Junction Transistor Questions in English

(C) In a transistor,the emitter region is heavily doped because its primary function is to inject a large number of charge carriers into the base.
The base region is very lightly doped and kept thin to minimize the recombination of charge carriers,allowing most of them to reach the collector.
The collector region is moderately doped compared to the emitter and base.
Therefore,the emitter is heavily doped and the base is lightly doped.

(B) The ratio of collector current $(I_C)$ to emitter current $(I_E)$ is known as the common-base current amplification factor,denoted by $\alpha$. Given: $\alpha = \frac{I_C}{I_E} = 0.98$.
The ratio of collector current $(I_C)$ to base current $(I_B)$ is known as the common-emitter current amplification factor,denoted by $\beta$.
The relationship between $\alpha$ and $\beta$ is given by the formula: $\beta = \frac{\alpha}{1 - \alpha}$.
Substituting the given value: $\beta = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49$.
Therefore,the ratio of collector current to base current is $49$.

(B) The current gain $\beta_{dc}$ for a common-emitter transistor configuration is defined as the ratio of the collector current $(i_c)$ to the base current $(i_b)$.
Mathematically,$\beta_{dc} = \frac{i_c}{i_b} = \frac{\text{Collector current}}{\text{Base current}}$.

(A) We know the relationship between the current gain parameters $\alpha_{dc}$ and $\beta_{dc}$ is given by $\beta_{dc} = \frac{\alpha_{dc}}{1 - \alpha_{dc}}$.
Rearranging this equation,we get $1 - \alpha_{dc} = \frac{\alpha_{dc}}{\beta_{dc}}$.
Now,we need to evaluate the expression $\frac{\beta_{dc} - \alpha_{dc}}{\alpha_{dc} \cdot \beta_{dc}}$.
This can be written as $\frac{\beta_{dc}}{\alpha_{dc} \cdot \beta_{dc}} - \frac{\alpha_{dc}}{\alpha_{dc} \cdot \beta_{dc}} = \frac{1}{\alpha_{dc}} - \frac{1}{\beta_{dc}}$.
Substituting $\frac{1}{\alpha_{dc}} = \frac{1 + \beta_{dc}}{\beta_{dc}}$ is not necessary here; instead,use $\frac{1}{\alpha_{dc}} - \frac{1}{\beta_{dc}} = \frac{\beta_{dc} - \alpha_{dc}}{\alpha_{dc} \cdot \beta_{dc}}$.
From the relation $\beta_{dc} = \frac{\alpha_{dc}}{1 - \alpha_{dc}}$,we have $\frac{1}{\beta_{dc}} = \frac{1 - \alpha_{dc}}{\alpha_{dc}} = \frac{1}{\alpha_{dc}} - 1$.
Therefore,$\frac{1}{\alpha_{dc}} - \frac{1}{\beta_{dc}} = 1$.
Thus,the value of the expression is $1$.

(A) For a transistor to operate as an amplifier,it must be 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 configuration is that the emitter-base junction is forward-biased and the base-collector junction is reverse-biased.

(D) Given that, the change in emitter current is $\Delta I_{E} = 8.0 \,mA$.
The change in collector current is $\Delta I_{C} = 7.9 \,mA$.
We know that the current gain $\alpha$ is defined as the ratio of change in collector current to the change in emitter current:
$\alpha = \frac{\Delta I_{C}}{\Delta I_{E}} = \frac{7.9}{8.0} = 0.9875 \approx 0.99$.
We also know the relationship between $\alpha$ and $\beta$ is given by $\beta = \frac{\alpha}{1 - \alpha}$.
Alternatively, using the currents directly: $\beta = \frac{\Delta I_{C}}{\Delta I_{B}}$.
Since $\Delta I_{E} = \Delta I_{C} + \Delta I_{B}$, we have $\Delta I_{B} = \Delta I_{E} - \Delta I_{C} = 8.0 \,mA - 7.9 \,mA = 0.1 \,mA$.
Therefore, $\beta = \frac{7.9 \,mA}{0.1 \,mA} = 79$.
Thus, the values are $\alpha = 0.99$ and $\beta = 79$.

