The output in the circuit shown in the figure | vedclass

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(D) The circuit acts as a half-wave rectifier with a filter capacitor. During the positive half-cycle of the input,the diode is forward-biased,and the

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(D) The circuit acts as a half-wave rectifier with a filter capacitor. During the positive half-cycle of the input,the diode is forward-biased,and the capacitor charges to the peak voltage of $5 \, V$. During the negative half-cycle,the diode is reverse-biased and does not conduct. The capacitor then discharges through the resistor $R$. The time constant of the circuit is $	au = RC = (1 	imes 10^3 \, \Omega) 	imes (10 	imes 10^{-6} \, F) = 0.01 \, s$. Assuming the input frequency is standard (e.g.,$50 \, Hz$,period $T = 0.02 \, s$),the time constant $	au$ is comparable to the period. However,in typical idealized problems of this type,the capacitor is assumed to hold its charge,resulting in a nearly constant $DC$ output voltage of $5 \, V$.

The output in the circuit shown in the figure is taken across a capacitor. The input signal is as shown in the figure.

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