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(A) The given waveform is a square wave with amplitude $V_0 = 10 \ V$.For a square wave that oscillates between $+V_0$ and $-V_0$,the instantaneous vo

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$10$

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(A) The given waveform is a square wave with amplitude $V_0 = 10 \ V$.For a square wave that oscillates between $+V_0$ and $-V_0$,the instantaneous voltage $V(t)$ is either $+V_0$ or $-V_0$.The $r.m.s.$ voltage is defined as $V_{rms} = \sqrt{\frac{1}{T} \int_0^T V^2(t) dt}$.Since $V^2(t) = V_0^2$ at all times,$V_{rms} = \sqrt{\frac{1}{T} \int_0^T V_0^2 dt} = \sqrt{\frac{V_0^2}{T} \cdot T} = V_0$.Given $V_0 = 10 \ V$,therefore $V_{rms} = 10 \ V$.

The $r.m.s.$ voltage of the waveform shown is......$V$

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