Question

In an ideal junction diode, the current flowing through PQ is

Choose the correct answer from the options given below

• A. $2 \times 10^{-3}\text{ A}$
• B. $2 \times 10^{-2}\text{ A}$
• C. $4 \times 10^{-3}\text{ A}$
• D. $10^{-3}\text{ A}$

Hint:

Be careful with negative potentials! A common mistake on exams is subtracting the values incorrectly as $3 - 5 = -2\text{ V}$ or $5 - 3 = 2\text{ V}$. Always use parentheses when subtracting negative values: $3 - (-5) = 8\text{ V}$ to find the true potential difference.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem presents an electronic circuit branch containing a p-n junction diode and a series resistor of resistance $R = 2\text{ k}\Omega = 2 \times 10^3\ \Omega$ spanning between two terminals $P$ and $Q$. The terminals are maintained at potentials $V_P = +3\text{ V}$ and $V_Q = -5\text{ V}$. We need to find the current flowing through the branch.

Step 2: Key Formula or Approach:
1.

Bias Check: Identify the operating state of the diode. The p-side of the diode is connected toward terminal $P$ ($+3\text{ V}$) and the n-side is connected toward terminal $Q$ (via the resistor to $-5\text{ V}$). Since the p-side potential is higher than the n-side potential ($3\text{ V} > -5\text{ V}$), the diode is

Forward Biased. 2.

Ideal Approximation: Since it is specified as an ideal junction diode, it has zero forward resistance and zero barrier voltage drop, acting simply as a closed switch (short circuit). 3.

Ohm's Law: The current $I$ through the branch is governed by the total potential difference across the series resistor: $$I = \frac{\Delta V}{R} = \frac{V_{\text{high}} - V_{\text{low}}}{R}$$

Step 3: Detailed Explanation:
Calculate the net potential difference ($\Delta V$) across the branch terminals from $P$ to $Q$: $$\Delta V = V_P - V_Q = 3\text{ V} - (-5\text{ V}) = 3 + 5 = 8\text{ V}$$ Now apply Ohm's law across the series resistor network to compute the total current: $$I = \frac{8\text{ V}}{2\text{ k}\Omega} = \frac{8}{2 \times 10^3\ \Omega}$$ $$I = 4 \times 10^{-3}\text{ A}$$ This matches option (C).

Step 4: Final Answer:
The current flowing through the circuit branch is $4 \times 10^{-3}\text{ A}$, corresponding to option (C).