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(B) $1$. Analyze the circuit: The battery is connected to a $100 \, \Omega$ resistor in series with a parallel combination of two branches.$2$. Branch

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$0.02$

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(B) $1$. Analyze the circuit: The battery is connected to a $100 \, \Omega$ resistor in series with a parallel combination of two branches.$2$. Branch $1$ contains diode $D_1$ (forward biased) and a $150 \, \Omega$ resistor. Total resistance of this branch = $R_{D1} + R_1 = 50 \, \Omega + 150 \, \Omega = 200 \, \Omega$.$3$. Branch $2$ contains diode $D_2$ (reverse biased). Since it has infinite backward resistance, no current flows through this branch.$4$. The total resistance of the circuit is the sum of the $100 \, \Omega$ resistor and the equivalent resistance of the parallel branches. Since only Branch $1$ is active, the total resistance $R_{eq} = 100 \, \Omega + 200 \, \Omega = 300 \, \Omega$.$5$. The current $I$ through the $100 \, \Omega$ resistor is given by Ohm's Law: $I = \frac{V}{R_{eq}} = \frac{6 \, V}{300 \, \Omega} = 0.02 \, A$.

The circuit shown in the following figure contains two diodes $D_1$ and $D_2$, each with a forward resistance of $50 \, \Omega$ and infinite backward resistance. If the battery voltage is $6 \, V$, find the current through the $100 \, \Omega$ resistor (in amperes).

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