Valve Electronics Questions in English

(B) The control grid in an electron gun is maintained at a negative potential with respect to the cathode.
Since electrons are negatively charged,this negative potential creates an electric field that opposes the motion of electrons.
By adjusting the magnitude of this negative potential,the grid can repel a portion of the electrons back toward the cathode.
Consequently,this mechanism allows for the precise control of the number of electrons (beam intensity) that pass through the aperture of the grid.

(D) In a discharge tube,gas molecules are ionized due to various factors like cosmic rays or high potential differences.
When a high potential difference is applied,the electric field accelerates these charged particles.
Collisions between these accelerated particles and neutral gas molecules lead to further ionization,creating a cascade of charge carriers.
Specifically,electrons are detached from molecules,forming positive ions.
At low pressures,these electrons can travel significant distances and may attach to other neutral molecules to form negative ions.
Therefore,the current conduction in a discharge tube is carried by positive ions,negative ions,and electrons.

(A) Semiconductor devices,such as transistors and diodes,are designed to operate at low power levels and low voltage ranges,typically between $0 \ V$ and $15 \ V$. They are highly sensitive to high voltage and can be permanently damaged if subjected to voltages beyond their rated limits. Therefore,their inability to handle high voltage is a significant limitation compared to vacuum tubes.

(A) In a vacuum tube,the $SCR$ (Space Charge Region) consists of a cloud of electrons that accumulate near the cathode.
These electrons create a negative potential barrier that repels other emitted electrons back toward the cathode.
As a result,the number of electrons reaching the plate (anode) is reduced.
Therefore,the plate current $(I_p)$ decreases.

(B) The plate resistance $({r_p})$ is defined as the ratio of the change in plate voltage $(\delta V)$ to the change in plate current $(\delta I)$,expressed as $r_p = \frac{\delta V}{\delta I}$.
At the saturation point,the current reaches its maximum value and does not change with further increases in voltage,meaning the change in current $\delta I = 0$.
Substituting this into the formula,we get $r_p = \frac{\delta V}{0} = \infty$.
Therefore,at saturation,the plate resistance is infinite.

(B) The mutual conductance $(g_m)$ of a triode valve is defined as the ratio of the change in plate current $(\Delta i_p)$ to the change in grid voltage $(\Delta v_g)$ at a constant plate voltage.
Formula: $g_m = \frac{\Delta i_p}{\Delta v_g}$
Given:
$g_m = 3 \times 10^{-4} \, \text{mho}$
$\Delta v_g = |(-3) - (-1)| = |-2| = 2 \, V$
Substituting the values:
$3 \times 10^{-4} = \frac{\Delta i_p}{2}$
$\Delta i_p = 3 \times 10^{-4} \times 2 = 6 \times 10^{-4} \, A$
Converting to milliamperes $(mA)$:
$\Delta i_p = 6 \times 10^{-4} \times 10^3 \, mA = 0.6 \, mA$
Therefore, the change in plate circuit current is $0.6 \, mA$.

(B) The voltage gain ${A_v}$ of a triode amplifier is given by the formula: ${A_v} = \frac{\mu R_L}{r_p + R_L} = \frac{\mu}{1 + \frac{r_p}{R_L}}$.
Given that $\mu = r_p \times g_m$,we can calculate the plate resistance ${r_p}$ as:
${r_p} = \frac{\mu}{g_m} = \frac{42}{2 \times 10^{-3}} = 21000 \, \Omega = 21 \, \text{k}\Omega$.
Now,substituting the values into the voltage gain formula:
${A_v} = \frac{42}{1 + \frac{21 \, \text{k}\Omega}{50 \, \text{k}\Omega}} = \frac{42}{1 + 0.42} = \frac{42}{1.42} \approx 29.57$.

