T V^2 dt}$Substituting $V(t)$:$V_{rms}^2 = \frac{1}{T} \int_0^T \left( \frac{2V_0}{T} t - V_0 \right)^2 dt$Let $u = \frac{2V_0}{T} t - V_0$,then $du = \frac{2V_0}{T} dt$,so $dt = \frac{T}{2V_0} du$.When $t=0, u=-V_0$; when $t=T, u=V_0$.$V_{rms}^2 = \frac{1}{T} \int_{-V_0}^{V_0} u^2 \left( \frac{T}{2V_0} \right) du = \frac{1}{2V_0} \left[ \frac{u^3}{3} \right]_{-V_0}^{V_0} = \frac{1}{2V_0} \left( \frac{V_0^3}{3} - \frac{(-V_0)^3}{3} \right) = \frac{1}{2V_0} \left( \frac{2V_0^3}{3} \right) = \frac{V_0^2}{3}$Therefore,$V_{rms} = \frac{V_0}{\sqrt{3}}$.

Find the $RMS$ value for the saw-tooth voltage of peak value $V_0$ from $t = 0$ to $t = 2T$ as shown in the figure.

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