(B) In a transistor,the current gain parameters are defined as $\beta = \frac{I_C}{I_B}$ and $\alpha = \frac{I_C}{I_E}$.
Since $I_E = I_B + I_C$,we can write $\alpha = \frac{I_C}{I_B + I_C}$.
Dividing the numerator and denominator by $I_C$,we get $\alpha = \frac{1}{\frac{I_B}{I_C} + 1} = \frac{1}{\frac{1}{\beta} + 1} = \frac{\beta}{1 + \beta}$.
From this,we can derive $\frac{1}{\alpha} = \frac{1 + \beta}{\beta} = \frac{1}{\beta} + 1$,which implies $\frac{1}{\beta} = \frac{1}{\alpha} - 1$.
Comparing these with the given options,the relation $\alpha = \frac{\beta}{1 - \beta}$ is incorrect.

(B) For a transistor,the $ \beta $-current gain and $ \alpha $-current gain are related by the formula:
$\beta = \frac{\alpha}{1 - \alpha}$
Given that the $ \alpha $-current gain is $ 0.98 $.
Substituting the value into the formula:
$\beta = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49$
Thus,the $ \beta $-current gain of the transistor is $ 49 $.

(A) Among the three parts of a transistor,namely emitter,base,and collector:
$\rightarrow$ Emitter is moderate in size and heavily doped.
$\rightarrow$ Base is thin in size and lightly doped.
$\rightarrow$ Collector is large in size and moderately doped.
Therefore,the emitter is of moderate size and heavily doped.

(A) The input characteristics of a transistor in common emitter $(CE)$ configuration describe the relationship between the input current and the input voltage.
Specifically,it is the curve obtained by plotting the base current $(I_{B})$ against the base-emitter voltage $(V_{BE})$ while keeping the collector-emitter voltage $(V_{CE})$ constant.
This graph shows how the base current changes as the input voltage increases for different fixed values of the output voltage.

(A) The frequency response curve of a transistor amplifier shows that the voltage gain is not uniform across all frequencies.
At low frequencies,the coupling and bypass capacitors have high reactance,which reduces the gain.
At high frequencies,the internal junction capacitances of the transistor and stray capacitances become significant,providing a low-impedance path that shunts the signal,thereby reducing the gain.
In the mid-frequency range,these effects are negligible,and the amplifier provides a stable and maximum voltage gain.
Therefore,the voltage gain is low at both low and high frequencies and remains constant in the mid-frequency range.

(A) In a Common Emitter $(CE)$ amplifier configuration,the transistor is biased such that the emitter-base junction is forward biased and the collector-base junction is reverse biased.
To amplify an input $ac$ signal,it is superimposed on the $dc$ bias voltage at the input side.
Therefore,the input $ac$ signal is applied across the forward biased emitter-base junction to control the flow of charge carriers from the emitter to the collector.

(A) Given: Current gain $\beta = 50$,$V_{CE} = 2 \ V$,and $R_{C} = 4 \ k\Omega$.
The collector current $I_{C}$ is determined by the output circuit loop: $I_{C} = \frac{V_{CE}}{R_{C}} = \frac{2 \ V}{4 \times 10^{3} \ \Omega} = 0.5 \times 10^{-3} \ A = 0.5 \ mA$.
The base current $I_{B}$ is related to the collector current by the current gain formula: $\beta = \frac{I_{C}}{I_{B}}$.
Rearranging for $I_{B}$: $I_{B} = \frac{I_{C}}{\beta} = \frac{0.5 \times 10^{-3} \ A}{50} = 0.01 \times 10^{-3} \ A = 10 \times 10^{-6} \ A = 10 \ \mu A$.
Thus,$I_{B} = 10 \ \mu A$ and $I_{C} = 0.5 \ mA$.

