PN Junction and Diode Questions in English

(D) In the given circuit,the diode $D_1$ is forward-biased because its p-side is connected to the positive terminal of the battery. The diode $D_2$ is reverse-biased because its p-side is connected to the negative terminal of the battery.
Therefore,no current flows through the branch containing $D_2$.
The circuit effectively consists of the battery $(5 \ V)$,the resistor $(20 \ \Omega)$,and the branch containing $D_1$ with a resistor $(30 \ \Omega)$ in series.
The total resistance of the circuit is $R_{eq} = 20 \ \Omega + 30 \ \Omega = 50 \ \Omega$.
Using Ohm's law,the current $I$ flowing through the circuit is $I = \frac{V}{R_{eq}} = \frac{5 \ V}{50 \ \Omega} = 0.1 \ A$.

(C) $p-n$ junction diode is reverse biased when the $p$-terminal (anode) is at a lower potential than the $n$-terminal (cathode).
Let us analyze each figure:
$(a)$ $p$-side is at $+6 \text{ V}$,$n$-side is at $-2 \text{ V}$. Since $V_p > V_n$,it is forward biased.
$(b)$ $p$-side is at $-5 \text{ V}$,$n$-side is at $+3 \text{ V}$. Since $V_p < V_n$,it is reverse biased.
$(c)$ $p$-side is at $0 \text{ V}$ (ground),$n$-side is at $-10 \text{ V}$. Since $V_p > V_n$,it is forward biased.
$(d)$ $p$-side is at $+6 \text{ V}$,$n$-side is at $0 \text{ V}$ (ground). Since $V_p > V_n$,it is forward biased.
Therefore,the diode in figure $(b)$ is reverse biased.

(B) In a $p-n$ junction,the potential barrier is created by the depletion region,which opposes the flow of majority charge carriers.
When a forward bias is applied,the positive terminal of the battery is connected to the $p$-side and the negative terminal to the $n$-side.
This external electric field opposes the internal electric field of the depletion region.
As a result,the width of the depletion region decreases,which leads to a reduction in the potential barrier height.
Therefore,the potential barrier decreases.

(B) When a $p-n$ junction diode is forward biased,the positive terminal of the external battery is connected to the $p$-side and the negative terminal to the $n$-side.
This configuration pushes the holes in the $p$-region and electrons in the $n$-region towards the junction.
As a result,the majority charge carriers move towards the junction,which effectively reduces the width of the depletion layer.
Consequently,the potential barrier height decreases,allowing current to flow easily through the diode.

(A) diode is forward-biased when the potential at the $p$-side (anode) is higher than the potential at the $n$-side (cathode).
In the given circuits,the triangle represents the $p$-side and the vertical bar represents the $n$-side.
For option $A$: $V_p = 0 \ V$,$V_n = -4 \ V$. Since $0 \ V > -4 \ V$,the diode is forward-biased.
For option $B$: $V_p = -4 \ V$,$V_n = -3 \ V$. Since $-4 \ V < -3 \ V$,the diode is reverse-biased.
For option $C$: $V_p = -2 \ V$,$V_n = +2 \ V$. Since $-2 \ V < +2 \ V$,the diode is reverse-biased.
For option $D$: $V_p = 3 \ V$,$V_n = 5 \ V$. Since $3 \ V < 5 \ V$,the diode is reverse-biased.
Therefore,the correct figure is $A$.

(D) When a $p-n$ junction is formed,electrons from the $n$-region diffuse into the $p$-region,and holes from the $p$-region diffuse into the $n$-region.
This diffusion leaves behind immobile ionized donor atoms (positive charges) in the $n$-side and immobile ionized acceptor atoms (negative charges) in the $p$-side near the junction.
This region,which is devoid of mobile charge carriers,is called the depletion region.
The accumulation of these fixed positive and negative charges creates an electric field that opposes further diffusion,resulting in a potential barrier.

(A) In the given circuit,the $p$-terminal of the diode is connected to the negative terminal of the $4 \ V$ battery,and the $n$-terminal is connected towards the $1 \ V$ battery.
Specifically,the potential at the $p$-side is $-4 \ V$ and the potential at the $n$-side is $-1 \ V$.
Since the potential at the $p$-side is lower than the potential at the $n$-side $(-4 \ V < -1 \ V)$,the diode is in a reverse-biased condition.
An ideal diode in reverse bias acts as an open circuit,meaning it offers infinite resistance.
Therefore,no current will flow through the circuit.
The magnitude of the current is $0 \ A$.

