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(B) The $r.m.s.$ value of a periodic function is given by $V_{rms} = \sqrt{\frac{1}{T} \int_0^T V^2 dt}$.From the figure,the voltage $V$ is $V_0$ for

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$\frac{V_0}{2}$

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(B) The $r.m.s.$ value of a periodic function is given by $V_{rms} = \sqrt{\frac{1}{T} \int_0^T V^2 dt}$.From the figure,the voltage $V$ is $V_0$ for $0 \le t Substituting these values into the formula:$V_{rms} = \sqrt{\frac{1}{T} \left( \int_0^{T/4} V_0^2 dt + \int_{T/4}^T 0^2 dt \right)}$$V_{rms} = \sqrt{\frac{1}{T} \left( V_0^2 [t]_0^{T/4} + 0 \right)}$$V_{rms} = \sqrt{\frac{1}{T} \cdot V_0^2 \cdot \frac{T}{4}}$$V_{rms} = \sqrt{\frac{V_0^2}{4}}$$V_{rms} = \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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