What will be the r.m.s. value of the given A. | vedclass

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(C) The given waveform represents a half-wave rectified $A.C.$ signal.For one complete cycle of period $T$,the voltage $V(t)$ is defined as:$V(t) = V_

Vedclass · Accepted Answer

$\frac{V_0}{2}$

Vedclass · Answer

(C) The given waveform represents a half-wave rectified $A.C.$ signal.For one complete cycle of period $T$,the voltage $V(t)$ is defined as:$V(t) = V_0 \sin(\omega t)$ for $0 \le t \le T/2$$V(t) = 0$ for $T/2 where $\omega = \frac{2\pi}{T}$.The $r.m.s.$ value is defined as $V_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} V^2(t) dt}$.$V_{rms}^2 = \frac{1}{T} \left[ \int_{0}^{T/2} (V_0 \sin(\omega t))^2 dt + \int_{T/2}^{T} 0^2 dt ight]$$V_{rms}^2 = \frac{V_0^2}{T} \int_{0}^{T/2} \sin^2(\frac{2\pi t}{T}) dt$Using the identity $\sin^2 	heta = \frac{1 - \cos(2	heta)}{2}$:$V_{rms}^2 = \frac{V_0^2}{T} \int_{0}^{T/2} \frac{1 - \cos(\frac{4\pi t}{T})}{2} dt = \frac{V_0^2}{2T} \left[ t - \frac{T}{4\pi} \sin(\frac{4\pi t}{T}) ight]_{0}^{T/2}$$V_{rms}^2 = \frac{V_0^2}{2T} \left[ (\frac{T}{2} - 0) - (0 - 0) ight] = \frac{V_0^2}{2T} \cdot \frac{T}{2} = \frac{V_0^2}{4}$Therefore,$V_{rms} = \sqrt{\frac{V_0^2}{4}} = \frac{V_0}{2}$.

What will be the $r.m.s.$ value of the given $A.C.$ voltage over one complete cycle?

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