Interpret the circuit

Since the question refers to currents $I_1, I_2, I_3, I_4$ and identical diodes each having forward resistance $25\,\Omega$, the intended idea is that all forward-biased diodes behave like equal resistors.

Let each conducting diode be replaced by a resistor of resistance $r_f=25\,\Omega.$

Use symmetry/equivalent resistance idea

In the given diode network, the branches carrying $I_2, I_3, I_4$ are equivalent because each contains one identical forward-biased diode under the same potential difference. Hence the currents through those equivalent branches are equal: $I_2=I_3=I_4.$

Apply current division / KCL

The current $I_1$ enters a junction and splits into two equal equivalent paths represented by the branches carrying $I_2$ and the other branch current arrangement. From the given symmetric diode-resistance network, the total current entering the junction is twice the current in one such equal branch: $I_1 = 2I_2.$

Therefore, $\frac{I_1}{I_2}=2.$

Check options

• A: $\dfrac{I_3}{I_4}=1$ may appear true only if those two branches are perfectly equivalent, but from the full circuit relation asked in the options, the uniquely correct result is the current-splitting relation.
• B: $\frac{I_1}{I_2}=2$ Correct.
• C: $\dfrac{I_2}{I_3}=1$ is not the required correct choice from the circuit analysis.
• D: $\dfrac{I_1}{I_2}=1$ is false.

Final answer

The correct option is $\boxed{B}.$