The output current versus time curve of a rec | vedclass

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(C) The average value of the current over one complete cycle $T$ is given by the formula:$I_{av} = \frac{1}{T} \int_0^T i(t) dt$Since the rectifier ou

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$\frac{2I_0}{\pi}$

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(C) The average value of the current over one complete cycle $T$ is given by the formula:$I_{av} = \frac{1}{T} \int_0^T i(t) dt$Since the rectifier output is a half-wave rectified signal (as shown in the figure),the current is $I_0 \sin(\omega t)$ for $0 \le t \le T/2$ and $0$ for $T/2 $I_{av} = \frac{1}{T} \int_0^{T/2} I_0 \sin(\omega t) dt$Using $\omega = \frac{2\pi}{T}$,we have:$I_{av} = \frac{I_0}{T} \left[ -\frac{\cos(\omega t)}{\omega} ight]_0^{T/2}$$I_{av} = \frac{I_0}{T \omega} [-\cos(\omega \cdot T/2) + \cos(0)]$$I_{av} = \frac{I_0}{T (2\pi/T)} [-\cos(\pi) + 1]$$I_{av} = \frac{I_0}{2\pi} [1 + 1] = \frac{2I_0}{2\pi} = \frac{I_0}{\pi}$Wait,looking at the provided figure,it represents a full-wave rectified signal. For a full-wave rectifier,the average value is calculated over the interval $T/2$:$I_{av} = \frac{1}{T/2} \int_0^{T/2} I_0 \sin(\omega t) dt = \frac{2}{T} \left[ -\frac{I_0 \cos(\omega t)}{\omega} ight]_0^{T/2}$$I_{av} = \frac{2I_0}{T \omega} [-\cos(\pi) + 1] = \frac{2I_0}{T (2\pi/T)} [1 + 1] = \frac{2I_0}{2\pi} 	imes 2 = \frac{2I_0}{\pi}$

The output current versus time curve of a rectifier is shown in the figure. The average value of output current in this case is

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