ROUTERA


Electric Current Through Gases

Class 12th Concepts Of Physics Part 2 HC Verma Solution



Short Answer
Question 1.

Why is conduction easier in gases if the pressure is low? Will the conduction continue to improve if the pressure is made as low as nearly zero?


Answer:

Conduction happens when an electron with enough kinetic energy collides with the electron of gas atoms. This process is called as ionization of gas atoms. This phenomenon is reported at low pressure, as at high pressure the mean free path of electron decrease which results in a collision of electron much before they get enough kinetic energy for ionization. Whereas at low pressure due to an increase in their mean free path they are able to gain enough kinetic energy for ionization process so conduction is easier. If the pressure is made too low, there won’t be enough atoms to ionize and hence conductivity will decrease.



Question 2.

An AC source is connected to a diode and a resistor in series. Is the current through the resistor AC or DC?


Answer:

When an AC source is connected to a diode half rectification of the source will happen and we will get a pulsating DC output.


The resistor in series would be getting a pulsating DC output. The given circuit in question is analogous to half rectification of AC source by diode, in that circuit we use resistor for measuring the load of DC.



Question 3.

How will the thermionic current vary if the filament current is increased?


Answer:

Thermionic current will increase if the filament current is increased as due to increase in temperature in filament. It is due to internal resistance through which the temperature of filament increases. This increase in temperature will give enough kinetic energy to electrons to leave the surface of filament and it will lead to increase in number of thermions number leaving per unit surface area of filament and hence thermionic current will increase.



Question 4.

Would you prefer a material having a high melting point or a low melting point to be used as a cathode in a diode?


Answer:

We know that electrons which are responsible for thermionic current will leave the surface at high temperature. Therefore, cathode material should be able to handle high temperature which will only be possible if the material has a high melting point.



Question 5.

Would you prefer a material having a high work function or a low work function to be used as a cathode in a diode?


Answer:

Work function’s magnitude accounts for how much energy we need to provide for the emission of electron from the surface. Therefore, we would prefer material having low work function as less energy will be required for emission.



Question 6.

An isolated metal sphere is heated to a high temperature. Will it become positively charged due to thermionic emission?


Answer:

When an isolated metal sphere is heated to high temperature it will give enough kinetic energy to electrons and hence will the surface. This will decrease the number of electrons on the surface of metal sphere. And due this reason the sphere will become positively charged.



Question 7.

A diode valve is connected to a battery and a load resistance. The filament is heated so that a constant current is obtained in the circuit. As the cathode continuously emits electrons, does it get more and more positively charged?


Answer:

The important thing to note in this question is that diode’s cathode is connected to battery’s negative terminal. When filament is heated it will continuously emit electrons from it but battery’s negative terminal will also provide electrons which makes the diode electrically neutral in nature



Question 8.

Why does thermionic emission not take place in non-conductors?


Answer:

No, non-conductors will not take part in thermionic emission due to a high work function. Also, Non-conductors do not have free electrons present on its surface hence no electrons are able to emit from it. Whereas in conductors, due to the presence of free-electrons on its surface it accounts for thermionic emission.



Question 9.

The cathode of a diode valve is replaced by another cathode of double the surface area. Keeping the voltage and temperature conditions the same, will the plate current decrease, increase or remain the same?


Answer:

The plate current will increase. We know the more the surface area of cathode more number of electrons will be present on it and hence probability of electrons leaving the surface also increases which in turn increase the plate’s current.



Question 10.

Why is the linear portion of the triode characteristic chosen to operate the triode as an amplifier?


Answer:

We know that in the linear portion of the triode characteristic graph, it has the least amount of distortion as comparted to other region of graph. This gives an advantage that voltage across the resistor will vary accordingly to input signal and amplification can be achieved easily without risking the life of vacuum tube.




Objective I
Question 1.

Cathode rays constitute a stream of
A. electrons

B. protons

C. positive ions

D. negative ions


Answer:

Cathode is responsible for thermionic emission and thermionic emission is due to electrons which leave its surface. Therefore, rays should constitute a stream of electrons.


Ions are present due to gases but they do not account for rays from cathode surface as only electrons are the one leaving the surface.


Question 2.

Cathode rays are passing through a discharge tube, in tube, there is
A. an electric field but no magnetic field.

B. a magnetic field but no electric field

C. an electric as well as a magnetic field

D. neither an electric nor a magnetic field


Answer:

We know that cathode rays constitute a stream of electrons which are responsible for thermionic emission. Therefore, field will be created hence option D should be ruled out. We know it is due to charge magnetic field is created. We also know that magnetic field and electric field complement each other. Magnetic and electric field concepts are relative to each other, that it depends on which frame of reference you are taking about hence both fields are created inside tube


Question 3.