(C) The voltage amplification (voltage gain) ${A_v}$ of an amplifier is given by the formula: ${A_v} = \frac{\mu}{1 + \frac{r_p}{R_L}}$.
Given that the load resistance ${R_L}$ is equal to the plate resistance ${r_p}$,i.e.,${R_L} = {r_p}$.
Substituting this into the formula: ${A_v} = \frac{\mu}{1 + \frac{r_p}{r_p}} = \frac{\mu}{1 + 1} = \frac{\mu}{2}$.
Therefore,the voltage amplification is equal to $\mu / 2$.

(B) In a triode,the plate current is determined by the number of electrons reaching the plate $P$ from the cathode $K$.
When the grid $G$ is given a positive potential,it exerts an attractive force on the electrons emitted from the cathode,helping them pass through the grid mesh towards the plate.
Since the plate $P$ must also be at a positive potential to attract these electrons,having both the grid and the plate at positive potentials results in the maximum flow of electrons to the plate,thereby maximizing the plate current.
Therefore,the correct option is $B$.

(A) The amplification factor $\mu$ is given by the product of plate resistance ${R_p}$ and transconductance ${g_m}$.
$\mu = {R_p} \times {g_m} = 7 \times 10^3 \Omega \times 2.5 \times 10^{-3} \text{ mho} = 17.5$.
Also, the amplification factor is defined as $\mu = - \frac{\Delta {V_p}}{\Delta {V_g}}$ for a constant plate current.
Given $\Delta {V_p} = 50 \text{ V}$, we have $17.5 = - \frac{50}{\Delta {V_g}}$.
Therefore, $\Delta {V_g} = - \frac{50}{17.5} \approx - 2.86 \text{ V}$.

(A) Given:
Amplification factor $\mu = 20$
Trans-conductance $g_m = 3 \text{ mS} = 3 \times 10^{-3} \ \Omega^{-1}$
Load resistance $R_L = 3 \times 10^4 \ \Omega$
First,we find the plate resistance $r_p$ using the relation $\mu = g_m \times r_p$:
$r_p = \frac{\mu}{g_m} = \frac{20}{3 \times 10^{-3}} = \frac{20000}{3} \ \Omega$
The formula for voltage gain $A_v$ is:
$A_v = \frac{\mu \times R_L}{r_p + R_L}$
Substituting the values:
$A_v = \frac{20 \times 3 \times 10^4}{\frac{20000}{3} + 30000}$
$A_v = \frac{600000}{\frac{20000 + 90000}{3}} = \frac{600000 \times 3}{110000}$
$A_v = \frac{180}{11} \approx 16.36$

(C) The voltage gain $A_v$ of a triode amplifier is given by the formula: $A_v = \frac{\mu R_L}{r_p + R_L} = \frac{\mu}{1 + \frac{r_p}{R_L}}$.
Given: $\mu = 25$,$r_p = 40 \text{ k}\Omega$,$R_L = 10 \text{ k}\Omega$,and $V_{\text{in}} = 0.5 \text{ V}$.
Substituting the values into the gain formula:
$A_v = \frac{25}{1 + \frac{40}{10}} = \frac{25}{1 + 4} = \frac{25}{5} = 5$.
Since voltage gain $A_v = \frac{V_{\text{out}}}{V_{\text{in}}}$,we have $V_{\text{out}} = A_v \times V_{\text{in}}$.
$V_{\text{out}} = 5 \times 0.5 \text{ V} = 2.5 \text{ V}$.

(B) The amplification factor $\mu$ of a triode is defined as the ratio of the change in plate voltage to the change in grid voltage for a constant plate current: $\mu = -\frac{\Delta V_p}{\Delta V_G}$.
Given: $\mu = 20$ and $\Delta V_G = -0.2\, V$ (since the potential is reduced).
To keep the plate current constant,we rearrange the formula: $\Delta V_p = -\mu \times \Delta V_G$.
Substituting the values: $\Delta V_p = -20 \times (-0.2\, V) = 4\, V$.
Therefore,the plate voltage must be increased by $4\, V$.

