Question

The circuit has two oppositely connected ideal diodes in parallel as shown in the figure. What is the current flowing in the circuit?

• A. 1.33A
• B. 1.71A
• C. 2.00A
• D. 2.31A

Hint:

To analyze diode circuits quickly, replace each ideal diode with a switch: open if the diode points against the conventional current flow (from positive to negative terminal), and closed if it points along the current flow. This simplifies the schematic into a standard resistor network instantly.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
An ideal diode acts as a perfect conductor (zero resistance) when forward biased and as a perfect insulator (infinite resistance) when reverse biased.
Forward bias occurs when the P-terminal is at a higher potential than the N-terminal.
In this parallel arrangement, one diode will be forward biased while the other will be reverse biased because they are connected in opposite directions.
The current will only flow through the branch containing the forward-biased diode.
Step 2: Key Formula or Approach:
1. Identify the bias of \(D_1\) and \(D_2\).
2. Replace the forward-biased diode with a wire and the reverse-biased diode with an open circuit.
3. Apply Ohm's Law: \(I = \frac{V}{R_{total}}\).
Step 3: Detailed Explanation:
In the diagram, the positive terminal of the 12V battery is connected to the top part of the parallel network.
Looking at the diode orientations:
The P-side of \(D_2\) is connected towards the positive terminal of the battery. Thus, \(D_2\) is forward biased.
The N-side of \(D_1\) is connected towards the positive terminal. Thus, \(D_1\) is reverse biased.
As a result, no current flows through the branch with \(D_1\) and the \(3 \Omega\) resistor.
The entire current flows through the series combination of the \(4 \Omega\) resistor and the \(2 \Omega\) resistor (the branch with \(D_2\)).
Total Resistance \(R_{total} = 4 \Omega + 2 \Omega = 6 \Omega\).
Circuit current \(I = \frac{V}{R_{total}} = \frac{12 \text{ V}}{6 \Omega} = 2.00 \text{ A}\).
Step 4: Final Answer:
Due to the parallel arrangement of opposite diodes, current is restricted to only one branch. Calculating the resistance of that active branch gives a total current of 2.00 A.