PN Junction and Diode Questions in English

(B) The width of the depletion region $(w)$ in a $p-n$ junction is inversely proportional to the doping concentration $(N)$ on that side,expressed as $w \propto 1/N$.
Given that the $p$-side has a lower doping concentration $(10^{15} \text{ atoms/cm}^3)$ compared to the $n$-side $(10^{18} \text{ atoms/cm}^3)$,the depletion region will extend further into the $p$-side.
Therefore,the depletion region width is greater on the $p$-side than on the $n$-side.

(C) The time constant of an $RC$ circuit is given by $\tau = R_{eq}C$.
In the given circuit,there is a resistor $R = 100 \Omega$ in series with a parallel combination of two identical diodes.
Each diode has a resistance of $10 \Omega$ in the forward bias condition.
The equivalent resistance of the two diodes in parallel is $R_d = \frac{10 \times 10}{10 + 10} = 5 \Omega$.
The total resistance of the circuit is $R_{total} = R + R_d = 100 \Omega + 5 \Omega = 105 \Omega$.
The capacitance is $C = 20 \mu\text{F} = 20 \times 10^{-6} \text{ F}$.
Thus,the time constant $\tau = R_{total} \times C = 105 \Omega \times 20 \times 10^{-6} \text{ F} = 2100 \times 10^{-6} \text{ s} = 2.1 \times 10^{-3} \text{ s}$.
Comparing this with $\alpha \times 10^{-3} \text{ s}$,we get $\alpha = 2.1$.

(C) Statement $A$ is correct: In a $p-n$ junction,forward bias reduces the potential barrier,allowing current to rise sharply after the threshold voltage.
Statement $B$ is incorrect: The current in forward bias is called forward current. The term 'reverse saturation current' refers to the very small,nearly constant current that flows in a $p-n$ junction diode when it is in reverse bias.
Therefore,Statement $A$ is true and Statement $B$ is false.

(B) The circuit consists of four parallel branches connected to a $10 \text{V}$ $DC$ source.
Each branch contains a resistor and a diode.
Analyzing the polarity of the diodes with respect to the $10 \text{V}$ battery:
$1$. The top branch ($4 \Omega$ resistor) has the diode in forward-biased condition.
$2$. The second branch ($3 \Omega$ resistor) has the diode in reverse-biased condition (it acts as an open circuit).
$3$. The third branch ($2 \Omega$ resistor) has the diode in forward-biased condition.
$4$. The bottom branch ($5 \Omega$ resistor) has the diode in reverse-biased condition (it acts as an open circuit).
Only the branches with $4 \Omega$ and $2 \Omega$ resistors are active.
These two resistors are in parallel,so the equivalent resistance $R_{\text{eq}}$ is:
$R_{\text{eq}} = \frac{4 \times 2}{4 + 2} = \frac{8}{6} = \frac{4}{3} \Omega$.
The total current $I$ drawn from the battery is:
$I = \frac{V}{R_{\text{eq}}} = \frac{10}{4/3} = \frac{30}{4} = 7.5 \text{A} = \frac{15}{2} \text{A}$.
Thus,the correct option is $B$.