(B) The voltage gain ${A_V}$ of a triode is given by the formula: ${A_V} = \frac{\mu}{1 + \frac{r_p}{R_L}}.$
First, we calculate the amplification factor $\mu$ using the relation $\mu = {r_p} \times {g_m}.$
Given ${r_p} = 10 \times 10^3 \ \Omega$ and ${g_m} = 3 \times 10^{-3} \ \mho$, we get $\mu = (10 \times 10^3) \times (3 \times 10^{-3}) = 30.$
It is given that the load resistance ${R_L} = 2{r_p}.$
Substituting these values into the voltage gain formula:
${A_V} = \frac{30}{1 + \frac{r_p}{2r_p}} = \frac{30}{1 + 0.5} = \frac{30}{1.5} = 20.$
Therefore, the voltage gain is $20$.

(C) The amplification produced by a triode is due to the action of the control grid. $A$ triode consists of three electrodes mounted inside an evacuated glass or metal container: a cathode,a control grid,and an anode (plate). The control grid is placed between the cathode and the anode. $A$ small change in the potential of the control grid causes a large change in the plate current,which is the fundamental principle behind the amplification factor of a triode.

(B) The voltage gain $(A_v)$ of a triode amplifier is given by the formula: $A_v = \frac{\mu R_L}{r_p + R_L}$,where $\mu$ is the amplification factor,$r_p$ is the plate resistance,and $R_L$ is the load resistance.
To find the maximum gain,we consider the limit as the load resistance $R_L$ approaches infinity $(R_L \to \infty)$.
$(A_v)_{max} = \lim_{R_L \to \infty} \frac{\mu R_L}{r_p + R_L} = \lim_{R_L \to \infty} \frac{\mu}{\frac{r_p}{R_L} + 1} = \frac{\mu}{0 + 1} = \mu$.
Therefore,the maximum gain is equal to the amplification factor $\mu$.

(B) The voltage gain $A_v$ of a triode amplifier is given by the formula: $A_v = \frac{\mu}{1 + \frac{r_p}{R_L}}$.
Given that the load resistance $R_L = 1.5 \, r_p$,where $r_p$ is the anode resistance.
Substituting the value of $R_L$ into the formula:
$A_v = \frac{\mu}{1 + \frac{r_p}{1.5 \, r_p}} = \frac{\mu}{1 + \frac{1}{1.5}} = \frac{\mu}{1 + \frac{2}{3}} = \frac{\mu}{\frac{5}{3}} = \frac{3}{5} \mu$.
Given $\mu = 20$,we calculate:
$A_v = \frac{3}{5} \times 20 = 3 \times 4 = 12$.

(C) The voltage gain $(A_v)$ of a triode is given by the formula $A_v = \mu \times \frac{R_L}{r_p + R_L}$,where $\mu$ is the amplification factor,$r_p$ is the plate resistance,and $R_L$ is the load resistance. Since the amplification factor $\mu$ and the plate resistance $r_p$ are intrinsic properties of the triode,the voltage gain depends directly on the plate resistance $r_p$ and the load resistance.

(C) The thermionic emission current $I$ is given by Richardson-Dushman equation: $I = AT^2 e^{-\phi/kT}$,where $A$ is a constant,$T$ is the absolute temperature,$\phi$ is the work function,and $k$ is the Boltzmann constant.
Since $I \propto T^2$,if the temperature $T$ is doubled $(T' = 2T)$,the new current $I'$ becomes $I' \propto (2T)^2 = 4T^2$.
Therefore,the thermionic current becomes nearly four times the original value.
Thus,option $C$ is correct.

(C) The amplification factor $\mu$ of a triode valve is defined as the ratio of the change in plate voltage to the change in grid voltage for a constant plate current.
Mathematically,$\mu = \left| \frac{\Delta V_p}{\Delta V_g} \right|$.
Given: $\mu = 15$ and $\Delta V_g = 0.3 \,V$.
To keep the plate current constant,the change in plate voltage $\Delta V_p$ is given by:
$\Delta V_p = \mu \times \Delta V_g$
$\Delta V_p = 15 \times 0.3 = 4.5 \,V$.
Therefore,the change in plate voltage is $4.5 \,V$.

