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(B) In the given circuit,the diode $D_1$ is in reverse bias,so it acts as an open circuit (infinite resistance). The diode $D_2$ is in forward bias,so

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(B) In the given circuit,the diode $D_1$ is in reverse bias,so it acts as an open circuit (infinite resistance). The diode $D_2$ is in forward bias,so it acts as a short circuit (zero resistance).Applying Kirchhoff's voltage law to the loop containing the battery,the $4 \ \Omega$ resistor,and the branch with $D_2$ and the $2 \ \Omega$ resistor:$12 \ V - I(4 \ \Omega) - I(2 \ \Omega) = 0$$12 - 6I = 0$$6I = 12$$I = 2 \ A$

In the given circuit,two ideal diodes are connected in parallel in opposite directions. What is the current $I$ flowing in the circuit (in $A$)?

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