In the circuit shown in the figure, determine | vedclass

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(C) For the positive half-cycle of the input, the resulting network is shown in figure $(b).$ Diode $D_1$ is in the "off" state and $D_2$ is in the "o

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(C) For the positive half-cycle of the input, the resulting network is shown in figure $(b).$ Diode $D_1$ is in the "off" state and $D_2$ is in the "on" state.The circuit can be simplified as a voltage divider consisting of two $2 	ext{ k}\Omega$ resistors in series across the input voltage. It is clear that:$V_0 = \frac{2 	ext{ k}\Omega}{2 	ext{ k}\Omega + 2 	ext{ k}\Omega} V_i = \frac{1}{2} V_i$Given the peak input voltage $V_{i \max} = 10 	ext{ V}$, the peak output voltage is:$V_{0 \max} = \frac{1}{2} (10 	ext{ V}) = 5 	ext{ V}$During the negative half-cycle, the roles of diodes $D_1$ and $D_2$ are interchanged, and the output $V_0$ will appear as a full-wave rectified signal with a peak of $5 	ext{ V}$, as shown in figure $(d).$For full-wave rectification, the d.c. level is given by:$V_{dc} = \frac{2 V_m}{\pi} = 0.636 V_m = 0.636 	imes 5 	ext{ V} = 3.18 	ext{ V}$

In the circuit shown in the figure, determine the output waveform and the output d.c. level of $V_0$.

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