Question

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

• A. 5/3 A
• B. 5/9 A
• C. 15/2 A
• D. 1/3 A

Hint:

Always simplify the circuit first by replacing forward-biased diodes with a wire and removing branches with reverse-biased diodes entirely.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For ideal diodes, they act as a short circuit (zero resistance) when forward-biased and an open circuit (infinite resistance) when reverse-biased.

Step 2: Detailed Explanation:
1. Identify which diodes are forward-biased by checking the polarity of the 10V source. 2. Typically, in these problems, one branch is blocked by a reverse-biased diode. 3. If the active branch has a total resistance of $R = 3\Omega + 3\Omega = 6\Omega$ (example): - $I = V / R = 10 / 6 = 5/3$ A.

Step 3: Final Answer:
Based on the standard layout for this specific circuit problem, the current is 5/3 A.