Junction Transistor Questions in English

(B) Given: Current gain $\beta = 100$. Change in collector current $\Delta I_C = 1 \ mA$.
We know that the relation between collector current and base current is $\Delta I_C = \beta \Delta I_B$.
Therefore,the change in base current is $\Delta I_B = \frac{\Delta I_C}{\beta} = \frac{1 \ mA}{100} = 0.01 \ mA$.
The relationship between emitter current,collector current,and base current is $\Delta I_E = \Delta I_C + \Delta I_B$.
Substituting the values: $\Delta I_E = 1 \ mA + 0.01 \ mA = 1.01 \ mA$.
Thus,the emitter current changes by $1.01 \ mA$.

(B) In a $P-N-P$ transistor,the structure consists of a $P$-type emitter,an $N$-type base,and a $P$-type collector.
Since the emitter is $P$-type,its majority charge carriers are holes.
Since the base is $N$-type,its majority charge carriers are electrons.
Therefore,the emitter has holes as majority carriers and the base has electrons as majority carriers.

(A) In a transistor symbol,the arrow is always placed on the emitter terminal.
By convention,the direction of the arrow represents the direction of the conventional current flow in the emitter.
For a $PNP$ transistor,the arrow points inward (towards the base),indicating conventional current flow from emitter to base.
For an $NPN$ transistor,the arrow points outward (away from the base),indicating conventional current flow from base to emitter.
Therefore,the arrow signifies the direction of the conventional current in the emitter.

(B) For an $N-P-N$ transistor to conduct (operate in the active region),the base-emitter junction must be forward-biased and the base-collector junction must be reverse-biased.
$1$. Forward bias for the base-emitter junction: Since the base is $P$-type and the emitter is $N$-type,the base must be at a higher potential than the emitter (or the emitter must be negative with respect to the base).
$2$. Reverse bias for the base-collector junction: Since the base is $P$-type and the collector is $N$-type,the collector must be at a higher potential than the base (or the collector must be positive with respect to the base).
Therefore,the collector is positive with respect to the base,and the emitter is negative with respect to the base.

(B) In an $N-P-N$ transistor,current flows when the collector-base junction is reverse-biased and the emitter-base junction is forward-biased.
For the emitter-base junction to be forward-biased,the emitter must be negative with respect to the base.
For the collector-base junction to be reverse-biased,the collector must be positive with respect to the base.
Therefore,the current flows when the collector is positive and the emitter is negative with respect to the base.

(C) Given: Base current $I_b = 100 \mu A = 0.1 \ mA$ and Collector current $I_c = 3 \ mA$.
$1$. Current gain $\beta$ is given by: $\beta = \frac{I_c}{I_b} = \frac{3 \ mA}{0.1 \ mA} = 30$.
$2$. Current gain $\alpha$ is given by: $\alpha = \frac{\beta}{1 + \beta} = \frac{30}{1 + 30} = \frac{30}{31} \approx 0.97$.
$3$. Emitter current $I_e$ is given by: $I_e = I_c + I_b = 3 \ mA + 0.1 \ mA = 3.1 \ mA$.
Thus,the values are $30, 0.97, 3.1 \ mA$.

(C) For a transistor to operate as an amplifier,it must be in the active region.
In the active region,the base-emitter junction must be forward-biased to allow current flow from the emitter to the base.
Therefore,statement $(1)$ is correct.
To amplify the input signal,the signal is superimposed on the $DC$ biasing voltage applied to the base-emitter junction.
Therefore,statement $(3)$ is correct.
Statements $(2)$ and $(4)$ are incorrect because the base-collector junction must be reverse-biased in the active region,and the input signal is not applied to the base-collector junction.

(A) For a transistor to function as an amplifier,it must operate in the active region.
In the active region,the emitter-base junction is forward-biased,which allows charge carriers to flow from the emitter into the base.
The base-collector junction is reverse-biased,which allows the collector to collect the majority of the charge carriers injected from the emitter.
Therefore,the correct condition is that the emitter-base junction is forward-biased and the base-collector junction is reverse-biased.

(C) In an $N-P-N$ transistor,the majority charge carriers are electrons. When used as an amplifier,the emitter is forward-biased,causing electrons to be injected from the $N$-type emitter into the $P$-type base. Due to the thin base region and the reverse-biased collector junction,these electrons then diffuse through the base and are collected by the collector.

