Physics - Semiconductors Question with Solution | TestHub

Let us analyze the potentials at the two ends of the given circuit branch.

From the circuit diagram, the left terminal is connected to a battery which provides a potential of $-4\text{V}$ at the p-side of the diode. Thus, the potential at the p-side is $V_p=-4\text{V}$.

The right terminal is connected to a battery which provides a potential of $-1\text{V}$ at the other end of the $3\Omega$ resistor. Thus, the potential at the n-side of the diode (if it were conducting) would be pulled towards $V_n=-1\text{V}$.

For a p-n junction diode to be forward biased and conduct current, the potential at the p-side must be greater than the potential at the n-side ($V_p>V_n$).

In this circuit, we have:
$V_p=-4\text{V}$
$V_n=-1\text{V}$

Comparing the two potentials, we see that $-4\text{V}<-1\text{V}$, which means $V_p<V_n$.

Since the p-side is at a lower potential than the n-side, the diode is reverse biased.

An ideal diode in reverse bias acts as an open circuit, offering infinite resistance. Therefore, it blocks the flow of current.

As a result, the current in the circuit is zero.

Answer: zero ampere