Physics - Semiconductors Question with Solution | TestHub

In the given circuit, the positive terminal of the $10\text{ V}$ battery is on the left and the negative terminal is on the right.

Diode $D_1$ has its p-side connected to the higher potential (positive terminal), so it is forward-biased. Being an ideal diode, it acts as a short circuit (zero resistance).

Diode $D_2$ has its n-side connected to the higher potential, so it is reverse-biased. It acts as an open circuit (infinite resistance).

Because $D_2$ is reverse-biased, no current flows through the lower branch. The entire current from the battery flows through the upper branch containing the $20\Omega$ resistor.

Using Ohm's law, the current supplied by the battery is:
$I=\frac{V}{R}=\frac{10}{20}=0.5\text{ A}$

Answer: 0.50 A