(A) Given,collector current variation,$ \Delta I_{C} = 0.49 \,mA $.
Emitter current variation,$ \Delta I_{E} = 0.50 \,mA $.
We know that the base current variation is given by $ \Delta I_{B} = \Delta I_{E} - \Delta I_{C} $.
Substituting the values,$ \Delta I_{B} = 0.50 \,mA - 0.49 \,mA = 0.01 \,mA $.
The current gain $ \beta $ is defined as the ratio of collector current change to base current change:
$ \beta = \frac{\Delta I_{C}}{\Delta I_{B}} $.
Substituting the values,$ \beta = \frac{0.49 \,mA}{0.01 \,mA} = 49 $.
Therefore,the current gain $ \beta $ is $ 49 $.

(D) To use a transistor as an amplifier,the input circuit (emitter-base junction) must be forward biased to allow current flow,and the output circuit (collector-base junction) must be reverse biased to provide high resistance and voltage gain.
Therefore,the emitter-base junction is forward biased,and the collector-base junction is reverse biased.

(C) The relationship between the current gain $\alpha$ (common-base) and $\beta$ (common-emitter) is given by the formula: $\alpha = \frac{\beta}{1 + \beta}$.
Given that $\beta = 100$,we substitute this value into the formula:
$\alpha = \frac{100}{1 + 100} = \frac{100}{101}$.
Performing the division,we get $\alpha \approx 0.99$.

(B) An $n-p-n$ transistor consists of two $p-n$ junctions.
In an $n-p-n$ transistor,the emitter-base junction is a $p-n$ junction and the collector-base junction is also a $p-n$ junction.
The base is the common $p$-type region.
For an $n-p-n$ transistor,the emitter is $n$-type,the base is $p$-type,and the collector is $n$-type.
This means the emitter-base diode has its $n$-side at the emitter and $p$-side at the base.
The collector-base diode has its $n$-side at the collector and $p$-side at the base.
Therefore,the two diodes are connected with their cathodes (n-sides) facing outwards and their anodes (p-sides) connected together at the base.
Looking at the provided figures,option $B$ correctly shows two diodes with their anodes connected at the base $B$ and cathodes pointing towards the emitter $E$ and collector $C$ respectively.

(D) In a common emitter $(CE)$ amplifier,the input signal $(V_i)$ is applied between the base $(B)$ and the emitter $(E)$ terminals of the transistor.
As shown in the circuit diagram,the input signal is coupled through a capacitor $(C_1)$ to the base,while the emitter is common to both the input and output circuits and is typically grounded.
Therefore,the input voltage is applied across the base-emitter junction.

(A) For a transistor to operate in the active region,which is required for amplification,the emitter-base junction must be forward biased and the base-collector junction must be reverse biased.
In the forward-biased emitter-base junction,charge carriers are injected from the emitter into the base.
In the reverse-biased base-collector junction,these carriers are collected by the collector,allowing for current control and signal amplification.

(D) Given:
Voltage gain $(A_v)$ = $300$
Current amplification factor $(\beta)$ = $60$
Collector resistance $(R_C)$ = $5 \, k\Omega$
We know that the voltage gain $(A_v)$ in a common emitter configuration is given by:
$A_v = \beta \times \frac{R_C}{R_B}$
Rearranging the formula to find the base resistance $(R_B)$:
$R_B = \beta \times \frac{R_C}{A_v}$
Substituting the given values:
$R_B = 60 \times \frac{5 \, k\Omega}{300}$
$R_B = 60 \times \frac{5}{300} \, k\Omega$
$R_B = \frac{300}{300} \, k\Omega$
$R_B = 1 \, k\Omega$
Therefore, the base resistance is $1 \, k\Omega$.

(D) For a common emitter amplifier,the voltage gain $(A_v)$ is given by the formula: $A_v = \beta \times \frac{R_C}{R_B}$,where $\beta$ is the current amplification factor,$R_C$ is the collector resistance,and $R_B$ is the base resistance.
Given that the ratio of voltage gain $(A_v)$ to the current amplification factor $(\beta)$ is $4$,we have: $\frac{A_v}{\beta} = 4$.
Substituting the expression for $A_v$ into the given ratio: $\frac{\beta \times (R_C / R_B)}{\beta} = 4$.
This simplifies to: $\frac{R_C}{R_B} = 4$.
Therefore,the ratio of the collector resistance to the base resistance is $4 : 1$.

