In one branch, 40 Ω resistance and a p-n junction diode are connected in series and in another branch, 40 Ω resistance is connected. Both the branches are connected in parallel to each other. The tota

Question

In one branch, 40 Ω resistance and a p-n junction diode are connected in series and in another branch, 40 Ω 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}

TestHub · Accepted Answer

When the ideal p-n junction diode is forward biased, it offers zero resistance. The resistance of the first branch is $40 Ω$ .

Since the two branches are in parallel, the equivalent resistance $R_{e q}$ is given by:

$R_{e q} = \frac{40 \times 40}{40 + 40} = 20 Ω$

When the ideal p-n junction diode is reverse biased, it offers infinite resistance. The first branch acts as an open circuit and no current flows through it.

The equivalent resistance of the circuit is simply the resistance of the second branch:

$R_{e q} = 40 Ω$

The total resistance of the circuit in forward and reverse bias respectively are $20 Ω$ and $40 Ω$ .

Answer: $20 Ω$ and $40 Ω$

Physics - Semiconductors Question with Solution | TestHub

Options:

Doubts & Discussion