Physics - Semiconductors Question with Solution | TestHub

The positive terminal of the $5\text{ V}$ battery is connected to the left side of the circuit, making diodes D1 and D2 forward-biased and diode D3 reverse-biased.

Since D3 has infinite backward resistance, it acts as an open circuit and no current flows through its branch.

The forward resistance of D1 and D2 is $50\Omega$ each. The resistance of the top branch is $R_1=50+150=200\Omega$.

The resistance of the middle branch is $R_2=50+50=100\Omega$.

These two branches are in parallel, so their equivalent resistance $R_p$ is:

$R_p=\frac{R_1{R}_2}{R_1+R_2}=\frac{200\times 100}{200+100}=\frac{20000}{300}=\frac{200}{3}\Omega$

The $100\Omega$ resistor is in series with this parallel combination. The total equivalent resistance of the circuit is:

$R_{eq}=R_p+100=\frac{200}{3}+100=\frac{500}{3}\Omega$

The total current $I$ from the battery, which is the current flowing through the $100\Omega$ resistor, is:

$I=\frac{V}{R_{eq}}=\frac{5}{\frac{500}{3}}=\frac{15}{500}\text{ A}$

$I=0.03\text{ A}=30\text{ mA}$