(A) The emitter section of a transistor is heavily doped to supply a large number of majority charge carriers to the base.
Therefore,the doping concentration is highest in the emitter.

(A) An oscillator is an electronic circuit that produces a periodic,oscillating electronic signal,often a sine wave or a square wave.
In the context of a transistor oscillator,it uses a $DC$ power source to generate an $AC$ signal.
This process involves feedback where a portion of the output signal is fed back into the input to maintain oscillations.
Therefore,a transistor oscillator converts $DC$ electrical energy into $AC$ electrical energy.

(B) transistor consists of three regions: the Emitter,the Base,and the Collector.
$1$. The Emitter is moderately doped and is of medium size.
$2$. The Collector is larger in size compared to the Emitter and Base to dissipate heat generated during operation.
$3$. The Base is the central region and is made very thin and lightly doped to allow the majority of charge carriers from the Emitter to pass into the Collector.
Therefore,the Base is the thinnest part of the transistor.

(D) In a properly biased transistor (e.g.,$NPN$ or $PNP$ in active mode),the emitter-base junction is forward-biased,which reduces the width of its depletion layer.
Conversely,the collector-base junction is reverse-biased,which increases the width of its depletion layer.
Therefore,the emitter-base junction is narrow,and the collector-base junction is wide.

(B) We know that the current gain in common base configuration is $\alpha = \frac{\Delta i_C}{\Delta i_E}$.
We know that the current gain in common emitter configuration is $\beta = \frac{\Delta i_C}{\Delta i_B}$.
Since the emitter current is the sum of collector and base currents,we have $\Delta i_E = \Delta i_C + \Delta i_B$,which implies $\Delta i_B = \Delta i_E - \Delta i_C$.
Substituting this into the expression for $\beta$:
$\beta = \frac{\Delta i_C}{\Delta i_E - \Delta i_C}$.
Dividing the numerator and denominator by $\Delta i_E$:
$\beta = \frac{\frac{\Delta i_C}{\Delta i_E}}{1 - \frac{\Delta i_C}{\Delta i_E}}$.
Since $\alpha = \frac{\Delta i_C}{\Delta i_E}$,we get:
$\beta = \frac{\alpha}{1 - \alpha}$.

(C) In a transistor,the emitter is heavily doped to provide a large number of charge carriers,the collector is moderately doped and larger in size to dissipate heat,and the base is very thin and lightly doped to allow most charge carriers to pass through to the collector.
Statement $(1)$ is false because the regions have different sizes and doping levels.
Statement $(2)$ is true because the base is designed to be thin and lightly doped to minimize recombination.
Statement $(3)$ is true because,for active operation (amplification),the emitter-base junction must be forward-biased and the base-collector junction must be reverse-biased.
Statement $(4)$ is false because this configuration corresponds to the saturation region,not the standard active mode of operation.
Therefore,statements $(2)$ and $(3)$ are correct.

(C) The relationship between current gain $\alpha$ (common base) and $\beta$ (common emitter) is given by the formula: $\beta = \frac{\alpha}{1 - \alpha}$.
Given $\alpha = 0.96$.
Substituting the value into the formula:
$\beta = \frac{0.96}{1 - 0.96}$
$\beta = \frac{0.96}{0.04}$
$\beta = \frac{96}{4} = 24$.
Therefore,the current gain $\beta$ is $24$.

(D) The current gain $A_i$ for a common-emitter transistor amplifier is given by the formula:
$A_i = - \left( \frac{h_{fe}}{1 + h_{oe} R_L} \right)$
Given values are $h_{fe} = 50$,$h_{oe} = 25 \ \mu A/V = 25 \times 10^{-6} \ S$,and $R_L = 1 \ k\Omega = 10^3 \ \Omega$.
Substituting these values into the formula:
$A_i = - \left( \frac{50}{1 + (25 \times 10^{-6} \times 10^3)} \right)$
$A_i = - \left( \frac{50}{1 + 0.025} \right)$
$A_i = - \left( \frac{50}{1.025} \right)$
$A_i \approx -48.78$