(C) The plate resistance $(r_p)$ of a vacuum tube diode is defined as the reciprocal of the slope of the plate characteristic curve at a given operating point.
Given slope = $10^{-3} \text{ mA/V} = 10^{-3} \times 10^{-3} \text{ A/V} = 10^{-6} \text{ S}$ (Siemens).
Plate resistance $r_p = \frac{1}{\text{slope}} = \frac{1}{10^{-6} \text{ S}} = 10^6 \, \Omega$.
$10^6 \, \Omega = 1000 \text{ k}\Omega$.
Since the resistance is calculated at a specific operating point on the curve,it represents the dynamic (or $AC$) resistance of the diode.

(B) Given:
Mutual conductance $(g_m)$ = $2 \times 10^{-3} \, \text{mho}$
Amplification factor $(\mu)$ = $50$
Load resistance $(R_L)$ = $25 \times 10^3 \, \Omega$
First,calculate the plate resistance $(r_p)$ using the relation $\mu = r_p \times g_m$:
$r_p = \frac{\mu}{g_m} = \frac{50}{2 \times 10^{-3}} = 25 \times 10^3 \, \Omega$
The voltage gain $(A_V)$ of a triode amplifier is given by the formula:
$A_V = \frac{\mu \times R_L}{r_p + R_L} = \frac{\mu}{1 + \frac{r_p}{R_L}}$
Substituting the values:
$A_V = \frac{50}{1 + \frac{25 \times 10^3}{25 \times 10^3}} = \frac{50}{1 + 1} = \frac{50}{2} = 25$
Therefore,the voltage gain is $25$.

(B) The current $i$ is given by the rate of flow of charge: $i = n \times e$,where $n = 14 \times 10^{15} \text{ electrons/s}$ and $e = 1.6 \times 10^{-19} \text{ C}$.
$i = (14 \times 10^{15}) \times (1.6 \times 10^{-19}) = 22.4 \times 10^{-4} \text{ A} = 2.24 \times 10^{-3} \text{ A}$.
The power consumed is $P = 448 \text{ mW} = 448 \times 10^{-3} \text{ W}$.
Using the formula $P = V \times i$,we find the voltage $V = \frac{P}{i}$.
$V = \frac{448 \times 10^{-3}}{2.24 \times 10^{-3}} = \frac{448}{2.24} = 200 \text{ V}$.

(C) The amplification factor $\mu$ is defined as the ratio of the change in plate potential to the change in grid potential for a constant plate current.
$\mu = -\left( \frac{\Delta V_p}{\Delta V_g} \right)_{I_p = \text{constant}}$
Given:
Initial plate potential $V_{p1} = 200 \ V$,Final plate potential $V_{p2} = 220 \ V$.
Change in plate potential $\Delta V_p = V_{p2} - V_{p1} = 220 - 200 = 20 \ V$.
Initial grid potential $V_{g1} = -0.5 \ V$,Final grid potential $V_{g2} = -1.3 \ V$.
Change in grid potential $\Delta V_g = V_{g2} - V_{g1} = -1.3 - (-0.5) = -0.8 \ V$.
Since the plate current remains constant,the amplification factor is:
$\mu = -\frac{20}{-0.8} = \frac{20}{0.8} = 25$.

(A) The relationship between the amplification factor $(\mu)$,plate resistance $(r_p)$,and mutual conductance $(g_m)$ is given by: $\mu = r_p \times g_m$.
Rearranging the formula to solve for $g_m$: $g_m = \frac{\mu}{r_p}$.
Given: $\mu = 22$ and $r_p = 6600 \, \Omega$.
Substituting the values: $g_m = \frac{22}{6600} = \frac{1}{300} \, \text{mho}$.

