Physics - Semiconductors Question with Solution | TestHub

From the circuit diagram, the positive terminal of the $3\text{ V}$ battery is on the left.

Diode $D_1$ is reverse-biased (n-side connected to the positive terminal) and acts as an open circuit.

Diode $D_2$ is forward-biased (p-side connected to the positive terminal) and acts as a closed circuit since it is an ideal diode.

Current flows only through the branch containing $D_2$. The circuit simplifies to the $30\Omega$ and $70\Omega$ resistors connected in series.

The equivalent resistance of the circuit is $R_{eq}=30+70=100\Omega$.

Using Ohm's law, the current through the $70\Omega$ resistance is:

$I=\frac{V}{R_{eq}}=\frac{3}{100}=0.03\text{ A}$