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(B) In the given circuit,diode $D_1$ has its anode connected to $+5 \ V$ and its cathode connected to $V_{	ext{out}}$. Diode $D_2$ has its anode conn

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$0 \ V$

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(B) In the given circuit,diode $D_1$ has its anode connected to $+5 \ V$ and its cathode connected to $V_{	ext{out}}$. Diode $D_2$ has its anode connected to $V_{	ext{out}}$ and its cathode connected to ground $(0 \ V)$.Assuming $D_2$ is forward biased,$V_{	ext{out}}$ will be clamped to $0 \ V$ (since the diode is ideal).If $V_{	ext{out}} = 0 \ V$,then for $D_1$,the anode is at $+5 \ V$ and the cathode is at $0 \ V$. Thus,$D_1$ is forward biased.However,if $D_1$ is forward biased,$V_{	ext{out}}$ would be $+5 \ V$. If $V_{	ext{out}} = +5 \ V$,then $D_2$ has its anode at $+5 \ V$ and cathode at $0 \ V$,making $D_2$ forward biased.Since $D_2$ is forward biased,it pulls the output voltage to the potential of its cathode,which is $0 \ V$.Therefore,the output voltage $V_{	ext{out}} = 0 \ V$.

The output voltage in the following circuit is (Consider ideal diode case)

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