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: What the circuit shows.
Between two points $P$ and $Q$ there is a diode and a resistor of $2\,\text{k}\Omega = 2 \times 10^3\ \Omega$ in a line. The point $P$ is at $+3$ V and the point $Q$ is at $-5$ V. We need the current in this path.
Step 2: Check the diode first.
A diode only lets current pass easily one way. Its p-side faces $P$ ($+3$ V) and its n-side faces the lower voltage. The p-side is at a higher voltage than the n-side, so the diode is forward biased. That means it is switched on.
Step 3: Use the ideal diode idea.
The diode is called ideal. So when it is on it acts like a plain wire with no voltage drop. All the voltage is then left for the resistor.
Step 4: Find the voltage across the path.
The voltage difference from $P$ to $Q$ is \[ \Delta V = V_P - V_Q = 3 - (-5) = 8\ \text{V} \]
Step 5: Apply Ohm's law.
Current equals voltage divided by resistance: \[ I = \frac{\Delta V}{R} = \frac{8}{2 \times 10^3} \]
Step 6: Get the answer.
\[ I = 4 \times 10^{-3}\ \text{A} \] So the current is $4$ mA, which is option (3). \[ \boxed{I = 4 \times 10^{-3}\ \text{A}} \]