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(C) In the given circuit,the positive terminal of the $12 \ V$ battery is connected such that diode $D_2$ is forward-biased and diode $D_1$ is reverse

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$2$

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(C) In the given circuit,the positive terminal of the $12 \ V$ battery is connected such that diode $D_2$ is forward-biased and diode $D_1$ is reverse-biased.Since $D_1$ is reverse-biased,it acts as an open circuit and no current flows through the $3 \ \Omega$ resistor branch.Since $D_2$ is forward-biased and ideal,it acts as a short circuit. The current flows through the $4 \ \Omega$ resistor and the $2 \ \Omega$ resistor branch.The total resistance of the circuit is $R = 4 \ \Omega + 2 \ \Omega = 6 \ \Omega$.Using Ohm's law,the current $i$ in the circuit is given by:$i = \frac{V}{R} = \frac{12 \ V}{6 \ \Omega} = 2 \ A$.

The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit (in $A$)?

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