(A) The current gain in a common-base $(CB)$ configuration is defined as $\alpha = \frac{\Delta I_C}{\Delta I_E}$.
The current gain in a common-emitter $(CE)$ configuration is defined as $\beta = \frac{\Delta I_C}{\Delta I_B}$.
Taking the ratio of $\alpha$ to $\beta$:
$\frac{\alpha}{\beta} = \frac{(\Delta I_C / \Delta I_E)}{(\Delta I_C / \Delta I_B)}$
$\frac{\alpha}{\beta} = \frac{\Delta I_C}{\Delta I_E} \times \frac{\Delta I_B}{\Delta I_C}$
$\frac{\alpha}{\beta} = \frac{\Delta I_B}{\Delta I_E}$

(B) The relationship between emitter current $(I_E)$,collector current $(I_C)$,and base current $(I_B)$ is given by $I_E = I_C + I_B$.
Therefore,$I_C = I_E - I_B$.
Given $I_E = 25 \, mA$ and $I_B = 1 \, mA$,we have $I_C = 25 \, mA - 1 \, mA = 24 \, mA$.
The ratio of collector current to emitter current is $\alpha = \frac{I_C}{I_E}$.
Substituting the values,$\alpha = \frac{24 \, mA}{25 \, mA} = \frac{24}{25}$.

(A) In a common base configuration,the current gain $\alpha$ is defined as the ratio of collector current to emitter current: $\alpha = \frac{I_C}{I_E} = 0.98$.
The current gain for a common emitter circuit,denoted by $\beta$,is related to $\alpha$ by the formula: $\beta = \frac{\alpha}{1 - \alpha}$.
Substituting the given value of $\alpha$ into the formula:
$\beta = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49$.
Therefore,the current gain for the common emitter circuit is $49$.

(C) Given: Base current $I_b = 100 \mu A$,Collector current $I_c = 3 \text{ mA}$.
Change in base current $\Delta I_b = 20 \mu A = 0.02 \text{ mA}$.
Change in collector current $\Delta I_c = 0.5 \text{ mA}$.
The $AC$ current gain $\beta_{ac}$ is defined as the ratio of the change in collector current to the change in base current.
$\beta_{ac} = \frac{\Delta I_c}{\Delta I_b}$
Substituting the values:
$\beta_{ac} = \frac{0.5 \text{ mA}}{0.02 \text{ mA}} = \frac{50}{2} = 25$.
Therefore,the value of $\beta_{ac}$ is $25$.

(C) The voltage gain $A_V$ is given by the formula: $A_V = \beta \times \left( \frac{R_o}{R_i} \right)$.
Here,$\beta$ is the current gain,which is defined as $\beta = \frac{I_C}{I_b}$.
Given values: $I_C = 1 \text{ mA} = 1 \times 10^{-3} \text{ A}$,$I_b = 20 \text{ }\mu\text{A} = 20 \times 10^{-6} \text{ A}$,$R_o = 4 \text{ k}\Omega$,and $R_i = 1 \text{ k}\Omega$.
First,calculate $\beta$: $\beta = \frac{1 \times 10^{-3}}{20 \times 10^{-6}} = \frac{1000}{20} = 50$.
Now,calculate the voltage gain: $A_V = 50 \times \left( \frac{4 \text{ k}\Omega}{1 \text{ k}\Omega} \right) = 50 \times 4 = 200$.

(D) Comparing the input voltage $v_i = 0.2 \sin(1000t)$ with the standard form $v = V_m \sin(\omega t)$,we get the peak input voltage $V_m = 0.2 \text{ V}$.
Given the voltage gain $A_v = 10$,the peak output voltage is $V_{out,m} = A_v \times V_m = 10 \times 0.2 = 2 \text{ V}$.
In a Common Emitter $(CE)$ amplifier,there is a phase shift of $\pi$ radians between the input and output signals.
Therefore,the equation for the output voltage is $v_o = 2 \sin(1000t + \pi) \text{ V}$.

(C) Given that the collector current $I_C = 20 \, mA$.
Since $90\%$ of the emitted electrons reach the collector,the collector current is $90\%$ of the emitter current $(I_E)$.
So,$I_C = 0.90 \times I_E$.
$I_E = \frac{I_C}{0.90} = \frac{20}{0.90} \approx 22.22 \, mA$.
The base current is given by $I_B = I_E - I_C$.
$I_B = 22.22 \, mA - 20 \, mA = 2.22 \, mA$.
Comparing this with the given options,the most appropriate choice is that the base current is $2 \, mA$ (approximately) or the emitter current is $22.22 \, mA$.

