The output current versus time curve of a rec | vedclass

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(C) From the given figure,it is evident that the rectifier is a full-wave rectifier.For a full-wave rectifier,the output current is given by $I = I_0

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

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(C) From the given figure,it is evident that the rectifier is a full-wave rectifier.For a full-wave rectifier,the output current is given by $I = I_0 \sin(\omega t)$ for the first half-cycle and $I = I_0 \sin(\omega t)$ for the second half-cycle (due to rectification).The average value of the current over one complete cycle $T$ is given by:$I_{	ext{avg}} = \frac{1}{T} \int_{0}^{T} I(t) dt$For a full-wave rectifier,the period is $T/2$. The average value is calculated as:$I_{	ext{avg}} = \frac{1}{T/2} \int_{0}^{T/2} I_0 \sin(\omega t) dt$$I_{	ext{avg}} = \frac{2}{T} \cdot I_0 \left[ -\frac{\cos(\omega t)}{\omega} ight]_{0}^{T/2}$Since $\omega = \frac{2\pi}{T}$,we have:$I_{	ext{avg}} = \frac{2 I_0}{T} \cdot \frac{T}{2\pi} [-\cos(\pi) + \cos(0)]$$I_{	ext{avg}} = \frac{I_0}{\pi} [1 + 1] = \frac{2 I_0}{\pi}$

List-$I$	List-$II$
$(a)$ Rectifier	$(i)$ Used for stepping up or stepping down the $\text{A.C.}$ voltage
$(b)$ Stabilizer	$(ii)$ Used to convert $\text{A.C.}$ voltage into $\text{D.C.}$ voltage
$(c)$ Transformer	$(iii)$ Used to remove any ripple in the rectified output voltage
$(d)$ Filter	$(iv)$ Used for constant output voltage

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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