(B) $p-n$ junction diode allows current to flow easily when it is forward biased.
When the terminals are reversed,the diode becomes reverse biased.
In a reverse biased state,the depletion region widens,offering very high resistance to the flow of charge carriers.
Consequently,the current drops to almost zero.
Therefore,the device $X$ is a $p-n$ junction diode.

(C) The correct option is $C$.
When a $p-n$ junction is formed,there is a high concentration of electrons in the $n$-region and a high concentration of holes in the $p$-region.
Due to this concentration gradient,electrons diffuse from the $n$-side to the $p$-side,and holes diffuse from the $p$-side to the $n$-side.
As these charge carriers cross the junction,they recombine near the junction interface.
This recombination leaves behind immobile ionized impurity atoms (positive ions on the $n$-side and negative ions on the $p$-side),which create an electric field that opposes further diffusion.
This region,depleted of mobile charge carriers,is known as the depletion layer.

(B) When the diode is forward biased,it acts as a short circuit (zero resistance). The two resistors $40 \ \Omega$ and $60 \ \Omega$ are in parallel.
$\therefore$ The effective resistance $R_1$ is given by:
$R_1 = \frac{40 \times 60}{40 + 60} = \frac{2400}{100} = 24 \ \Omega$
When the diode is reverse biased,it acts as an open circuit (infinite resistance). No current flows through the upper branch.
$\therefore$ The effective resistance $R_2$ is simply the resistance of the lower branch,which is $60 \ \Omega$.
$\therefore \frac{R_1}{R_2} = \frac{24}{60} = \frac{2}{5}$

(C) The reading of the ammeter represents the current flowing through the circuit.
According to Ohm's law for a circuit containing a diode,the current $I$ is given by:
$I = \frac{V - V_{\text{diode}}}{R}$
Here,the supply voltage $V = 5 \text{ V}$,the resistance $R = 200 \text{ } \Omega$,and for a silicon diode,the forward voltage drop $V_{\text{diode}} = 0.7 \text{ V}$.
Substituting these values into the formula:
$I = \frac{5 \text{ V} - 0.7 \text{ V}}{200 \text{ } \Omega}$
$I = \frac{4.3 \text{ V}}{200 \text{ } \Omega} = 0.0215 \text{ A}$
Converting the current to milliamperes $(1 \text{ A} = 1000 \text{ mA})$:
$I = 0.0215 \times 1000 \text{ mA} = 21.5 \text{ mA}$
Thus,the reading in the ammeter is $21.5 \text{ mA}$.

(B) In the given circuit,the diode $D_2$ is connected in reverse bias because its $n$-terminal is connected to the positive terminal of the battery.
Therefore,no current flows through the branch containing the $40 \Omega$ resistor.
The circuit simplifies to a single loop with the $40 \Omega$ resistor and the diode $D_1$ in forward bias.
Using Ohm's law,the current $I$ supplied by the battery is $I = \frac{V}{R} = \frac{10 \text{ V}}{40 \Omega} = 0.25 \text{ A}$.

(B) For the given circuit, the diode is forward-biased because the potential at the anode $(3 \text{ V})$ is higher than the potential at the cathode $(1 \text{ V})$.
Since the diode is ideal, its resistance in the forward-biased state is $0 \Omega$.
The potential difference across the resistor is $\Delta V = 3 \text{ V} - 1 \text{ V} = 2 \text{ V}$.
Using Ohm's law, the current $i$ flowing through the circuit is:
$i = \frac{\Delta V}{R} = \frac{2 \text{ V}}{100 \Omega} = 0.02 \text{ A}$.
Converting this to milliamperes:
$i = 0.02 \times 1000 \text{ mA} = 20 \text{ mA}$.

