The r.m.s. voltage of the waveform shown is: | vedclass

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(A) The given waveform is a square wave with amplitude $V_0 = 10 \ V$ and $-V_0 = -10 \ V$.For a square wave,the instantaneous voltage $V(t)$ is $10 \

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$10 \ V$

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(A) The given waveform is a square wave with amplitude $V_0 = 10 \ V$ and $-V_0 = -10 \ V$.For a square wave,the instantaneous voltage $V(t)$ is $10 \ V$ for half the period and $-10 \ V$ for the other half.The $r.m.s.$ voltage is defined as $V_{r.m.s.} = \sqrt{\frac{1}{T} \int_{0}^{T} V^2(t) \ dt}$.Since $V^2(t) = (10)^2 = 100$ throughout the entire period $T$,we have:$V_{r.m.s.} = \sqrt{\frac{1}{T} \int_{0}^{T} 100 \ dt} = \sqrt{\frac{1}{T} \cdot 100 \cdot T} = \sqrt{100} = 10 \ V$.Thus,the correct option is $A$.

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

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