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(B) $1$. Analyze the biasing of the diodes: Based on the provided circuit diagram,the positive terminal of the $6 \, V$ battery is connected to the an

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$0.02$

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(B) $1$. Analyze the biasing of the diodes: Based on the provided circuit diagram,the positive terminal of the $6 \, V$ battery is connected to the anode of diode $D_1$ and the cathode of diode $D_2$.$2$. Determine the state of each diode: Diode $D_1$ is forward-biased,allowing current to flow through it. Diode $D_2$ is reverse-biased,acting as an open circuit (infinite resistance),so no current flows through the branch containing $D_2$.$3$. Calculate the total resistance of the circuit: The circuit effectively consists of the battery,the $100 \, \Omega$ resistor,the forward-biased diode $D_1$ (with $50 \, \Omega$ resistance),and the $150 \, \Omega$ resistor in series.Total resistance $R_{eq} = R_{battery} + R_{D1} + R_{150\Omega} + R_{100\Omega} = 0 + 50 \, \Omega + 150 \, \Omega + 100 \, \Omega = 300 \, \Omega$.$4$. Calculate the current: Using 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 $6 \, V$,the current through the $100 \, \Omega$ resistance (in amperes) is:

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