The current $I$ in the circuit shown below is: (All diodes are ideal and identical)

![](https://res.cloudinary.com/ddr1kuyb3/image/upload/v1786187229/questions/xd0jy9krhhvfwo59zivz.png)

Question

The current $I$ in the circuit shown below is: (All diodes are ideal and identical)

![](https://res.cloudinary.com/ddr1kuyb3/image/upload/v1786187229/questions/xd0jy9krhhvfwo59zivz.png)

Accepted Answer

D. 1. Recall how an **ideal diode** behaves in a circuit.

An ideal diode offers **zero resistance** when forward biased (acts like a plain wire) and **infinite resistance** when reverse biased (acts like an open circuit, blocking all current through that branch).

2. Simplify the given circuit using this rule.

By checking the polarity of the battery relative to each diode, the forward-biased diodes are replaced by simple wires (their resistance branches remain), while any reverse-biased diode branch is removed as an open circuit. Given the figure, this redraws the circuit into two parallel resistive branches of $2\ \Omega$ and $4\ \Omega$, both connected across the same $10$ V source.

3. Use **Ohm's Law** to find the current through each simplified branch.

Since each branch independently sees the full $10$ V:
$$
I_1 = \frac{10}{2}, \quad I_2 = \frac{10}{4}
$$

4. Add the branch currents to get the total current supplied by the battery, since the branches are in parallel.

Therefore:
$$
I = \frac{10}{2} + \frac{10}{4} = \frac{30}{4} = \frac{15}{2}	ext{ A}
$$

Hence, the current $I$ is $\frac{15}{2}$ A, which is **option D**.

The current III in the circuit shown below is: (All diodes are ideal and identical)

The current $I$ in the circuit shown below is: (All diodes are ideal and identical)