(B) In the given circuit,the $P$-terminal of the diode is connected to $-5 \text{ V}$ and the $N$-terminal is connected to $-2 \text{ V}$ through a resistor.
For a diode to be forward biased,the potential at the $P$-terminal must be higher than the potential at the $N$-terminal.
Here,the potential at the $P$-terminal $(V_P = -5 \text{ V})$ is less than the potential at the $N$-terminal $(V_N = -2 \text{ V})$.
Since $V_P < V_N$,the diode is in reverse bias.
In an ideal diode,no current flows in reverse bias. Therefore,the current in the circuit is zero.

(A) When a $p-n$ junction is formed,electrons diffuse from the $n$-region to the $p$-region,and holes diffuse from the $p$-region to the $n$-region.
This diffusion creates a depletion region with a built-in electric field directed from the $n$-side to the $p$-side.
Due to this electric field,the $p$-side acquires a higher potential relative to the $n$-side.
Therefore,the $n$-type side has a lower potential than the $p$-type side.

(B) When a $p-n$ junction diode is reverse biased,the positive terminal of the battery is connected to the $n$-region and the negative terminal to the $p$-region.
This configuration pulls the majority charge carriers (holes in $p$-region and electrons in $n$-region) away from the junction.
As a result,the depletion layer is depleted of charge carriers even further,causing its width to increase.
Therefore,the correct option is $B$.

(D) When the diode $D$ is forward biased,it acts as a short circuit (zero resistance). The circuit consists of two resistors of $40 \ \Omega$ and $400 \ \Omega$ in parallel.
Therefore,the effective resistance $R_1$ is given by:
$R_1 = \frac{40 \times 400}{40 + 400} = \frac{16000}{440} = \frac{1600}{44} = \frac{400}{11} \ \Omega$
When the diode $D$ is reverse biased,it acts as an open circuit (infinite resistance). No current flows through the upper branch.
Therefore,the effective resistance $R_2$ is simply the resistance of the lower branch:
$R_2 = 400 \ \Omega$
Now,calculating the ratio $R_1: R_2$:
$\frac{R_1}{R_2} = \frac{400/11}{400} = \frac{1}{11}$
Thus,the ratio $R_1: R_2$ is $1: 11$.

(D) In a $p-n$ junction diode,when it is in forward bias,the positive terminal of the battery is connected to the $p$-region and the negative terminal to the $n$-region.
This external electric field opposes the internal barrier electric field.
As a result,the majority charge carriers are pushed towards the junction,which reduces the width of the depletion layer.
Therefore,the barrier potential decreases,allowing current to flow easily through the diode.

(D) When a $p-n$ junction is forward biased,the positive terminal of the external battery is connected to the $p$-type region and the negative terminal to the $n$-type region.
This external electric field opposes the internal electric field of the depletion region.
As a result,the effective potential barrier $(V_B)$ decreases.
Due to the reduction in the potential barrier,the majority charge carriers are pushed towards the junction,which causes the width $(X)$ of the depletion region to decrease.
Therefore,both the potential barrier and the width of the depletion region decrease.

(B) junction diode is forward biased when the potential at the $P$-terminal $(V_P)$ is higher than the potential at the $N$-terminal $(V_N)$,i.e.,$V_P > V_N$.
Let us analyze each option:
$A$: $V_P = +5 \ V$,$V_N = +10 \ V$. Here $V_P < V_N$,so it is reverse biased.
$B$: $V_P = -1.0 \ V$,$V_N = -1.5 \ V$. Here $V_P > V_N$ (since $-1.0 > -1.5$),so it is forward biased.
$C$: $V_P = 0 \ V$,$V_N = +1 \ V$. Here $V_P < V_N$,so it is reverse biased.
$D$: $V_P = -2 \ V$,$V_N = 0 \ V$. Here $V_P < V_N$,so it is reverse biased.
Therefore,the correct option is $B$.

(C) In a $P-N$ junction diode,when the reverse bias voltage is increased to a large value,it eventually reaches the breakdown voltage.
At this point,the covalent bonds in the crystal lattice break,leading to a rapid generation of charge carriers.
Consequently,the current through the diode suddenly increases.

(A) In a $p-n$ junction diode,forward bias occurs when the $p$-type semiconductor is connected to the positive terminal of the external battery and the $n$-type semiconductor is connected to the negative terminal.
This configuration reduces the width of the depletion layer and lowers the potential barrier,allowing current to flow easily through the diode.