(C) In a standard bipolar junction transistor $(BJT)$,the three regions are designed with specific physical dimensions to optimize performance.
$1$. The $Base$ $(Y)$ is made very thin and lightly doped to allow most charge carriers to pass through to the collector.
$2$. The $Collector$ $(Z)$ is made the largest in size to handle the heat dissipation generated by the power collected from the emitter.
$3$. The $Emitter$ $(X)$ is of moderate size,larger than the base but smaller than the collector,and is heavily doped to provide a large number of charge carriers.
Therefore,the order of sizes is $Collector > Emitter > Base$,which corresponds to $Z > X > Y$.

(D) The power gain $(A_p)$ of a transistor amplifier is given by the product of the current gain $(\beta)$ and the voltage gain $(A_v)$.
$A_p = \beta \times A_v$
Voltage gain $(A_v)$ is defined as the product of the current gain $(\beta)$ and the ratio of output resistance $(R_c)$ to input resistance $(R_b)$:
$A_v = \beta \times \frac{R_c}{R_b}$
Given:
$\beta = 100$
$R_c = 3 \ k\Omega$
$R_b = 2 \ k\Omega$
Calculating $A_v$:
$A_v = 100 \times \frac{3 \ k\Omega}{2 \ k\Omega} = 100 \times 1.5 = 150$
Now,calculating power gain $(A_p)$:
$A_p = \beta \times A_v = 100 \times 150 = 15000$
Therefore,the power gain is $15000$.

(C) Given that the collector current $I_C = 0.98 I_E$.
We know that the emitter current is the sum of the base current and the collector current: $I_E = I_B + I_C$.
Substituting the value of $I_C$,we get $I_E = I_B + 0.98 I_E$.
Rearranging the terms to find $I_B$: $I_B = I_E - 0.98 I_E = 0.02 I_E$.
Now,we need to find the ratio of the base current to the collector current: $\frac{I_B}{I_C} = \frac{0.02 I_E}{0.98 I_E}$.
Simplifying the ratio: $\frac{I_B}{I_C} = \frac{2}{98} = \frac{1}{49}$.
Therefore,the ratio is $1: 49$.

(B) In a common-emitter $(CE)$ configuration of a transistor,the output signal is phase-shifted by $180^{\circ}$ relative to the input signal. This occurs because the input signal is applied to the base-emitter junction,and the output is taken from the collector-emitter junction,resulting in an inversion of the signal polarity.

(C) For a common emitter $(CE)$ configuration of a transistor:
Given: Voltage gain $(A_v)$ = $150$, Current gain $(\beta)$ = $50$, Base resistance $(R_B)$ = $850 \Omega$.
The formula for voltage gain in a $CE$ amplifier is given by:
$A_v = \beta \times \left( \frac{R_C}{R_B} \right)$
Substituting the given values into the formula:
$150 = 50 \times \left( \frac{R_C}{850} \right)$
Rearranging to solve for collector resistance $(R_C)$:
$\frac{R_C}{850} = \frac{150}{50}$
$\frac{R_C}{850} = 3$
$R_C = 3 \times 850 = 2550 \Omega$
Therefore, the resistance in the collector circuit is $2550 \Omega$.

(D) For a common emitter $(CE)$ amplifier,the voltage gain $(A_v)$ is defined as the product of the current gain $(\beta)$ and the ratio of the output load resistance $(R_o)$ to the input resistance $(R_i)$.
Given:
$\beta = 40$
$R_i = 2 \ k\Omega$
$R_o = 10 \ k\Omega$
Using the formula:
$A_v = \beta \times \left( \frac{R_o}{R_i} \right)$
$A_v = 40 \times \left( \frac{10 \ k\Omega}{2 \ k\Omega} \right)$
$A_v = 40 \times 5$
$A_v = 200$

(D) In an $n-p-n$ transistor,the relationship between emitter current $(I_E)$,collector current $(I_C)$,and base current $(I_B)$ is given by $I_E = I_B + I_C$.
Given that $95\%$ of the emitted electrons reach the collector,we have $I_C = 0.95 \times I_E$.
Given $I_C = 10 \text{ mA}$,we can find $I_E$ as:
$I_E = \frac{I_C}{0.95} = \frac{10}{0.95} \approx 10.53 \text{ mA}$.
Now,calculating the base current:
$I_B = I_E - I_C = 10.53 \text{ mA} - 10 \text{ mA} = 0.53 \text{ mA}$.

