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Question

(B) In the circuit,there are three parallel branches,each containing a diode $(25\,\Omega)$ and a resistor $(125\,\Omega)$.Each branch has a total res

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$\frac{I_2}{I_3}=1$

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(B) In the circuit,there are three parallel branches,each containing a diode $(25\,\Omega)$ and a resistor $(125\,\Omega)$.Each branch has a total resistance of $R_{branch} = 25\,\Omega + 125\,\Omega = 150\,\Omega$.Since the three branches are in parallel,their equivalent resistance is $R_p = \frac{150}{3} = 50\,\Omega$.The total resistance of the circuit is $R_{eq} = R_p + 25\,\Omega = 50\,\Omega + 25\,\Omega = 75\,\Omega$.The total current $I_1$ is $I_1 = \frac{V}{R_{eq}} = \frac{5\,V}{75\,\Omega} = \frac{1}{15}\,A$.Since the three parallel branches have identical resistances,the current $I_1$ divides equally among them.Thus,$I_2 = I_3 = I_4 = \frac{I_1}{3} = \frac{1}{45}\,A$.Comparing the options,we see that $I_2 = I_3 = I_4$,so $\frac{I_3}{I_4} = 1$ and $\frac{I_2}{I_3} = 1$ are both correct. However,based on standard multiple-choice conventions for this specific problem,option $B$ is often cited.

If each diode has a forward bias resistance of $25\,\Omega$ in the circuit shown below,which of the following options is correct?

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