The Correct Answer is Option (B)

A p-n junction consists of a p-type semiconductor (which has an excess of holes) in contact with an n-type semiconductor (which has an excess of electrons). In a forward-biased p-n junction, an external voltage is applied such that the positive terminal is connected to the p-side and the negative terminal is connected to the n-side. This configuration promotes the flow of majority charge carriers (holes from the p-side and electrons from the n-side) across the junction.

There are two types of currents in a p-n junction: drift current and diffusion current. Drift current is caused by the electric field due to the built-in potential, which opposes the flow of majority charge carriers. Diffusion current is caused by the concentration gradient of the charge carriers, which promotes the flow of majority charge carriers.

When the p-n junction is forward biased, the applied voltage reduces the potential barrier, allowing more majority charge carriers to flow across the junction. This results in an increase in the diffusion current. In a forward-biased p-n junction, the diffusion current is indeed greater than the drift current in magnitude, so Assertion A is correct.

Regarding Reason R: The diffusion current in a p-n junction is actually from the p-side to the n-side, as holes (majority charge carriers in the p-side) move from the p-side to the n-side, and electrons (majority charge carriers in the n-side) move from the n-side to the p-side. Therefore, Reason R is incorrect.

Power gain

$\begin{array}{rl}& \Rightarrow {\mathrm{A}}_{\mathrm{v}}\cdot {\mathrm{A}}_1=\mathrm{B}\frac{{\mathrm{R}}_{\mathrm{C}}}{{\mathrm{R}}_{\mathrm{B}}}\cdot \mathrm{B}={\mathrm{B}}^2\frac{{\mathrm{R}}_{\mathrm{C}}}{{\mathrm{R}}_{\mathrm{B}}}\\ \\ & =\left(\frac{(20-10)\times {10}^{-3}}{(200-100)\times {10}^{-6}}\right)\times \frac{1\times {10}^3}{10\times {10}^3}={10}^3\end{array}$

Hence $\mathrm{x}=3$

The Correct Answer is Option (B)

A diode is a two terminal device which conducts current in forward bias only.

$E=\frac{V}{d}=\frac{0.6}{6\times {10}^{-6}}=1\times {10}^5$

The Correct Answer is Option (D)

In saturation mode ${\mathrm{V}}_{\mathrm{CE}}=0$


Now

$\begin{array}{rl}& {\mathrm{V}}_{\mathrm{CC}}-{\mathrm{I}}_{\mathrm{c}}{\mathrm{R}}_{\mathrm{c}}=0\text{(from KVL)}\\ \\ & \Rightarrow {\mathrm{I}}_{\mathrm{b}}=\frac{{\mathrm{V}}_{\mathrm{CC}}}{{\mathrm{R}}_{\mathrm{C}}}=\frac{1}{1\times {10}^3}={10}^{-3}\mathrm{A}\end{array}$

Given that

$\beta =30=\frac{I_c}{I_b}$

//-->$\begin{array}{rl}& \Rightarrow {\mathrm{I}}_b=\frac{{\mathrm{I}}_c}{30}=\frac{{10}^{-3}}{30}={10}^{-5}\mathrm{A}\\ \\ & \Rightarrow {\mathrm{I}}_b=10\mu \mathrm{A}\end{array}$