(C) The plate resistance ${r_p}$ of a triode is defined as the ratio of the change in plate voltage to the change in plate current at a constant grid voltage.
${r_p} = \left( \frac{\Delta V_p}{\Delta I_p} \right)_{V_g = \text{constant}}$
Given:
${V_{p1}} = 75 \text{ V}, {I_{p1}} = 2 \text{ mA} = 2 \times 10^{-3} \text{ A}$
${V_{p2}} = 100 \text{ V}, {I_{p2}} = 4 \text{ mA} = 4 \times 10^{-3} \text{ A}$
Calculating the change:
$\Delta V_p = 100 - 75 = 25 \text{ V}$
$\Delta I_p = (4 - 2) \text{ mA} = 2 \text{ mA} = 2 \times 10^{-3} \text{ A}$
Substituting the values:
${r_p} = \frac{25}{2 \times 10^{-3}} = 12.5 \times 10^3 \Omega = 12.5 \text{ k}\Omega$.

(D) triode is an electronic device with three electrodes: a cathode, an anode, and a control grid. The performance of a triode is characterized by three fundamental constants:
$1$. Plate resistance $(r_p)$: The ratio of change in plate voltage to the change in plate current at a constant grid voltage.
$2$. Amplification factor $(\mu)$: The ratio of change in plate voltage to the change in grid voltage at a constant plate current.
$3$. Mutual conductance $(g_m)$: The ratio of change in plate current to the change in grid voltage at a constant plate voltage.
These three parameters are related by the equation $\mu = g_m \times r_p$. Therefore, all the listed options are triode constants.

(A) Mutual conductance $(g_m)$ is defined as the ratio of the change in plate current $(\Delta I_p)$ to the change in grid voltage $(\Delta V_g)$ while keeping the plate voltage constant.
Mathematically, $g_m = \frac{\Delta I_p}{\Delta V_g}$.
Since current is measured in Amperes $(A)$ and voltage in Volts $(V)$, the unit is $A/V$, which is equivalent to Siemens $(S)$ or Mho $(\Omega^{-1})$.
Therefore, the correct unit is Siemen.

(A) The voltage amplification $A_v$ of a triode amplifier is given by the formula: $A_v = \frac{\mu}{1 + \frac{r_p}{R_L}}$, where $\mu$ is the amplification factor, $r_p$ is the plate resistance, and $R_L$ is the load resistance.
For the first case: $25 = \frac{\mu}{1 + \frac{r_p}{50}}$ (where $r_p$ and $R_L$ are in $k\Omega$).
$\Rightarrow \mu = 25 \left(1 + \frac{r_p}{50}\right) = 25 + 0.5 r_p$......$(i)$
For the second case: $30 = \frac{\mu}{1 + \frac{r_p}{100}}$.
$\Rightarrow \mu = 30 \left(1 + \frac{r_p}{100}\right) = 30 + 0.3 r_p$......$(ii)$
Equating $(i)$ and $(ii)$:
$25 + 0.5 r_p = 30 + 0.3 r_p$
$0.2 r_p = 5$
$r_p = \frac{5}{0.2} = 25 \, k\Omega$.

(D) In a triode valve,a cloud of electrons forms near the cathode due to thermionic emission,known as the space charge. This negative space charge repels other electrons emitted from the cathode,thereby limiting the plate current. When a grid is introduced between the cathode and the plate,it can be biased to control the electric field. By applying a positive potential to the grid,it neutralizes the negative space charge,allowing more electrons to reach the plate and thus increasing the plate current.

(C) In a vacuum diode,before reaching the saturation state,the current follows the Child-Langmuir law,which states that the plate current $i_p$ is proportional to the power of the plate voltage $V_p$ (specifically $i_p \propto V_p^{3/2}$).
However,in many textbook contexts for this specific problem type,it is assumed that the diode operates in the space-charge limited region where the relationship is often simplified or evaluated based on the given parameters.
Given $V_1 = 400 \, V$ and $V_2 = 200 \, V$,using the relation $i \propto V^{3/2}$:
$\frac{i_1}{i_2} = \left( \frac{V_1}{V_2} \right)^{3/2} = \left( \frac{400}{200} \right)^{3/2} = (2)^{3/2} = 2\sqrt{2}$.
Thus,the correct ratio is $2\sqrt{2}$.