(C) The forward bias characteristic of a $PN$ junction diode is non-linear and exponential in nature.
As the forward voltage $V$ increases,the current $I$ increases slowly at first and then rises rapidly after the knee voltage.
Among the given curves,curve $D$ shows an exponential increase in current with voltage,which is characteristic of the forward bias region of a diode.
Curve $A$ represents a linear decrease,$B$ represents a slow linear increase,and $C$ represents a linear increase.
Therefore,graph $D$ correctly represents the forward bias characteristic.

(C) In the given circuit, the $10 \,V$ source is connected to the circuit.
Looking at the orientation of the diodes:
- Diode $D_1$ has its cathode connected to the positive potential side (through $R_1$), making it reverse-biased. Thus, $D_1$ acts as an open circuit (no current flows through $R_2$).
- Diode $D_2$ has its anode connected to the positive potential side, making it forward-biased. Thus, $D_2$ acts as a closed switch (short circuit).
Therefore, the circuit simplifies to a series combination of the $10 \,V$ battery, resistor $R_1 = 2 \,\Omega$, and resistor $R_3 = 2 \,\Omega$.
The total resistance in the circuit is $R_{eq} = R_1 + R_3 = 2 \,\Omega + 2 \,\Omega = 4 \,\Omega$.
The current flowing through $R_1$ is given by Ohm's law: $I = \frac{V}{R_{eq}} = \frac{10 \,V}{4 \,\Omega} = 2.5 \,A$.

(D) When a $p-n$ junction diode is reverse biased,the positive terminal of the battery is connected to the $n$-region and the negative terminal to the $p$-region.
This configuration pulls the majority charge carriers (electrons in the $n$-region and holes in the $p$-region) away from the junction.
As a result,the width of the depletion layer increases.
Because the depletion layer width increases,the potential barrier height also increases,making it more difficult for majority charge carriers to cross the junction.

(D) The diode is connected such that the p-side is at $+5 \ V$ and the n-side is at $+3 \ V$.
Since the potential at the p-side is higher than the potential at the n-side,the diode is forward-biased.
Assuming an ideal diode,the potential difference across the resistor is $V = 5 \ V - 3 \ V = 2 \ V$.
The resistance $R$ is $200 \ \Omega$.
Using Ohm's law,the current $I$ is given by $I = \frac{V}{R} = \frac{2 \ V}{200 \ \Omega} = 0.01 \ A = 10^{-2} \ A$.

(C) The diode is connected in forward bias because the $p$-side is at a higher potential $(+3 \text{ V})$ than the $n$-side $(-5 \text{ V})$.
For an ideal junction diode in forward bias, the resistance of the diode is zero.
The potential difference across the resistor is $V = V_P - V_Q = 3 \text{ V} - (-5 \text{ V}) = 8 \text{ V}$.
The resistance $R = 2 \text{ k}\Omega = 2000 \text{ }\Omega$.
Using Ohm's law, the current $I$ is given by $I = \frac{V}{R} = \frac{8 \text{ V}}{2000 \text{ }\Omega} = 4 \times 10^{-3} \text{ A}$.

(B) In circuit $(a)$,the two diodes are connected in series such that they oppose each other. Specifically,one diode is forward-biased while the other is reverse-biased. Since an ideal reverse-biased diode acts as an open circuit (infinite resistance),no current flows through the circuit. Therefore,the ammeter will not show any deflection.
In circuit $(b)$,both diodes are forward-biased,allowing current to flow.
In circuits $(c)$ and $(d)$,the diodes are in parallel,and at least one path is forward-biased,allowing current to flow.

(C) In circuit $X$,both diodes $D_1$ and $D_2$ are forward biased,so both conduct current.
The two resistors of $4 \ \Omega$ each are connected in parallel.
The equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$,so $R_{eq} = 2 \ \Omega$.
The total current $I_X = \frac{V}{R_{eq}} = \frac{8 \ V}{2 \ \Omega} = 4 \ A$.
In circuit $Y$,diode $D_1$ is forward biased,but diode $D_2$ is reverse biased.
Therefore,only diode $D_1$ conducts current.
The effective resistance in the circuit is $4 \ \Omega$.
The total current $I_Y = \frac{V}{R} = \frac{8 \ V}{4 \ \Omega} = 2 \ A$.
Thus,the currents are $4 \ A$ and $2 \ A$ respectively.