Let i0 be the thermionic current from a metal surface when the absolute temperature of the surface is T0. The temperature is slowly increased and the thermionic current is measured as a function of temperature. Which of the following plots may represent the variation in (i/i0) against (T/T0)?




Answer:

We know that current in thermionic emission varies to square of Temperature. of cathode with exponential term



The best suited curve from the figures which satisfy the equation is the parabolic one.


Question 4.

When the diode shows saturated current, dynamic plate resistance is
A. zero

B. infinity

C. indeterminate

D. different for different diodes


Answer:

We know the dynamic plate resistance is given by



Now when diode is at saturated current region the change in ∆IP will be very small or it will be near to zero, therefore dynamic resistance will at infinity.


It can’t be indeterminate as although it will saturate current but still changes in decimal places will be still be visible and hence ∆IP will not be equal to zero.


Question 5.

The anode of a thermionic diode is connected to the negative terminal of a battery and the cathode to its positive terminal.
A. No appreciable current will pass through the diode.

B. A large current will pass through the diode from the anode to the cathode.

C. A large current will pass through the diode from the cathode to the anode.

D. The diode will be damaged.


Answer:

When cathode is connected to positive terminal and anode to negative terminal electrons will not able to leave the surface of cathode and hence no thermionic emission will be accounted. This will result in no current passing through diode.


Question 6.

A diode, a resistor and a 50 Hz AC source are connected in series. The number of current pulses per second through the resistor is
A. 25

B. 50

C. 100

D. 200


Answer:


We can see from figure that in one second 1 pulse will pass through when 1Hztherefore total of 50 pulses will pass through in one second


Question 7.

A triode is operated in the linear region of its characteristics. If the plate voltage is slightly increased, the dynamic plate resistance will
A. increase

B. decrease

C. remain almost the same

D. become zero


Answer:

We know the formula for dynamic plate resistance is



When triode is operating in linear region, and plate voltage is slightly changed it will lead to change in current of plate and hence both of these effect will cancel each other and no change in resistance will be observed.


Question 8.

The plate current in a triode valve is maximum when the potential of the grid is
A. positive

B. zero

C. negative

D. non-positive


Answer:


We know electrons get attracted to positive potential. When potential of the grid is made positive it makes easier for us to push electrons from cathode to anode which accounts for increase in current. Hence plate current in a triode valve is maximum when the potential of the grid is made positive


Question 9.

The amplification factor of a triode operating in the linear region depends strongly on
A. the temperature of the cathode

B. the plate potential

C. the grid potential

D. the separations of the grid from the cathode and the anode.


Answer:

When the triode operating in the linear region, it indirectly


implies that the grid voltage, plate voltage and plate current is


specified and hence plays no contribution to the amplification factor in this question


Now if the grid is too near the cathode it will partially or completely lie in space charge region which comprises of electrons from the cathode. And hence it will pick electrons from cathode more rapidly which will increase the plate current and hence amplification factor will get affected. As amplification factors depend on the plate’s current.


And when the grid is far away from cathode, it will able to pick as many electrons compared to when it was near to cathode and hence amplification factor will get affected



Objective Ii
Question 1.

Electric conduction takes place in a discharge tube due to the movement of
A. positive ions

B. negative ions

C. electrons

D. protons


Answer:

Cathode ray which comprises of electrons is mainly responsible for electric conduction. But ions are also present in gases which play a role in conduction by getting ionized by colliding with electrons which are present in cathode rays. This process is called ionization which also contributes to electric conduction. Therefore, both positive and negative ions with electrons are responsible for conduction


Question 2.

Which of the following are true for cathode ray?
A. It travels along straight lines.

B. It emits X-ray when strikes a metal

C. It is an electromagnetic wave

D. It is not deflected by magnetic field.


Answer:

Cathode rays are wave but due to their massless property they are not able to travel at speed which is equal to light hence they are not categorized under Electromagnetic wave.


Cathode consists of Electrons and they will travel from cathode to anode by the shortest route which is obliviously a straight line until any other forces are applied across the tube. In presence of magnetic field, they will deviate from their usual path.


When cathode accelerated to the anode they produce X-rays, which is in accordance to Maxwell's principle that accelerating charge emit radiation when it strikes the surface of metal.


Question 3.

