Question

In one branch, \(40\Omega\) resistance and a p-n junction diode are connected in series and in another branch, \(40\Omega\) resistance is connected. Both the branches are connected in parallel to each other. The total resistance of the circuit when a) diode is forward biased and when b) diode is reverse biased respectively are {consider the p-n junction diode as an ideal diode}

• A. \(40\Omega\) and \(20\Omega\)
• B. \(60\Omega\) and \(30\Omega\)
• C. \(20\Omega\) and \(40\Omega\)
• D. \(30\Omega\) and \(60\Omega\)

Hint:

An ideal diode is a short circuit in forward bias and an open circuit in reverse bias.

Answer 1

The Correct Option is C

Step 1: Conductance Method
In forward bias, conductance adds: $\dfrac{1}{R}=\dfrac{1}{40}+\dfrac{1}{40}=\dfrac{1}{20}$.

Step 2: Reverse Bias
The diode branch has infinite resistance, so $\dfrac{1}{R}=\dfrac{1}{40}+0$.

Step 3: Result
$R=20\,\Omega$ and $R=40\,\Omega$ respectively, which is option (C).

Final Answer:
Adding conductances gives 20 ohm in forward bias and 40 ohm in reverse bias. \[ \boxed{\text{(C) }\text{20}\,\Omega\text{ and 40}\,\Omega} \]