The circuit shown contains a 12 , V battery,a | vedclass

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(B) In the given circuit,the battery provides a potential difference of $12 \, V$. Diode $D_1$ is connected in reverse bias relative to the positive t

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$2$

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(B) In the given circuit,the battery provides a potential difference of $12 \, V$. Diode $D_1$ is connected in reverse bias relative to the positive terminal of the battery,so it acts as an open circuit (no current flows through the $3 \, \Omega$ branch). Diode $D_2$ is connected in forward bias,so it acts as a short circuit (ideal diode). Therefore,the total resistance of the circuit is the sum of the $4 \, \Omega$ resistor and the $2 \, \Omega$ resistor in the active branch: $R_{eq} = 4 \, \Omega + 2 \, \Omega = 6 \, \Omega$. Using Ohm's law,the current $I$ is given by $I = \frac{V}{R_{eq}} = \frac{12 \, V}{6 \, \Omega} = 2 \, A$.

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