(D) The voltage gain $(A_v)$ of a common-emitter amplifier is given by the formula:
$A_v = \beta \times \frac{R_{out}}{R_{in}}$
Given:
Input resistance $(R_{in})$ = $3 \ \Omega$
Load resistance $(R_{out})$ = $24 \ \Omega$
Current gain $(\beta)$ = $6$
Substituting the values:
$A_v = 6 \times \frac{24}{3}$
$A_v = 6 \times 8$
$A_v = 48$

(B) The voltage gain $A_V$ is given as $50$.
Input resistance $R_i = 100 \ \Omega$.
Output resistance $R_o = 200 \ \Omega$.
The power gain $A_P$ of an amplifier is defined as the product of voltage gain and current gain, or $A_P = A_V \times A_I$.
Since $A_V = \beta \times \frac{R_o}{R_i}$, we can express power gain as $A_P = A_V \times \beta$.
However, a more direct formula using the given parameters is $A_P = \frac{A_V^2 \times R_i}{R_o}$ is incorrect; the standard relation is $A_P = \frac{V_{out} \times I_{out}}{V_{in} \times I_{in}} = A_V \times A_I$.
Given $A_V = 50$, and assuming the current gain $\beta$ is related to the voltage gain by $A_V = \beta \frac{R_o}{R_i}$, we find $\beta = A_V \times \frac{R_i}{R_o} = 50 \times \frac{100}{200} = 25$.
Thus, power gain $A_P = A_V \times \beta = 50 \times 25 = 1250$.

(D) Given: Current gain $\beta = 50$,input resistance $r_i = 1 \ k\Omega = 10^3 \ \Omega$,and peak $AC$ input voltage $v_S = 0.01 \ V$.
The base current $I_B$ is given by the relation $I_B = \frac{v_S}{r_i}$.
Substituting the values: $I_B = \frac{0.01 \ V}{10^3 \ \Omega} = 10^{-5} \ A$.
In a common-emitter transistor amplifier,the current gain $\beta$ is defined as $\beta = \frac{I_C}{I_B}$.
Therefore,the collector current $I_C = \beta \times I_B$.
Substituting the values: $I_C = 50 \times 10^{-5} \ A = 500 \times 10^{-6} \ A = 500 \ \mu A$.

(B) Given: Base current $I_b = 20 \ \mu A = 20 \times 10^{-6} \ A$ and current gain $\beta = 100$.
First,calculate the collector current $I_c$ using the formula $I_c = \beta I_b$:
$I_c = 100 \times 20 \times 10^{-6} \ A = 2000 \times 10^{-6} \ A = 2 \times 10^{-3} \ A = 2 \ mA$.
Now,calculate the emitter current $I_e$ using the relation $I_e = I_b + I_c$:
$I_e = 20 \ \mu A + 2000 \ \mu A = 2020 \ \mu A$.
Convert the emitter current to $mA$:
$I_e = 2020 \times 10^{-6} \ A = 2.02 \times 10^{-3} \ A = 2.02 \ mA$.

(A) The current gain $\beta$ is defined as the ratio of output current $(I_o)$ to input current $(I_{in})$, so $I_o = \beta \times I_{in}$.
Given that the input current $I_{in} = \frac{V_{in}}{R_{in}}$, we substitute the values:
$I_{in} = \frac{5 \ V}{1 \ k\Omega} = \frac{5}{1000} \ A = 5 \ mA$.
Now, calculating the output current:
$I_o = 50 \times 5 \ mA = 250 \ mA$.

(B) Given that the collector current $I_C = 10 \ mA$.
Since $90\%$ of the electrons emitted from the emitter reach the collector,we have the relation: $I_C = 0.90 \times I_E$.
Rearranging for $I_E$: $I_E = \frac{I_C}{0.90} = \frac{10 \ mA}{0.9} = 11.11 \ mA \approx 11 \ mA$.
Using the Kirchhoff's current law for a transistor: $I_E = I_B + I_C$.
Therefore,the base current is $I_B = I_E - I_C = 11 \ mA - 10 \ mA = 1 \ mA$.
Thus,$I_E = 11 \ mA$ and $I_B = 1 \ mA$.

