Consider the following circuit. The waveform | vedclass

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(A) The circuit is a double-diode clipper circuit.$1$. When the input voltage $V_i$ is positive and exceeds the battery voltage $V_{B_1}$,the diode $D

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(A) The circuit is a double-diode clipper circuit.$1$. When the input voltage $V_i$ is positive and exceeds the battery voltage $V_{B_1}$,the diode $D_1$ becomes forward-biased and conducts. This clamps the output voltage $V_o$ to the value $V_{B_1}$,clipping the positive peak.$2$. When the input voltage $V_i$ is negative and its magnitude exceeds the battery voltage $V_{B_2}$,the diode $D_2$ becomes forward-biased and conducts. This clamps the output voltage $V_o$ to the value $-V_{B_2}$,clipping the negative peak.$3$. Since both $V_{B_1}$ and $V_{B_2}$ are less than the peak value of the input,both the positive and negative peaks of the input waveform will be clipped.$4$. Therefore,the output waveform will be clipped at both the positive and negative peaks,as shown in option $A$.

Consider the following circuit. The waveform given in input is as shown. If ${V_{B_1}}$ and ${V_{B_2}}$ are less than the peak value of the input,then the output will look like:

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