Question

What are the values of the currents flowing in each of the following diode circuits X and Y respectively? (Assume that the diodes are ideal)

Choose the correct answer from the options given below

• A. 1 A, 2 A
• B. 2 A, 1 A
• C. 4 A, 2 A
• D. 2 A, 4 A

Hint:

To quickly find the current through any forward-biased ideal diode branch, find the net potential difference between the two outer ends of the branch and divide it by the total resistance: $I = \frac{\Delta V}{R}$. For Y, this gives $\frac{4 - 2}{2} = 1\ \text{A}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem presents two different circuit configurations, labeled X and Y, containing ideal diodes connected to resistors and voltage sources. We need to determine the stable forward current passing through each individual system.

Step 2: Key Formula or Approach:
An ideal diode acts as a perfect structural switch based on its biasing state:
1.

Forward Bias: When the p-side potential is higher than the n-side potential, the diode acts as a closed switch with zero resistance ($R=0$).
2.

Reverse Bias: When the n-side potential is higher than the p-side potential, the diode acts as an open circuit with infinite resistance ($R=\infty$), blocking all current.
Once the bias state is determined, use Ohm's Law ($I = \frac{V}{R}$) to find the current.

Step 3: Detailed Explanation:
Let's analyze each circuit individually using the values from the circuit diagram:

Circuit X: The p-side of the diode is connected to the $+4\ \text{V}$ terminal, and the n-side is connected through a $2\ \Omega$ resistor to the $0\ \text{V}$ ground terminal. Since the p-side is at a higher potential than the n-side, the diode is

forward-biased and acts as a short circuit.
Applying Ohm's law to find the current $I_X$:
$$I_X = \frac{V_{\text{p}} - V_{\text{n}}}{R} = \frac{4\ \text{V} - 0\ \text{V}}{2\ \Omega} = 2\ \text{A}$$

Circuit Y: The circuit contains a $+4\ \text{V}$ source connected through a $2\ \Omega$ resistor to the p-side of a diode, while its n-side is connected to a $+2\ \text{V}$ bias terminal. Since $+4\ \text{V} > +2\ \text{V}$, the p-side is at a higher potential, so the diode is

forward-biased and acts as a closed switch.
Applying Ohm's law to find the current $I_Y$:
$$I_Y = \frac{V_{\text{high}} - V_{\text{low}}}{R} = \frac{4\ \text{V} - 2\ \text{V}}{2\ \Omega} = \frac{2\ \text{V}}{2\ \Omega} = 1\ \text{A}$$ Combining our results yields a current of 2 A for circuit X and 1 A for circuit Y.

Step 4: Final Answer:
The currents flowing in the circuits are 2 A and 1 A respectively, matching option (B).