If each diode has a forward bias resistance o | vedclass

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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 r

Vedclass · Accepted Answer

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

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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 $R_{	ext{branch}} = 25\,\Omega + 125\,\Omega = 150\,\Omega$.Since the three branches are in parallel, their equivalent resistance is $R_p = \frac{150\,\Omega}{3} = 50\,\Omega$.The total resistance of the circuit is $R_{	ext{eq}} = R_p + 25\,\Omega = 50\,\Omega + 25\,\Omega = 75\,\Omega$.The total current $I_1$ is $I_1 = \frac{V}{R_{	ext{eq}}} = \frac{5\,V}{75\,\Omega} = \frac{1}{15}\,A$.Since the three parallel branches have identical resistances, the current $I_1$ divides equally among them: $I_2 = I_3 = I_4 = \frac{I_1}{3}$.Therefore, $\frac{I_1}{I_2} = 3$ and $\frac{I_2}{I_3} = 1$. Since only option $B$ is correct among the given choices, the answer is $B$.

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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