(C) In the given circuit,the total current flowing from the battery is $1.125 \, A$.
Part of this current flows through the filament circuit,which is given as $1.112 \, A$.
The plate current $I_P$ is the difference between the total current and the filament current.
$I_P = 1.125 \, A - 1.112 \, A = 0.013 \, A$.
To convert this into milliamperes $(mA)$,we multiply by $1000$:
$I_P = 0.013 \times 1000 \, mA = 13 \, mA$.

(A) The coating of strontium oxide on a tungsten cathode reduces the work function of the surface.
According to Richardson-Dushman equation,the thermionic emission current density $J$ is given by $J = AT^2 e^{-\phi/kT}$,where $\phi$ is the work function.
Since the work function $\phi$ decreases due to the oxide coating,the thermionic emission increases significantly at a given temperature.

(A) In a triode valve,the three parameters are the amplification factor $(\mu)$,the mutual conductance $(g_m)$,and the plate resistance $(r_p)$.
These parameters are related by the fundamental equation: $\mu = g_m \times r_p$.
Here,$g_m$ is defined as the change in plate current for a change in grid voltage at a constant plate voltage,and $r_p$ is the change in plate voltage for a change in plate current at a constant grid voltage.

(D) The dynamic plate resistance $(r_p)$ is defined as the ratio of the change in plate voltage $(\Delta V_p)$ to the change in plate current $(\Delta i_p)$.
Given:
$\Delta V_p = 150 \, V - 100 \, V = 50 \, V$
$\Delta i_p = 12 \, mA - 7.5 \, mA = 4.5 \, mA = 4.5 \times 10^{-3} \, A$
Using the formula:
$r_p = \frac{\Delta V_p}{\Delta i_p} = \frac{50}{4.5 \times 10^{-3}} \, \Omega$
$r_p = \frac{50}{4.5} \times 10^3 \, \Omega$
$r_p \approx 11.1 \times 10^3 \, \Omega = 11.1 \, k\Omega$.

(B) In a diode valve,the saturation current is limited by the number of electrons emitted from the cathode per second.
Saturation occurs when all electrons emitted by the cathode are collected by the anode (plate).
This state is reached more easily when the plate voltage is high (to pull all electrons) and the filament current is relatively low (limiting the total emission to a level that the plate can easily collect).

(B) The voltage amplification $A_V$ for a triode valve is given by the formula: $A_V = \frac{\mu R_L}{r_p + R_L}$,where $\mu$ is the amplification factor,$r_p$ is the plate resistance,and $R_L$ is the load resistance.
Given $\mu = 40$ and $R_L = 4 \, k\Omega$.
For the first valve,$r_{p1} = 2 \, k\Omega$,so $A_1 = \frac{40 \times 4}{2 + 4} = \frac{160}{6} = \frac{80}{3}$.
For the second valve,$r_{p2} = 4 \, k\Omega$,so $A_2 = \frac{40 \times 4}{4 + 4} = \frac{160}{8} = 20$.
The ratio of voltage amplification is $\frac{A_1}{A_2} = \frac{80/3}{20} = \frac{80}{3 \times 20} = \frac{4}{3}$.
Wait,re-evaluating the ratio: $\frac{A_1}{A_2} = \frac{r_{p2} + R_L}{r_{p1} + R_L} = \frac{4 + 4}{2 + 4} = \frac{8}{6} = \frac{4}{3}$.