(C) Given: Current gain $\beta = 80$, Collector resistance $R_C = 5 \text{ k}\Omega$, Base resistance $R_B = 1 \text{ k}\Omega$, Input voltage $V_I = 2 \text{ mV} = 2 \times 10^{-3} \text{ V}$.
The voltage gain $A_V$ for a common emitter transistor is given by the formula:
$A_V = \frac{V_O}{V_I} = \beta \times \frac{R_C}{R_B}$
Substituting the given values:
$V_O = V_I \times \beta \times \frac{R_C}{R_B}$
$V_O = (2 \times 10^{-3} \text{ V}) \times 80 \times \frac{5 \text{ k}\Omega}{1 \text{ k}\Omega}$
$V_O = 2 \times 10^{-3} \times 80 \times 5$
$V_O = 800 \times 10^{-3} \text{ V}$
$V_O = 0.8 \text{ V}$.

(D) The voltage gain $A_V$ of a transistor amplifier in common emitter configuration is defined as the ratio of output voltage $V_O$ to input voltage $V_I$.
$A_V = \frac{V_O}{V_I} = \frac{V_C}{V_B}$
Since $V_C = I_C R_C$ and $V_B = I_B R_B$,we can write:
$A_V = \frac{I_C R_C}{I_B R_B}$
Rearranging the terms,we get:
$A_V = \left(\frac{I_C}{I_B}\right) \left(\frac{R_C}{R_B}\right)$
Given that the current amplification factor $\beta = \frac{I_C}{I_B}$,substituting this into the equation gives:
$A_V = \beta \cdot \frac{R_C}{R_B}$

(A) In a $p-n-p$ transistor,the emitter is heavily doped to provide a large number of charge carriers for current flow. The base is very thin and lightly doped to allow most carriers to pass through to the collector. The collector is moderately doped and has a larger physical area compared to the emitter to dissipate heat generated during operation.

(D) The relationship between emitter current $(I_E)$,collector current $(I_C)$,and base current $(I_B)$ in a transistor is given by the equation:
$I_E = I_B + I_C$
Given:
$I_E = 9.85 \,mA$
$I_C = 9.5 \,mA$
To find the base current $(I_B)$,we rearrange the formula:
$I_B = I_E - I_C$
Substituting the given values:
$I_B = 9.85 \,mA - 9.5 \,mA = 0.35 \,mA$
Therefore,the base current is $0.35 \,mA$.

(A) Given: Emitter current, $I_{E} = 6.6 \, mA$.
Current amplification factor, $\beta = 59$.
We know that the relationship between emitter current $(I_{E})$, collector current $(I_{C})$, and base current $(I_{B})$ is $I_{E} = I_{C} + I_{B}$.
Since $I_{C} = \beta I_{B}$, we can write $I_{E} = \beta I_{B} + I_{B} = I_{B}(\beta + 1)$.
Therefore, the base current is $I_{B} = \frac{I_{E}}{\beta + 1}$.
Substituting the values: $I_{B} = \frac{6.6 \, mA}{59 + 1} = \frac{6.6 \, mA}{60} = 0.11 \, mA$.

(B) Given: Load resistance $R_C = 5 \ k\Omega = 5 \times 10^3 \ \Omega$.
Base-emitter voltage $V_{BE} = 0.02 \ V$.
Collector current $I_C = 2 \ mA = 2 \times 10^{-3} \ A$.
The output voltage $V_{CE}$ is given by $V_{CE} = I_C \times R_C$.
Substituting the values: $V_{CE} = (2 \times 10^{-3} \ A) \times (5 \times 10^3 \ \Omega) = 10 \ V$.
The voltage gain $A_v$ is defined as the ratio of output voltage to input voltage: $A_v = \frac{V_{CE}}{V_{BE}}$.
Calculating the gain: $A_v = \frac{10 \ V}{0.02 \ V} = 500$.