Because of the space charge in a diode valve,
A. the plate current decreases

B. the plate voltage increases

C. the rate of emission of thermions increases

D. the saturation current increases.


Answer:

Space charge is three dimensional concept. It is a cloud of electrons which are distributed evenly over the 3D space rather than being at one place. As we know cathode emits electrons and this electron take some finite amount of time to travel and due to this a region near cathode is created which create repulsion for electrons which are coming cathode. This phenomenon is called space charge effect. This decrease the plate’s current.


This Space charge has no effect on the rate of emission of thermions, as rate depend on how much energy we are providing to it. Also it is not related to saturation current or plate’s voltage.


Question 4.

The saturation current in a triode valve can be changed by changing
A. the grid voltage

B. the plate voltage

C. the separation between the grid and the cathode

D. the temperature of the cathode


Answer:


Grid Voltage when made negative, it will not allow the electrons from cathode to pass through, as electrons will start repealing from it hence this will decrease the electric conductivity. Whereas when grid voltage is made positive it will start attracting electrons and make it easier for electrons to leave the surface of cathode. Therefore, through grid voltage we are indirectly controlling the plate’s current. Also in triode valve we know that the maximum current is the saturation current, hence by controlling the grid voltage we are in a way controlling the saturation current. Whereas separation won’t make any difference as it is the voltage which is actually plays an important role in controlling the plate’s current. Changing the separation will only effect the amplification.


No matter what is the temperature of cathode is, at the end it depends whether grid allows the electrons to pass through to it or not. Therefore, it is the grid voltage which effects the saturation current.


Question 5.

Mark the correct options.
A. A diode valve can be used as a rectifier

B. A triode valve can be used as a rectifier

C. A diode valve can be used an amplifier

D. A triode valve can be used as an amplifier


Answer:

We know triode has a property through which output can be controlled and it is due to this property triode is used as amplifier. This property of manipulating output is not present in diode therefore it can’t be used as amplifier.


Also, it is well known fact that diode and triode allow only in single direction flow of current through it and hence they act as rectifier for AC input and convert it to pulsating DC.


Question 6.

The plate current in a diode is zero. It is possible that
A. the plate voltage is zero

B. the plate voltage is slightly negative

C. the plate voltage is slightly positive

D. the temperature of the filament is low


Answer:

A. the plate voltage is zero


B. the plate voltage is slightly negative


C. the plate voltage is slightly positive


D. the temperature of the filament is low


When the temperature of the filament is low, then there is a possibility that not enough kinetic energy is being produced and due to these electrons may not able to leave the surface of the cathode and hence zero current in a diode. When plate voltage is zero it means indirectly that current is zero as we know both are in direct relation to each other. And when the plate’s voltage is negative with respect to cathode no electrons will leave the surface and hence zero current. Now when the voltage is slightly positive, it may be that it is not greater than forward voltage and hence electrons are not able to accelerate towards the anode. As we know around cathode a cloud of electrons formed and when voltage is not enough for electrons, they will not able to reach the anode.


Question 7.

The plate current in a triode valve is zero. The temperature of the filament is high. It is possible that
A. Vg > 0, Vp > 0

B. Vg > 0, Vp < 0

C. Vg < 0, Vp > 0

D. Vg < 0, Vp < 0


Answer:

B. Vg > 0, Vp < 0


C. Vg < 0, Vp > 0


D. Vg < 0, Vp < 0


Here Vg means voltage grid and Vp means positive plate’s voltage. Now it is already given in question that temperature of the filament is high therefore electrons have enough kinetic energy to leave the surface of cathode.


But still even if grid voltage is negative, electrons will not able to reach the surface of anode and hence no current will be visible Also if positive plate’s voltage is negative, then also electrons won’t be able to reach the anode’s surface. Therefore, current will be zero even if one of the above conditions are met.



Exercises
Question 1.

A discharge tube contains helium at a low pressure. A large potential difference is applied across the tube. Consider a helium atom that has just been ionized due to the detachment of an atomic electron. Find the ratio of the distance travelled by the free electron to that by the positive ion in a short time dt after the ionization.


Answer:

Let the distance travelled by free electron and positive ion be SE and SH respectively. Also, mass of the electron and positive helium ion be me and mh where,

me=9.1×10-31kg


mh=4×mass of proton


=4×1.6×10-27kg


We know,



where s is the distance travelled, u is the initial velocity, t is the time and a is the magnitude of acceleration


Here, initial velocity of electron and helium ion is zero




Now, magnitude of force experienced by electron,





where F is the magnitude of force, E is the electric field strength and q is the magnitude of charge of the particle.


∴ Distance travelled by free electron for dt duration is


Magnitude of force experienced by positive helium ion,





∴ Distance travelled by positive helium ion for dt duration is


The required ratio is








Question 2.