(A) Given: Current gain $\alpha = 0.98$, Load resistance $R_L = 5 \text{ k}\Omega = 5000 \Omega$, Input resistance $R_{in} = 70 \Omega$.
The voltage gain $A_v$ is given by the formula: $A_v = \alpha \times \frac{R_L}{R_{in}}$.
Substituting the values: $A_v = 0.98 \times \frac{5000}{70} = 0.98 \times 71.428 \approx 70$.
The power gain $A_p$ is given by the product of current gain and voltage gain: $A_p = \alpha \times A_v$.
Substituting the values: $A_p = 0.98 \times 70 = 68.6$.
Therefore, the voltage gain is $70$ and the power gain is $68.6$.

(D) The voltage gain $A_v$ of a $CE$ amplifier is given by the formula:
$A_v = \beta \times \frac{R_C}{R_{in}}$
Given:
$\beta = 50$
$R_C = 5 \, k\Omega$
$R_{in} = 1 \, k\Omega$
$V_{in} = 0.01 \, V$
Substituting the values:
$A_v = 50 \times \frac{5 \, k\Omega}{1 \, k\Omega} = 250$
The output voltage $V_{out}$ is:
$V_{out} = A_v \times V_{in} = 250 \times 0.01 \, V = 2.5 \, V$

(A) Given that the collector current $I_c = 10 \ mA$.
Since $95\%$ of the emitted electrons reach the collector,we have $I_c = 0.95 \ I_e$.
Therefore,the emitter current $I_e = \frac{I_c}{0.95} = \frac{10 \ mA}{0.95} \approx 10.53 \ mA$.
Using the relation for transistor currents,$I_e = I_c + I_b$,we can find the base current $I_b$.
$I_b = I_e - I_c = 10.53 \ mA - 10 \ mA = 0.53 \ mA$.

(C) The voltage gain $A_v$ for a common-emitter amplifier is given by $A_v = \beta \times \frac{R_L}{R_{in}}$.
Given: $\beta = 100$, $R_L = 10 \text{ k}\Omega$, $R_{in} = 1 \text{ k}\Omega$, and $V_{in} = 1 \text{ mV}$.
The output voltage $V_{out}$ is calculated as:
$V_{out} = A_v \times V_{in} = \beta \times \frac{R_L}{R_{in}} \times V_{in}$
$V_{out} = 100 \times \frac{10 \text{ k}\Omega}{1 \text{ k}\Omega} \times 1 \text{ mV}$
$V_{out} = 100 \times 10 \times 1 \text{ mV} = 1000 \text{ mV} = 1.0 \text{ V}$.

(C) From the collector circuit,applying Kirchhoff's Voltage Law $(KVL)$:
$V_{CC} = I_C R_C + V_{CE}$
Given $V_{CC} = 10 \text{ V}$,$V_{CE} = 5 \text{ V}$,and $R_C = 1 \text{ k}\Omega = 1000 \Omega$.
$10 = I_C (1000) + 5$
$I_C = \frac{5}{1000} = 5 \times 10^{-3} \text{ A} = 5 \text{ mA}$.
Now,for the base circuit:
$V_{CC} = I_B R_b + V_{BE}$
Given $V_{BE} = 0$ and $V_{CC} = 10 \text{ V}$.
Since $\beta = \frac{I_C}{I_B}$,we have $I_B = \frac{I_C}{\beta} = \frac{5 \times 10^{-3}}{100} = 5 \times 10^{-5} \text{ A}$.
Substituting into the base circuit equation:
$10 = (5 \times 10^{-5}) R_b + 0$
$R_b = \frac{10}{5 \times 10^{-5}} = 2 \times 10^5 \Omega = 200 \times 10^3 \Omega$.

(C) Given: Current gain $\beta = 60$,Load resistance $R_L = 5 \times 10^3 \ \Omega$,Input resistance $r_i = 5 \times 10^2 \ \Omega$.
The voltage gain $A_V$ for a common emitter amplifier is given by the formula:
$A_V = \beta \times \frac{R_L}{r_i}$
Substituting the given values into the formula:
$A_V = 60 \times \frac{5 \times 10^3}{5 \times 10^2}$
$A_V = 60 \times 10 = 600$
Therefore,the voltage gain is $600$.

