Physics - Semiconductors Question with Solution | TestHub

From the given circuit diagram, the positive terminal of the $6\text{ V}$ battery is connected to the top wire and the negative terminal is connected to the bottom wire through the $100\Omega$ resistor. Therefore, the top wire is at a higher potential than the bottom wire.

Based on this polarity:
Diode $D_1$ has its n-side connected to the higher potential, so it is reverse-biased. Its resistance is infinite, and no current flows through this branch.
Diodes $D_2$ and $D_3$ have their p-sides connected to the higher potential, so they are forward-biased. Each offers a forward resistance of $40\Omega$.

The resistance of the branch containing $D_2$ is:
$R_2=40\Omega +60\Omega =100\Omega$

The resistance of the branch containing $D_3$ is:
$R_3=40\Omega +160\Omega =200\Omega$

These two branches are in parallel. Their equivalent resistance $R_p$ is:
$\frac{1}{R_p}=\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{100}+\frac{1}{200}=\frac{3}{200}{\Omega }^{-1}$
$R_p=\frac{200}{3}\Omega$

This parallel combination is in series with the $100\Omega$ resistor and the $6\text{ V}$ battery. The total equivalent resistance of the circuit is:
$R_{eq}=100+R_p=100+\frac{200}{3}=\frac{500}{3}\Omega$

The current through the $100\Omega$ resistor is the total current $I$ drawn from the battery:
$I=\frac{V}{R_{eq}}=\frac{6}{\frac{500}{3}}=\frac{18}{500}\text{ A}$

Converting the current to milliamperes (mA):
$I=\frac{18}{500}\times 1000\text{ mA}=18\times 2\text{ mA}=36\text{ mA}$

Answer: 36 mA