Physics - Semiconductors Question with Solution | TestHub

From the given circuit diagram, the positive terminal of the $12\text{ V}$ battery is connected to the top wire and the negative terminal to the bottom wire.

Diode $D_1$ has its cathode (n-side) connected to the higher potential and anode (p-side) to the lower potential. Therefore, $D_1$ is reverse-biased and acts as an open circuit, meaning no current flows through the $3\Omega$ resistor branch.

Diode $D_2$ has its anode connected to the higher potential and cathode to the lower potential. Therefore, $D_2$ is forward-biased. Since it is an ideal diode, it acts as a short circuit (zero resistance).

The effective circuit consists of the $12\text{ V}$ battery, the $4\Omega$ resistor, and the $2\Omega$ resistor in series.

The equivalent resistance of the circuit is:
$R_{\text{eq}}=4\Omega +2\Omega =6\Omega$

The current flowing in the circuit is:
$I=\frac{V}{R_{\text{eq}}}=\frac{12}{6}=2.00\text{ A}$

Answer: $2.00$A