In the circuit shown in the figure,determine | vedclass

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(A) In the given circuit,both silicon $(Si)$ and germanium $(Ge)$ diodes are connected in parallel. The cut-in voltage for a silicon diode is $0.7 \,

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$11.7 \, V$

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(A) In the given circuit,both silicon $(Si)$ and germanium $(Ge)$ diodes are connected in parallel. The cut-in voltage for a silicon diode is $0.7 \, V$ and for a germanium diode is $0.3 \, V$.As the input voltage increases from $0 \, V$ to $12 \, V$,the voltage across the diodes increases. When the voltage reaches $0.3 \, V$,the germanium diode starts conducting and acts as a battery of $0.3 \, V$. Since the diodes are in parallel,the voltage across the silicon diode is clamped at $0.3 \, V$. Because $0.3 \, V Therefore,the circuit behaves as a series combination of a $12 \, V$ source,a $0.3 \, V$ battery (representing the conducting $Ge$ diode),and a $2.2 \, k\Omega$ resistor.The voltage $V_0$ across the resistor is given by the potential difference between the input and the diode drop:$V_0 = 12 \, V - 0.3 \, V = 11.7 \, V$.

In the circuit shown in the figure,determine the voltage $V_0$ for the circuit network.

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