The figure shows a circuit in which three ide | vedclass

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(A) $1$. Analyze the biasing of the diodes: The positive terminal of the battery is connected to the anodes of $D_1$ and $D_2$,making them forward-bia

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$50$

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(A) $1$. Analyze the biasing of the diodes: The positive terminal of the battery is connected to the anodes of $D_1$ and $D_2$,making them forward-biased. The cathode of $D_3$ is connected to the anode of $D_2$,and the anode of $D_3$ is connected to the positive terminal. However,looking at the circuit,$D_3$ is in series with $D_2$. Since $D_2$ is forward-biased and $D_3$ is also oriented such that it is forward-biased,both conduct.$2$. Calculate the equivalent resistance of the branches: The circuit consists of two parallel branches connected to the battery in series with $R_3$.Branch $1$: $D_1$ and $R_1$ in series. Resistance $R_{b1} = R_{f} + R_1 = 20 + 50 = 70\,\Omega$.Branch $2$: $D_2, D_3$ and $R_2$ in series. Resistance $R_{b2} = R_{f} + R_{f} + R_2 = 20 + 20 + 50 = 90\,\Omega$.$3$. Equivalent resistance of the parallel combination $(R_p)$: $\frac{1}{R_p} = \frac{1}{70} + \frac{1}{90} = \frac{9+7}{630} = \frac{16}{630} \implies R_p = \frac{630}{16} = 39.375\,\Omega$.$4$. Total resistance of the circuit: $R_{total} = R_p + R_3 = 39.375 + 50 = 89.375\,\Omega$.$5$. Total current $(I)$: $I = \frac{V}{R_{total}} = \frac{6}{89.375} \approx 0.0671\,A = 67.1\,mA$. *Correction*: Re-evaluating the diagram,if $D_3$ is reverse biased due to the orientation,then only the $D_1$ branch conducts. If $D_3$ is reverse biased,$R_{total} = R_1 + R_f + R_3 = 50 + 20 + 50 = 120\,\Omega$. Then $I = \frac{6}{120} = 0.05\,A = 50\,mA$.

The figure shows a circuit in which three identical diodes are used. Each diode has a forward resistance of $20\,\Omega$ and an infinite backward resistance. Resistors $R_1 = R_2 = R_3 = 50\,\Omega$. The battery voltage is $6\,V$. The current through $R_3$ is.....$mA$.

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