(B) Given: Collector current $I_C = 5.488 \ mA$ and Emitter current $I_E = 5.6 \ mA$.
First,calculate the base current $I_B$ using the relation $I_E = I_C + I_B$.
$I_B = I_E - I_C = 5.6 \ mA - 5.488 \ mA = 0.112 \ mA$.
The current gain $\beta$ is defined as the ratio of collector current to base current: $\beta = \frac{I_C}{I_B}$.
Substituting the values: $\beta = \frac{5.488}{0.112} = 49$.
Therefore,the current gain $\beta$ is $49$.

(B) The relationship between common base current gain $\alpha$ and common emitter current gain $\beta$ is given by the formula: $\beta = \frac{\alpha}{1 - \alpha}$.
Given $\alpha = 0.95$.
Substituting the value: $\beta = \frac{0.95}{1 - 0.95} = \frac{0.95}{0.05}$.
Calculating the result: $\beta = \frac{95}{5} = 19$.

(D) When the negative lead of the multimeter is connected to terminal $R$ and the positive lead is connected to either $P$ or $Q,$ the meter shows some resistance. This indicates that both junctions (base-collector and base-emitter) are forward-biased,which implies that $R$ must be the base terminal.
Since the meter shows resistance when the negative lead is connected to the base $(R)$,it means the base is of $N$-type material. Therefore,the transistor must be a $P-N-P$ transistor.

(A) Given: Current gain $\beta = 49$,Change in base current $\Delta I_B = 5.0 \ \mu A$.
The change in collector current is given by $\Delta I_C = \beta \times \Delta I_B$.
Substituting the values: $\Delta I_C = 49 \times 5.0 \ \mu A = 245 \ \mu A$.
The change in emitter current is given by $\Delta I_E = \Delta I_B + \Delta I_C$.
Substituting the values: $\Delta I_E = 5.0 \ \mu A + 245 \ \mu A = 250 \ \mu A$.
Therefore,the change in collector current is $245 \ \mu A$ and the change in emitter current is $250 \ \mu A$.

(A) In a common base $(CB)$ amplifier configuration,the input signal is applied between the emitter and base,and the output is taken between the collector and base.
Since the input and output signals are in phase with each other,the phase difference between them is $0$.

(C) Given: Current gain $\beta = 50$ and emitter current $I_E = 6.6 \ mA$.
We know that the relationship between collector current $I_C$ and base current $I_B$ is given by $\beta = \frac{I_C}{I_B}$,which implies $I_C = 50 I_B$.
Also,the emitter current is the sum of collector current and base current: $I_E = I_C + I_B$.
Substituting $I_C = 50 I_B$ into the equation: $6.6 \ mA = 50 I_B + I_B = 51 I_B$.
Therefore,$I_B = \frac{6.6}{51} \ mA \approx 0.129 \ mA$.

(D) The change in base current is given by $\Delta I_B = 150 \ \mu A - 100 \ \mu A = 50 \ \mu A = 50 \times 10^{-6} \ A$.
The change in collector current is given by $\Delta I_C = 10 \ mA - 5 \ mA = 5 \ mA = 5 \times 10^{-3} \ A$.
The current gain $\beta$ is defined as the ratio of the change in collector current to the change in base current:
$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{5 \times 10^{-3} \ A}{50 \times 10^{-6} \ A} = \frac{5000 \times 10^{-6}}{50 \times 10^{-6}} = 100$.

(C) From the given circuit,applying Kirchhoff's voltage law to the base-emitter loop:
$V_{source} = I_B R_b + V_{BE}$
Given $V_{source} = 7 \ V$,$I_B = 35 \ \mu A = 35 \times 10^{-6} \ A$,and $V_{BE} \approx 0 \ V$.
Substituting the values:
$7 = (35 \times 10^{-6}) \times R_b + 0$
$R_b = \frac{7}{35 \times 10^{-6}} \ \Omega$
$R_b = \frac{1}{5} \times 10^6 \ \Omega = 0.2 \times 10^6 \ \Omega = 200 \ \times 10^3 \ \Omega = 200 \ k\Omega$.