(B) In an unbiased $p-n$ junction,the diffusion of electrons from the $n$-region to the $p$-region and holes from the $p$-region to the $n$-region occurs near the junction.
When these charge carriers cross the junction,they recombine and neutralize each other.
As a result,the region near the junction becomes devoid of mobile charge carriers (free electrons and holes).
This region is called the depletion layer or depletion region.
Therefore,the layer is depleted of mobile charge carriers,which are electrons and holes.

(D) In the given circuit,the two diodes $D_{1}$ and $D_{2}$ are connected in series with a $DC$ voltage source.
For the circuit to conduct,both diodes must be forward-biased.
However,looking at the orientation of the diodes,$D_{1}$ is forward-biased while $D_{2}$ is reverse-biased.
In a series circuit,the current $I$ is the same through all components.
Since $D_{2}$ is reverse-biased,it offers a very high resistance,effectively acting as an open switch.
Consequently,the entire potential difference of the battery appears across the reverse-biased diode $D_{2}$,while the potential difference across the forward-biased diode $D_{1}$ is negligible (approximately $0.7 \ V$ for Silicon).
If the potential differences across $D_{1}$ and $D_{2}$ are stated to be equal,it implies that the circuit configuration or the assumption of identical characteristics must be re-evaluated,or the diodes are operating in a specific breakdown region. Given the standard interpretation,the potential difference across a reverse-biased diode is much larger than that across a forward-biased one.

(D) In a $PN$ junction diode,the diode is forward biased if the potential at the anode (p-side) is higher than the potential at the cathode (n-side). It is reverse biased if the potential at the anode is lower than the potential at the cathode.
In diagram $(A)$,the anode is at $-4 \ V$ and the cathode is at $-3 \ V$. Since $-4 \ V < -3 \ V$,the anode is at a lower potential than the cathode,so diagram $(A)$ is reverse biased.
In diagram $(B)$,assuming the standard configuration where the anode is at $-2 \ V$ and the cathode is at $-4 \ V$ (based on typical textbook problems of this type),the anode is at a higher potential than the cathode,so diagram $(B)$ is forward biased.
Therefore,diagram $(A)$ is reverse biased and diagram $(B)$ is forward biased.

(B) In the given circuit,the $p$-side of the diode is connected to $+2 V$ and the $n$-side is connected to $-2 V$ through a $400 \Omega$ resistor.
Since the potential of the $p$-side is higher than the potential of the $n$-side,the diode is forward biased.
For an ideal diode,the forward resistance is $0 \Omega$.
Therefore,the total resistance in the circuit is $R = 400 \Omega$.
The potential difference across the circuit is $V = 2 V - (-2 V) = 4 V$.
Using Ohm's law,the current $I$ is given by $I = V / R$.
$I = 4 V / 400 \Omega = 1 / 100 A = 0.01 A$.
Converting to milliamperes,$I = 0.01 \times 1000 mA = 10 mA$.

(D) $(i)$ When a $p-n$ junction is forward biased,the external electric field opposes the internal electric field of the depletion region. This reduces the potential barrier and consequently decreases the width of the depletion layer.
$(ii)$ When a $p-n$ junction is reverse biased,the external electric field supports the internal electric field of the depletion region. This increases the potential barrier and consequently increases the width of the depletion layer.

(D) When a $p-n$ junction diode is reverse biased,the positive terminal of the battery is connected to the $n$-region and the negative terminal to the $p$-region.
This configuration pulls the majority charge carriers away from the junction.
As a result,the width of the depletion layer increases.
Consequently,the potential barrier also increases,which opposes the flow of current,making electrical conduction negligible.

(A) In a $p-n$ junction diode,when reverse bias is applied,the positive terminal of the battery is connected to the $n$-region and the negative terminal to the $p$-region.
This causes the majority charge carriers (electrons in $n$-region and holes in $p$-region) to move away from the junction.
As a result,the depletion layer widens,and the barrier potential increases.
Due to the increased width of the depletion layer,the flow of majority charge carriers is blocked,which means the diode offers high resistance to the current.