(D) The frequency response curve of a transistor amplifier indicates that the voltage gain is constant only within the mid-frequency range.
At very low frequencies,the gain drops due to the reactance of coupling capacitors.
At very high frequencies,the gain drops due to the internal junction capacitances of the transistor.
Therefore,the gain is low at both high and low frequencies and remains constant at mid-frequencies.

(B) Given: Current amplification factor, $\beta = 100$
Collector current, $I_C = 2.2 \text{ mA} = 2.2 \times 10^{-3} \text{ A}$
We know the relation between collector current and base current in a $CE$ configuration is given by:
$\beta = \frac{I_C}{I_B}$
Rearranging the formula to solve for base current $(I_B)$:
$I_B = \frac{I_C}{\beta}$
Substituting the given values:
$I_B = \frac{2.2 \times 10^{-3} \text{ A}}{100} = 2.2 \times 10^{-5} \text{ A}$
Converting to microamperes $(\mu A)$:
$I_B = 2.2 \times 10^{-5} \times 10^6 \mu A = 22 \mu A$
Therefore, the base current is $22 \mu A$.

(B) In a transistor, the emitter is heavily doped to inject charge carriers into the base, while the collector is lightly doped and has a larger area to collect them.
If the emitter and collector connections are interchanged, the collector (which is lightly doped) acts as the emitter and the emitter (which is heavily doped) acts as the collector.
This mismatch in doping levels and physical structure significantly reduces the current gain $(\beta)$ of the transistor.
Consequently, for a given base current, the collector current drops significantly, and the base current itself decreases due to the change in the effective operation of the junction.
Therefore, the correct observation is that the base current decreases.

(D) The input resistance $R_{\text{in}}$ of a transistor is defined as the ratio of the change in base-emitter voltage $(\Delta V_{BE})$ to the change in base current $(\Delta I_b)$.
Given:
$\Delta I_b = 60 \mu A = 60 \times 10^{-6} \ A$
$\Delta V_{BE} = 1.2 \ V$
Using the formula:
$R_{\text{in}} = \frac{\Delta V_{BE}}{\Delta I_b}$
$R_{\text{in}} = \frac{1.2}{60 \times 10^{-6}}$
$R_{\text{in}} = \frac{1.2 \times 10^6}{60}$
$R_{\text{in}} = \frac{1200000}{60} = 20000 \ \Omega$
Therefore,the input resistance is $20000 \ \Omega$.

(C) The voltage gain $A_v$ is defined as the ratio of output signal voltage to input signal voltage:
$A_v = \frac{V_{out}}{V_{in}} = \frac{2.5 \ V}{0.02 \ V} = 125$.
We also know that the voltage gain for a $CE$ amplifier is given by $A_v = \beta \times \frac{R_c}{R_b}$,where $\beta$ is the current amplification factor,$R_c$ is the collector resistance,and $R_b$ is the base resistance.
Substituting the given values:
$125 = \beta \times \frac{2.5 \ k\Omega}{1.5 \ k\Omega}$.
$125 = \beta \times \frac{5}{3}$.
$\beta = 125 \times \frac{3}{5} = 25 \times 3 = 75$.
Thus,the current amplification factor is $75$.

(A) Given,collector resistance $R_C = 800 \Omega$ and voltage drop across it $V_R = 0.5 V$.
Collector current $I_C = \frac{V_R}{R_C} = \frac{0.5}{800} = 6.25 \times 10^{-4} A = 0.625 \times 10^{-3} A$.
We know that the current gain $\alpha = \frac{I_C}{I_E}$,so emitter current $I_E = \frac{I_C}{\alpha} = \frac{0.625 \times 10^{-3}}{0.96} \approx 6.51 \times 10^{-4} A$.
The base current $I_B = I_E - I_C = I_C \left( \frac{1}{\alpha} - 1 \right) = I_C \left( \frac{1 - \alpha}{\alpha} \right)$.
Substituting the values: $I_B = 0.625 \times 10^{-3} \times \left( \frac{1 - 0.96}{0.96} \right) = 0.625 \times 10^{-3} \times \left( \frac{0.04}{0.96} \right) = 0.625 \times 10^{-3} \times \frac{1}{24} \approx 0.02604 \times 10^{-3} A = 2.6 \times 10^{-5} A$.

