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(C) In the given circuit,diode $D_{1}$ is connected to $12.7\; V$ and diode $D_{2}$ is connected to $4\; V$.Since the potential at the anode of $D_{1}

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$12$

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(C) In the given circuit,diode $D_{1}$ is connected to $12.7\; V$ and diode $D_{2}$ is connected to $4\; V$.Since the potential at the anode of $D_{1}$ $(12.7\; V)$ is higher than the potential at the anode of $D_{2}$ $(4\; V)$,diode $D_{1}$ will be forward biased and diode $D_{2}$ will be reverse biased.For a forward-biased diode with a built-in potential of $0.7\; V$,the output voltage at point $A$ is given by $V_{A} = V_{in} - V_{barrier}$.Substituting the values,we get $V_{A} = 12.7\; V - 0.7\; V = 12\; V$.Thus,the voltage at point $A$ is $12\; V$.

Both the diodes used in the circuit shown are assumed to be ideal and have negligible resistance when these are forward biased. The built-in potential in each diode is $0.7\; V$. For the input voltages shown in the figure,the voltage (in $Volts$) at point $A$ is

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