Physics - Semiconductors Question with Solution | TestHub

From the given circuit, the positive terminal of the $10\text{ V}$ battery is connected to the left side of the parallel branches.

Therefore, the left side is at a higher potential compared to the right side.

The diodes in the branches with $4\Omega$ and $2\Omega$ resistors have their anodes connected to the higher potential side, so they are forward-biased and act as short circuits (since they are ideal).

The diodes in the branches with $3\Omega$ and $5\Omega$ resistors have their cathodes connected to the higher potential side, so they are reverse-biased and act as open circuits.

The effective circuit consists of the $4\Omega$ and $2\Omega$ resistors connected in parallel across the $10\text{ V}$ battery.

The equivalent resistance of the circuit is:

$R_{eq}=\frac{4\times 2}{4+2}=\frac{8}{6}=\frac{4}{3}\Omega$

The total current $I$ in the circuit is:

$I=\frac{V}{R_{eq}}=\frac{10}{\frac{4}{3}}=\frac{30}{4}=\frac{15}{2}\text{ A}$

Answer: $\frac{15}{2}$ A