(B) Given: $\Delta I_b = 20 \ \mu A = 20 \times 10^{-6} \ A$,$\Delta I_c = 2 \ mA = 2 \times 10^{-3} \ A$,and $\Delta V_{BE} = 0.04 \ V$.
The input resistance $R_{\text{input}}$ is defined as the ratio of the change in base-emitter voltage to the change in base current:
$R_{\text{input}} = \frac{\Delta V_{BE}}{\Delta I_b} = \frac{0.04}{20 \times 10^{-6}} = \frac{0.04 \times 10^6}{20} = 2 \times 10^3 \ \Omega = 2 \ k\Omega$.
The $AC$ current gain $\beta$ is defined as the ratio of the change in collector current to the change in base current:
$\beta = \frac{\Delta I_c}{\Delta I_b} = \frac{2 \times 10^{-3}}{20 \times 10^{-6}} = \frac{2000}{20} = 100$.
Thus,the input resistance is $2 \ k\Omega$ and the $AC$ current gain is $100$.

(B) Given that, $V_{CE}=7 \, V$, $\beta=100$, $R_C=2 \, k\Omega = 2000 \, \Omega$, and $V_{CC}=15 \, V$.
We know that the current gain in a common emitter circuit is given by:
$\beta = \frac{I_C}{I_B} \dots (i)$
Applying Kirchhoff's Voltage Law $(KVL)$ in the output loop (collector-emitter loop):
$V_{CC} - I_C R_C - V_{CE} = 0$
Substituting the given values:
$15 - I_C(2000) - 7 = 0$
$8 = I_C(2000)$
$I_C = \frac{8}{2000} \, A = 4 \times 10^{-3} \, A = 4 \, mA \dots (ii)$
Now, substituting the value of $I_C$ from equation $(ii)$ into equation $(i)$:
$100 = \frac{4 \, mA}{I_B}$
$I_B = \frac{4}{100} \, mA = 0.04 \, mA$.

(A) Given, current gain, $\alpha = 0.96$.
Emitter current, $I_E = 7.2 \,mA = 7.2 \times 10^{-3} \,A$.
We know that, $\alpha = I_C / I_E$.
Therefore, $I_C = \alpha \times I_E = 0.96 \times 7.2 \,mA = 6.912 \,mA$.
Also, the relation between emitter, collector, and base current is $I_E = I_C + I_B$.
Thus, $I_B = I_E - I_C = 7.2 \,mA - 6.912 \,mA = 0.288 \,mA \approx 0.29 \,mA$.

(C) In a common base amplifier for a $p-n-p$ transistor, the current gain is given by $\alpha = 0.96$.
Given emitter current, $I_E = 7.2 \,mA$.
We know the relation for current gain: $\alpha = \frac{I_C}{I_E}$.
Substituting the values: $0.96 = \frac{I_C}{7.2}$.
Calculating collector current: $I_C = 0.96 \times 7.2 = 6.912 \,mA$.
We also know that the emitter current is the sum of collector current and base current: $I_E = I_C + I_B$.
Therefore, base current $I_B = I_E - I_C$.
$I_B = 7.2 \,mA - 6.912 \,mA = 0.288 \,mA$.
Rounding to two decimal places, we get $I_B \simeq 0.29 \,mA$.

(C) In a transistor,the common base current gain $\alpha$ is related to the common emitter current gain $\beta$ by the formula: $\beta = \frac{\alpha}{1 - \alpha}$.
Case $I$: When $\alpha = \frac{20}{21}$,
$\beta = \frac{20/21}{1 - 20/21} = \frac{20/21}{1/21} = 20$.
Case $II$: When $\alpha = \frac{100}{101}$,
$\beta = \frac{100/101}{1 - 100/101} = \frac{100/101}{1/101} = 100$.
Therefore,the value of $\beta$ varies between $20$ and $100$.

