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

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(B) $1$. Analyze the diode states: The battery positive terminal is connected to the anode of $D_2$ and the cathode of $D_1$. Therefore,diode $D_2$ is

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$2$

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(B) $1$. Analyze the diode states: The battery positive terminal is connected to the anode of $D_2$ and the cathode of $D_1$. Therefore,diode $D_2$ is forward-biased and diode $D_1$ is reverse-biased.$2$. Simplify the circuit: Since $D_1$ is reverse-biased,it acts as an open circuit (infinite resistance),so no current flows through the $3 \ \Omega$ resistor branch.$3$. Calculate the total resistance: The circuit effectively consists of the $12 \ V$ battery,the $4 \ \Omega$ resistor,and the $2 \ \Omega$ resistor in series.$R_{total} = 4 \ \Omega + 2 \ \Omega = 6 \ \Omega$.$4$. Calculate the current: Using Ohm's law,$I = \frac{V}{R_{total}} = \frac{12 \ V}{6 \ \Omega} = 2 \ A$.

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