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

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(B) $1$. Analyze the orientation of the diodes: The positive terminal of the $12 \, V$ battery is connected to the top rail. Diode $D_1$ is connected

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$2$

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(B) $1$. Analyze the orientation of the diodes: The positive terminal of the $12 \, V$ battery is connected to the top rail. Diode $D_1$ is connected such that its cathode is towards the top rail,making it reverse-biased. Diode $D_2$ is connected such that its anode is towards the top rail,making it forward-biased.$2$. Since the diodes are ideal,the branch with $D_1$ acts as an open circuit (no current flows through the $3 \, \Omega$ resistor).$3$. The branch with $D_2$ acts as a short circuit (zero resistance),so the current flows through the $4 \, \Omega$ resistor and the $2 \, \Omega$ resistor in series.$4$. The total resistance of the circuit is $R_{eq} = 4 \, \Omega + 2 \, \Omega = 6 \, \Omega$.$5$. Using Ohm's law,the total current $I$ is $I = \frac{V}{R_{eq}} = \frac{12 \, V}{6 \, \Omega} = 2 \, A$.

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