(C) In a vacuum triode,the amplification factor $\mu$ is defined as the ratio of the change in plate voltage to the change in grid voltage for a constant plate current: $\mu = -\left( \frac{\Delta V_p}{\Delta V_g} \right)_{I_p = \text{constant}}$.
This parameter $\mu$ is primarily determined by the physical geometry of the triode (the spacing between the cathode,grid,and plate).
While the transconductance $g_m$ and the plate resistance $R_p$ are dependent on the operating point (plate voltage and grid voltage),the amplification factor $\mu$ remains approximately constant over a wide range of operating conditions.
Therefore,$\mu$ does not vary significantly with plate or grid voltages.

(B) triode valve consists of three electrodes: the cathode,the anode (plate),and the control grid. The control grid is placed between the cathode and the anode. By applying a negative potential to the grid relative to the cathode,it creates an electric field that opposes the flow of electrons from the cathode to the anode. Thus,the grid acts as a gate to control the magnitude of the plate current.

(C) The relationship between amplification factor $(\mu)$,plate resistance $(r_p)$,and mutual conductance $(g_m)$ is given by the formula: $\mu = r_p \times g_m$.
Given: $\mu = 20$ and $g_m = 10^{-3} \text{ mho}$.
Rearranging the formula to solve for plate resistance $(r_p)$: $r_p = \frac{\mu}{g_m}$.
Substituting the values: $r_p = \frac{20}{10^{-3}} = 20 \times 10^{3} \ \Omega = 2 \times 10^{4} \ \Omega$.
Therefore,the plate resistance is $2 \times 10^{4} \ \Omega$.

(B) The amplification factor $\mu$ of a triode is defined as the ratio of the change in plate potential to the change in grid potential for a constant plate current,with a negative sign indicating that the potentials must change in opposite directions to maintain the same current.
$\mu = - \frac{\Delta V_p}{\Delta V_g}$
Given: $\mu = 50$ and $\Delta V_g = -0.20\, V$.
To keep the plate current unchanged,we need to find the required increase in plate potential $\Delta V_p$.
$\Delta V_p = - \mu \times \Delta V_g$
$\Delta V_p = -50 \times (-0.20\, V) = 10\, V$.
Therefore,an increase of $10\, V$ in plate potential is required.

(B) The plate resistance $(r_p)$ of a vacuum diode is defined as the reciprocal of the slope of its plate characteristic curve.
Given slope = $2 \times 10^{-2} \, mA/V$.
To convert the slope to $A/V$ (or $S$): $2 \times 10^{-2} \times 10^{-3} \, A/V = 2 \times 10^{-5} \, A/V$.
Therefore,$r_p = \frac{1}{\text{slope}} = \frac{1}{2 \times 10^{-5}} \, \Omega$.
$r_p = 0.5 \times 10^5 \, \Omega = 50,000 \, \Omega = 50 \, k\Omega$.
Thus,the correct option is $B$.

(A) The voltage amplification $A_v$ of a triode amplifier is given by the formula: $A_v = \frac{\mu R_L}{r_p + R_L}$, where $\mu = g_m \times r_p$.
Given: Transconductance $g_m = 2.5 \times 10^{-3} \, \Omega^{-1}$, Plate resistance $r_p = 20 \times 10^3 \, \Omega$, and Voltage amplification $A_v = 10$.
First, calculate the amplification factor $\mu$: $\mu = g_m \times r_p = (2.5 \times 10^{-3}) \times (20 \times 10^3) = 50$.
Now, substitute the values into the amplification formula: $10 = \frac{50 \times R_L}{20 \times 10^3 + R_L}$.
$10(20 \times 10^3 + R_L) = 50 R_L$.
$200 \times 10^3 + 10 R_L = 50 R_L$.
$40 R_L = 200 \times 10^3$.
$R_L = \frac{200 \times 10^3}{40} = 5 \times 10^3 \, \Omega = 5 \, \text{k}\Omega$.