(B) In a $p-n$ junction diode,when it is reverse biased,the applied voltage supports the barrier potential,which causes the width of the depletion region to increase.
In forward biasing,the width of the depletion region decreases because the forward voltage opposes the potential barrier.
Additionally,the width of the depletion region decreases with heavy doping because the increased concentration of charge carriers leads to a narrower space charge region.

(C) The diode is connected in forward bias because the $p$-side is at $+3 \text{ V}$ and the $n$-side is at $+1 \text{ V}$.
Since the diode is ideal, its resistance in forward bias is zero.
The potential difference across the resistor $R = 100 \ \Omega$ is $V = 3 \text{ V} - 1 \text{ V} = 2 \text{ V}$.
Using Ohm's law, the current $I$ is given by $I = \frac{V}{R} = \frac{2 \text{ V}}{100 \ \Omega} = 0.02 \text{ A}$.
Converting to milliamperes, $I = 0.02 \times 1000 \text{ mA} = 20 \text{ mA}$.

(C) When a $p-n$ junction is in forward bias,the positive terminal of the battery is connected to the $p$-side and the negative terminal to the $n$-side.
This reduces the width of the depletion layer and lowers the potential barrier.
As a result,the diode offers very low resistance and allows current to flow through it.
Therefore,in forward bias,the $p-n$ junction diode acts as a closed switch.

(B) When a $pn$-junction diode is reverse-biased,the positive terminal of the external battery is connected to the $n$-region and the negative terminal to the $p$-region.
This configuration pulls the majority charge carriers (holes in the $p$-region and electrons in the $n$-region) away from the junction.
As a result,the concentration of immobile ions near the junction increases,which causes the depletion layer width to increase.

(C) When a diode is forward biased,the positive terminal of the battery is connected to the $p$-type region and the negative terminal to the $n$-type region.
This external electric field opposes the internal electric field of the depletion region.
As the forward voltage increases,the potential barrier is lowered,and the majority charge carriers are pushed towards the junction.
This results in a reduction of the width of the depletion region.
Therefore,the correct option is $C$.

(A) The space charge region,also known as the depletion layer,is formed at the interface of $p$-type and $n$-type semiconductors.
In a typical $pn$ junction diode,the width of this depletion layer is very small.
It is generally on the order of $10^{-6} \ m$,which is equivalent to $1 \ \mu m$.
Specifically,the width of the depletion region is typically in the range of $0.1 \ \mu m$ to $1 \ \mu m$.
Among the given options,$0.5 \ \mu m$ is the most appropriate value representing the typical width of the space charge region.

(C) The potential barrier $V$ is related to the electric field $E$ and the width of the depletion region $d$ by the formula $V = E \cdot d$.
Given:
Electric field $E = 1 \times 10^{6} \text{ V/m}$
Width $d = 5000 \text{ Å} = 5000 \times 10^{-10} \text{ m} = 5 \times 10^{-7} \text{ m}$
Substituting the values:
$V = (1 \times 10^{6} \text{ V/m}) \times (5 \times 10^{-7} \text{ m})$
$V = 5 \times 10^{-1} \text{ V}$
$V = 0.5 \text{ V}$
Therefore,the correct option is $C$.

(A) For an ideal diode,the forward bias resistance is $0 \ \Omega$ and the reverse bias resistance is $\infty \ \Omega$.
$(i)$ Case $V_A > V_B$:
In this case,both diodes $D_1$ and $D_2$ are forward biased.
Therefore,the resistance of each branch is $50 \ \Omega + 0 \ \Omega = 50 \ \Omega$.
Since the two branches are in parallel,the equivalent resistance $R_{AB}$ is given by:
$\frac{1}{R_{AB}} = \frac{1}{50} + \frac{1}{50} = \frac{2}{50} = \frac{1}{25}$
$R_{AB} = 25 \ \Omega$.
(ii) Case $V_B > V_A$:
In this case,both diodes $D_1$ and $D_2$ are reverse biased.
Therefore,the resistance of each branch is $50 \ \Omega + \infty \ \Omega = \infty \ \Omega$.
Since the two branches are in parallel,the equivalent resistance $R_{AB}$ is given by:
$\frac{1}{R_{AB}} = \frac{1}{\infty} + \frac{1}{\infty} = 0 + 0 = 0$
$R_{AB} = \infty \ \Omega$.
Thus,the equivalent resistances are $25 \ \Omega$ and $\infty \ \Omega$ respectively.

