Semiconductor Electronics, on the exam screen.

An ideal diode has zero forward resistance and infinite reverse resistance, so only the branches whose diodes are forward biased conduct; here that leaves the 4 $\Omega$ and 2 $\Omega$ resistors in parallel across the 10 V battery. The current is $I = \frac{10}{2} + \frac{10}{4} = \frac{15}{2}$ A, giving (4). The other options wrongly include the reverse-biased 3 $\Omega$ and 5 $\Omega$ branches that actually carry no current.

The applied alternating current (ac) voltage drives current through the diode and R in series. During the half-cycle that reverse-biases the diode, almost the entire source voltage appears across the non-conducting diode, whereas in the forward-biased half-cycle the diode drop is nearly zero. So $v_D$ consists of one-sided humps with flat (near-zero) gaps, matching option (4). A full sine or a double-hump trace would wrongly imply the diode never blocks the current.

Statement A is correct: beyond the threshold (knee) voltage in forward bias, the diode current rises steeply, as seen in the forward V-I characteristic. Statement B is wrong: this rapidly increasing current is the forward current, whereas reverse saturation current is the very small, nearly constant current that flows under reverse bias.

Once an ideal Zener diode is driven past its breakdown voltage it CLAMPS the voltage across itself at 3 V and behaves like a constant-voltage source. The remaining $5 - 3 = 2\ \text{V}$ of the supply must therefore appear across the series resistor, which is the element between $B$ and $A$. NCERT Class 12, Semiconductor Electronics (Zener diode as a voltage regulator).

Why the other options are wrong:

- (1) reports the voltage across the ZENER rather than across the resistor between $B$ and $A$
- (3) subtracts twice, or assumes the diode drops an extra 1 V forward-bias contribution
- (4) assumes the reverse-biased diode blocks all current, so no drop appears across the resistor - but breakdown means current does flow

Depletion width is set by the net potential barrier at the junction. Forward bias lowers the barrier and NARROWS the depletion layer; reverse bias raises it and WIDENS the layer; an unbiased junction sits at its equilibrium width in between. In this circuit $D_1$ is forward biased (narrowest), $D_2$'s branch carries no bias so it keeps its equilibrium width, and $D_3$ is reverse biased (widest), giving $W_3 > W_2 > W_1$. NCERT Class 12, Semiconductor Electronics (p-n junction).

Why the other options are wrong:

- (1) reverses the bias effect, assuming forward bias widens and reverse bias narrows the depletion region
- (2) assumes forward and reverse bias shift the width by equal and opposite amounts so that $W_1$ and $W_3$ coincide
- (4) treats the unbiased diode as the extreme case rather than the intermediate one

The width of the depletion region increases under reverse bias, as the applied voltage enhances the electric field and repels more majority carriers away from the junction. Chapter: Semiconductor Electronics.

For transistor action, the base region must be very thin and lightly doped to ensure that most of the charge carriers injected from the emitter reach the collector. Chapter: Semiconductor Electronics.

Each input passes through a NOT gate, and the two inverted inputs feed a NOT-OR (NOR) gate, so Y = NOT(NOT A OR NOT B), which simplifies to A AND B. The circuit is therefore an AND gate, whose output is high only when both inputs are high:

Chapter: Semiconductor Electronics.

Why the other options are wrong:

- (1) This is the OR truth table: it makes Y high whenever either input is high, which the inverters plus NOR does not do.
- (2) This is the NOT-AND (NAND) truth table, the inverse of the correct one. It would follow from an AND gate with an inverted output.
- (3) This is the NOR truth table, which is what a single NOR gate gives for A and B directly, without the two inverters.