(C) The voltage gain $A_V$ of a triode amplifier is given by the formula:
$A_V = \frac{\mu R_L}{r_p + R_L} = \frac{\mu}{1 + \frac{r_p}{R_L}}$
Given:
Amplification factor $\mu = 18$
Plate resistance $r_p = 8 \times 10^{3} \ \Omega$
Load resistance $R_L = 10^{4} \ \Omega = 10 \times 10^{3} \ \Omega$
Substituting the values:
$A_V = \frac{18}{1 + \frac{8 \times 10^{3}}{10 \times 10^{3}}} = \frac{18}{1 + 0.8} = \frac{18}{1.8} = 10$
Therefore,the voltage gain is $10$.

(B) In a triode valve, the transconductance $(g_m)$ is defined as the ratio of the change in plate current $(\Delta I_p)$ to the change in grid voltage $(\Delta V_g)$ while keeping the plate voltage $(V_p)$ constant.
Mathematically, this is expressed as: $g_m = \left. \frac{\Delta I_p}{\Delta V_g} \right|_{V_p = \text{const.}}$
Therefore, option $B$ is the correct relation.

(B) In a diode valve,thermionic emission is used to release electrons from the cathode.
To facilitate easier emission of electrons,the cathode must be heated to a high temperature to provide sufficient thermal energy.
Additionally,the work function $\phi$ (the minimum energy required to remove an electron from the surface) should be as low as possible.
Therefore,the cathode temperature must be high and $\phi$ should be low.

(A) The amplification factor $(\mu)$ of a triode is defined as the product of plate resistance $(r_p)$ and mutual conductance $(g_m)$.
Given:
Plate resistance $(r_p)$ = $2.5 \times 10^{4} \ \Omega$
Mutual conductance $(g_m)$ = $2 \times 10^{-3} \ \text{mho}$
Formula:
$\mu = r_p \times g_m$
Calculation:
$\mu = (2.5 \times 10^{4}) \times (2 \times 10^{-3})$
$\mu = 5.0 \times 10^{1} = 50$
Therefore,the amplification factor is $50$.

(C) The amplification factor $\mu$ of a triode is defined as the ratio of the change in plate voltage to the change in grid voltage required to keep the plate current constant.
$\mu = \left( \frac{\Delta V_p}{\Delta V_g} \right)_{i_p = \text{constant}}$
Given:
$\Delta V_p = 225 \, V - 200 \, V = 25 \, V$
$\Delta V_g = 5.75 \, V - 5 \, V = 0.75 \, V$
Substituting the values:
$\mu = \frac{25}{0.75} = \frac{2500}{75} = 33.33$
Thus,the amplification factor is $33.3$.

(A) Transconductance $(g_m)$ is defined as the ratio of the change in anode current $(\Delta I_p)$ to the change in grid potential $(\Delta V_g)$ at a constant anode potential $(V_p)$.
$g_m = \left( \frac{\Delta I_p}{\Delta V_g} \right)_{V_p = \text{constant}}$
Given:
Initial current $I_1 = 7.5 \, mA$, Initial grid potential $V_{g1} = -1.2 \, V$
Final current $I_2 = 5.5 \, mA$, Final grid potential $V_{g2} = -2.2 \, V$
Change in current $\Delta I_p = I_1 - I_2 = 7.5 \, mA - 5.5 \, mA = 2.0 \, mA$
Change in grid potential $\Delta V_g = V_{g1} - V_{g2} = -1.2 \, V - (-2.2 \, V) = 1.0 \, V$
$g_m = \frac{2.0 \, mA}{1.0 \, V} = 2 \, mA/V = 2 \, m \, mho$.

(D) The relationship between amplification factor $\mu$, plate resistance $r_p$, and mutual conductance $g_m$ is given by the formula: $\mu = r_p \times g_m$.
Given: $\mu = 20$ and $r_p = 10 \, k\Omega = 10 \times 10^3 \, \Omega = 10^4 \, \Omega$.
Rearranging the formula to solve for $g_m$: $g_m = \frac{\mu}{r_p}$.
Substituting the values: $g_m = \frac{20}{10 \times 10^3} = \frac{20}{10000} = 2 \times 10^{-3} \, \text{mho}$ (or Siemens).
Therefore, the correct option is $D$.