(A) In the given circuit,the diode is connected in reverse bias because the positive terminal of the battery is connected to the $n$-side (cathode) and the negative terminal is connected to the $p$-side (anode) of the diode.
Since the diode has infinite reverse bias resistance,the current $I_1$ flowing through the diode branch is $I_1 = 0.0 \ A$.
The resistor of $50 \ \Omega$ is connected in parallel to the battery of $10 \ V$. Therefore,the current $I_2$ flowing through the resistor is given by Ohm's law:
$I_2 = \frac{V}{R} = \frac{10 \ V}{50 \ \Omega} = 0.2 \ A$.
Thus,$I_1 = 0.0 \ A$ and $I_2 = 0.2 \ A$.

(B) When a $p-n$ junction is forward biased,the positive terminal of the external battery is connected to the $p$-side and the negative terminal to the $n$-side.
This external electric field opposes the internal electric field of the depletion region.
As a result,the width of the depletion layer decreases,and the potential barrier height is lowered.
Therefore,the correct option is $B$.

(B) Given current $I = 0.2 \, mA = 0.2 \times 10^{-3} \, A$.
Voltage drop across the diode $V_d = 0.2 \, V$.
Resistors $R_1 = R_2 = 5 \, k\Omega = 5 \times 10^3 \, \Omega$.
The circuit consists of a resistor $R_1$, a diode, and a resistor $R_2$ in series.
The voltage drop across resistor $R_1$ is $V_{R1} = I \times R_1 = (0.2 \times 10^{-3} \, A) \times (5 \times 10^3 \, \Omega) = 1 \, V$.
The voltage drop across resistor $R_2$ is $V_{R2} = I \times R_2 = (0.2 \times 10^{-3} \, A) \times (5 \times 10^3 \, \Omega) = 1 \, V$.
The total voltage difference between $A$ and $B$ is the sum of the voltage drops across the components: $V_{AB} = V_{R1} + V_d + V_{R2} = 1 \, V + 0.2 \, V + 1 \, V = 2.2 \, V$.

(C) In the given circuit,the positive terminal of the $3 \text{ V}$ battery is connected to the cathode of diode $D_1$ and the anode of diode $D_2$.
$1$. Diode $D_1$: The positive terminal of the battery is connected to the n-side (cathode) of $D_1$. Thus,$D_1$ is reverse-biased and acts as an open circuit (no current flows through this branch).
$2$. Diode $D_2$: The positive terminal of the battery is connected to the p-side (anode) of $D_2$. Thus,$D_2$ is forward-biased and acts as a closed switch (ideal diode has zero resistance).
$3$. Equivalent Resistance: Since $D_1$ is an open circuit,the current flows only through the branch containing $D_2$ and the $30 \Omega$ resistor,which is in series with the $70 \Omega$ resistor.
$R_{eq} = 30 \Omega + 70 \Omega = 100 \Omega$
$4$. Current Calculation: Using Ohm's law,$I = \frac{V}{R_{eq}}$
$I = \frac{3 \text{ V}}{100 \Omega} = 0.03 \text{ A}$

(B) Given:
Battery emf,$V = 5.7 \ V$
Barrier potential of the diode,$V_B = 0.7 \ V$
Series resistance,$R_S = 5 \ k\Omega = 5 \times 10^3 \ \Omega$
In a forward-biased $p-n$ junction diode,the effective voltage across the resistor is the difference between the battery emf and the barrier potential.
Effective voltage,$V_{eff} = V - V_B = 5.7 \ V - 0.7 \ V = 5.0 \ V$
Using Ohm's law,the current $I$ in the circuit is:
$I = \frac{V_{eff}}{R_S} = \frac{5.0 \ V}{5 \times 10^3 \ \Omega} = 1 \times 10^{-3} \ A$
$I = 1 \ mA$

(A) In a $p-n$ junction diode,the current in the semiconductor is due to both electrons and holes.
However,in the external connecting wires,the current is purely electronic.
When the diode is forward biased,the potential difference causes free electrons to flow through the external circuit (connecting wires) from the negative terminal to the positive terminal.
Therefore,the charge carriers flowing in the connecting wire are free electrons.