(C) The power gain $(A_P)$ of an amplifier is defined as the product of the voltage gain $(A_V)$ and the current gain $(\beta)$.
$A_P = A_V \times \beta$
Dividing both sides by the voltage gain $(A_V)$, we get:
$\frac{A_P}{A_V} = \beta$
Given that $\beta = 62$, the ratio of power gain to voltage gain is $62$.

(A) The output resistance $(R_o)$ in common base $(CB)$ configuration is defined as the ratio of the change in collector-base voltage $(\Delta V_{CB})$ to the change in collector current $(\Delta I_C)$ while keeping the emitter current constant.
Formula: $R_o = \frac{\Delta V_{CB}}{\Delta I_C}$
Given:
$\Delta V_{CB} = 0.5 \ V$
$\Delta I_C = 0.05 \ mA$
Calculation:
$R_o = \frac{0.5 \ V}{0.05 \ mA} = \frac{0.5}{0.05} \ k\Omega = 10 \ k\Omega$
Therefore, the output resistance is $10 \ k\Omega$.

(C) The emitter current $I_e$ is given by $I_e = \frac{n \cdot e}{t} = \frac{10^{10} \times 1.6 \times 10^{-19}}{10^{-6}} = 1.6 \times 10^{-3} \ A = 1.6 \ mA$.
Since $2\%$ of electrons recombine in the base,the base current $I_b = 0.02 \times I_e = 0.02 \times 1.6 \ mA = 0.032 \ mA$.
The collector current $I_c = I_e - I_b = 1.6 \ mA - 0.032 \ mA = 1.568 \ mA$.
The current gain $\alpha = \frac{I_c}{I_e} = \frac{1.568}{1.6} = 0.98$.
The current gain $\beta = \frac{I_c}{I_b} = \frac{1.568}{0.032} = 49$.
Thus,the values are $0.98$ and $49$.

(A) The relationship between current gain $\alpha$ and $\beta$ is given by the formula: $\beta = \frac{\alpha}{1 - \alpha}$.
Given that $\beta = 50$,we substitute this into the equation:
$50 = \frac{\alpha}{1 - \alpha}$.
Multiplying both sides by $(1 - \alpha)$,we get:
$50(1 - \alpha) = \alpha$.
$50 - 50\alpha = \alpha$.
$50 = 51\alpha$.
$\alpha = \frac{50}{51} \approx 0.98$.

(A) Given: Load resistance $R_L = 2 \text{ k}\Omega = 2 \times 10^3 \text{ }\Omega$.
Change in base current $\Delta I_B = 10 \text{ }\mu\text{A} = 10^{-5} \text{ A}$.
Change in collector current $\Delta I_C = 1 \text{ mA} = 10^{-3} \text{ A}$.
The change in collector-emitter voltage is given by the formula $\Delta V_{CE} = R_L \times \Delta I_C$.
Substituting the values: $\Delta V_{CE} = (2 \times 10^3 \text{ }\Omega) \times (10^{-3} \text{ A}) = 2 \text{ V}$.
Therefore,the change in collector-emitter voltage is $2 \text{ V}$.

(A) Given $I_C = 4mA = 4 \times 10^{-3} A$. Applying Kirchhoff's voltage law to the collector circuit:
$V_{CC} = I_C R_L + V_{CE}$
$8V = (4 \times 10^{-3} A) R_L + 4V$
$4V = (4 \times 10^{-3} A) R_L$
$R_L = \frac{4}{4 \times 10^{-3}} \Omega = 1000 \Omega = 1 k\Omega$
Now,for the base circuit,the base current $I_B$ is given by:
$I_B = \frac{I_C}{\beta} = \frac{4mA}{100} = 0.04mA = 4 \times 10^{-5} A$
Applying Kirchhoff's voltage law to the base circuit:
$V_{CC} = I_B R_B + V_{BE}$
$8V = (4 \times 10^{-5} A) R_B + 0.6V$
$7.4V = (4 \times 10^{-5} A) R_B$
$R_B = \frac{7.4}{4 \times 10^{-5}} \Omega = 1.85 \times 10^5 \Omega = 185 k\Omega$
Thus,$R_L = 1 k\Omega$ and $R_B = 185 k\Omega$.