(B) For the output loop of the common emitter circuit,applying Kirchhoff's voltage law $(KVL)$:
$15 \text{ V} - I_C \times (2 \text{ k}\Omega) - V_{CE} = 0$
Given $V_{CE} = 7 \text{ V}$,we have:
$15 - I_C \times 2000 - 7 = 0$
$8 = I_C \times 2000$
$I_C = \frac{8}{2000} \text{ A} = 4 \times 10^{-3} \text{ A} = 4 \text{ mA}$
Now,using the current gain relation $\beta = \frac{I_C}{I_B}$:
$100 = \frac{4 \text{ mA}}{I_B}$
$I_B = \frac{4 \text{ mA}}{100} = 0.04 \text{ mA}$

(B) In the given circuit,the base loop equation is $V_{CC} = i_B R_1 + V_{BE}$.
Given $V_{CC} = 8 \text{ V}$,$V_{BE} = 0.6 \text{ V}$,and $i_C = 5 \text{ mA}$.
The base current is $i_B = \frac{i_C}{\beta} = \frac{5 \times 10^{-3} \text{ A}}{50} = 1 \times 10^{-4} \text{ A}$.
Substituting the values into the base loop equation:
$8 = (1 \times 10^{-4}) R_1 + 0.6$
$R_1 = \frac{8 - 0.6}{1 \times 10^{-4}} = \frac{7.4}{10^{-4}} = 74 \times 10^3 \Omega = 74 \text{ k}\Omega$.
Now,for the collector loop,the $KVL$ equation is $V_{CC} = i_C R_2 + V_{CE}$.
Substituting the values:
$8 = (5 \times 10^{-3}) R_2 + 3$
$5 = (5 \times 10^{-3}) R_2$
$R_2 = \frac{5}{5 \times 10^{-3}} = 10^3 \Omega = 1 \text{ k}\Omega$.
Thus,$R_1 = 74 \text{ k}\Omega$ and $R_2 = 1 \text{ k}\Omega$.

(D) Given,current amplification factor in common-base configuration,$\alpha = 0.95$.
We know that the relationship between $\alpha$ and $\beta$ (current amplification factor in common-emitter configuration) is given by $\beta = \frac{\alpha}{1 - \alpha}$.
Substituting the value of $\alpha$:
$\beta = \frac{0.95}{1 - 0.95} = \frac{0.95}{0.05} = 19$.
In common-emitter configuration,the current gain $\beta$ is defined as the ratio of change in collector current $(\Delta I_C)$ to the change in base current $(\Delta I_B)$:
$\beta = \frac{\Delta I_C}{\Delta I_B}$.
Given $\Delta I_B = 2 \mu A$,we can calculate $\Delta I_C$:
$\Delta I_C = \beta \times \Delta I_B = 19 \times 2 \mu A = 38 \mu A$.

(C) The input current $i_i$ is given by $i_i = \frac{V_i}{R_i}$,where $V_i = 100 \ mV = 0.1 \ V$ and $R_i = 2000 \ \Omega$.
$i_i = \frac{0.1}{2000} = 5 \times 10^{-5} \ A$.
The output current $i_o$ is given by $i_o = \beta \times i_i$.
$i_o = 50 \times 5 \times 10^{-5} = 250 \times 10^{-5} = 2.5 \times 10^{-3} \ A$.
The output voltage $V_o$ is given by $V_o = i_o \times R_o$,where $R_o = 5000 \ \Omega$.
$V_o = 2.5 \times 10^{-3} \times 5000 = 12.5 \ V$.
Thus,the peak value of the output voltage is $12.5 \ V$.

(A) In a Common Emitter $(CE)$ amplifier configuration,the output signal is inverted with respect to the input signal,meaning there is a phase shift of $\pi$ radians $(180^{\circ})$ between them.
For an oscillator to sustain oscillations,the Barkhausen criterion requires the total phase shift around the loop to be $0$ or $2n\pi$.
Since the $CE$ amplifier provides a phase shift of $\pi$,the feedback network must provide an additional phase shift of $\pi$ to satisfy the condition for oscillation.
Therefore,the phase difference between the input and output signal of the $CE$-transistor stage itself is $\pi$.