In the circuit, all three diodes D_1, D_2, D_ | vedclass

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(B) $1$. Analyze the circuit: The diodes $D_1$ and $D_2$ are forward-biased, while $D_3$ is reverse-biased.$2$. Since $D_3$ is reverse-biased, it acts

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$30$

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(B) $1$. Analyze the circuit: The diodes $D_1$ and $D_2$ are forward-biased, while $D_3$ is reverse-biased.$2$. Since $D_3$ is reverse-biased, it acts as an open circuit (infinite resistance), so no current flows through the branch containing $D_3$.$3$. The circuit simplifies to two parallel branches connected in series with the $100 	ext{ } \Omega$ resistor.$4$. Branch $1$ (containing $D_1$): Total resistance $R_1 = R_{D1} + 150 	ext{ } \Omega = 50 	ext{ } \Omega + 150 	ext{ } \Omega = 200 	ext{ } \Omega$.$5$. Branch $2$ (containing $D_2$): Total resistance $R_2 = R_{D2} + 50 	ext{ } \Omega = 50 	ext{ } \Omega + 50 	ext{ } \Omega = 100 	ext{ } \Omega$.$6$. The equivalent resistance of the parallel combination of $R_1$ and $R_2$ is $R_p = \frac{R_1 	imes R_2}{R_1 + R_2} = \frac{200 	imes 100}{200 + 100} = \frac{20000}{300} = \frac{200}{3} 	ext{ } \Omega$.$7$. The total resistance of the circuit is $R_{total} = R_p + 100 	ext{ } \Omega = \frac{200}{3} + 100 = \frac{500}{3} 	ext{ } \Omega$.$8$. The total current $I$ flowing from the battery is $I = \frac{V}{R_{total}} = \frac{5}{500/3} = \frac{15}{500} = 0.03 	ext{ A} = 30 	ext{ mA}$.

In the circuit, all three diodes $D_1, D_2, D_3$ have a forward resistance of $50 \text{ } \Omega$ each and infinite backward resistance. If the battery voltage is $5 \text{ V}$, find the current through the $100 \text{ } \Omega$ resistance. (in $\text{mA}$)

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