A molecule of a gas, filled in a discharge tube, gets ionized when an electron is detached from it. An electric field of 5.0 kV m–1 exists in the vicinity of the event.

(a) Find the distance travelled by the free electron in 1 μs assuming no collision.

(b) If the mean free path of the electron is 1.0 mm, estimate the time of transit of the free electron between successive collisions


Answer:

a) Given electric field E= 5 kV m-1 = 5×103 V m-1


Time t=1 μs = 1×10-6 s.


Let m be the mass of electron and q be the charge of electron where m=9.1×10-31 kg


q =1.6×10-19 C


We know,





where F is the force, E is the electric field strength and q is the


magnitude of charge.


Also, distance travelled by an electron is


(as initial velocity is 0)





b) Here mean free path S travelled by electron is given as S=1mm


=10-3 m


We know,



Also,








Question 3.

The mean free path of electrons in the gas in a discharge tube is inversely proportional to the pressure inside it. The Crookes dark space occupies half the length of the discharge tube when the pressure is 0.02 mm of mercury. Estimate the pressure at which the dark space will fill the whole tube.


Answer:

Let the mean free path be L and pressure be P


Given,
where L = half of the tube length and P = 0.02 mm of Hg


∴ When P becomes half, L doubles, i.e. the whole tube is filled with Crook’s dark space.


Hence the required pressure



Question 4.

Two discharge tubes have identical material structure and the same gas is filled in them. The length of one tube is 10 cm and that of the other tube is 20 cm. Sparking starts in both the tubes when the potential difference between the cathode and the anode is 100 V. If the pressure in the shorter tube is 1.0 mm of mercury, what is the pressure in the longer tube?


Answer:

Let d1 and p1 be the length and pressure of short tube. Also d2 and p2 be the length and pressure of long tube.i.e. p1=1.0 mm d1=10 cm, d2=20 cm

According to Paschen’s law,


, i.e. the sparking potential of a gas in a discharge tube is the function of the product of pressure of the gas and separation between the electrodes.





Question 5.

Calculate n(T)/n (1000 K) for tungsten emitter at T = 300 K, 2000 K and 3000 K where n(T) represents the number of thermions emitted per second by the surface at temperature T. Work function of tungsten is 4.52 eV.


Answer:

According to Richardson-Dushman Equation thermionic current i is


where, n is the thermions emitted, S is the surface area, T is the absolute temperature, k is the Boltzmann constant, A is the constant depend on nature of metal and ϕ is the work function.


Case 1


T=300 K


Now,


Case 2


T=2000 K


Now,


Case 3


T=3000 K


Now,



Question 6.

The saturation current from a thoriated-tungsten cathode at 2000 K is 100 mA. What will be the saturation current for a pure-tungsten cathode of the same surface area operating at the same temperature? The constant A in the Richardson-Dushman equation is 60 × 104 A m-2K-2 pure tungsten and 3.0 × 104 A m-2K-2 for thoriated tungsten. The work function of pure tungsten is 4.5 eV and that of thoriated tungsten is 2.6 eV.


Answer:

According to Richardson-Dushman Equation thermionic current i is


where, n is the thermions emitted, S is the surface area, T is the absolute temperature, k is the Boltzmann constant, A is the constant depend on nature of metal and ϕ is the work function.


Given,


Ittungsten=100 mA


T=2000 K


Attungsten=3 × 104 A m-2 K-2


Aptungsten=60× 104 A m-2 K-2


Φttungsten =2.6 eV


Φptungsten =4.5 eV


Now,







Question 7.

A tungsten cathode and a thoriated-tungsten cathode have the same geometrical dimensions and are operated at the same temperature. The thoriated-tungsten cathode gives 5000 times more current than the other one. Find the operating temperature. Take relevant data from the previous problem.


Answer:

According to Richardson-Dushman Equation thermionic current i is


where, n is the thermions emitted, S is the surface area, T is the absolute temperature, k is the Boltzmann constant, A is the constant depend on nature of metal and ϕ is the work function.


Given,


Attungsten=3 × 104 A m-2 K-2


Aptungsten=60× 104 A m-2 K-2


Φttungsten =2.6 eV


Φptungsten =4.5 eV


Ittungsten=5000Iptungsten





Taking log on both sides,



K



Question 8.

If the temperature of a tungsten filament is raised from 2000 K to 2010 K, by what factor does the emission current change? Work function of tungsten is 4.5 eV.


Answer:

According to Richardson-Dushman Equation thermionic current i is


where, n is the thermions emitted, S is the surface area, T is the absolute temperature, k is the Boltzmann constant, A is the constant depend on nature of metal and ϕ is the work function


Given,T1=2000 K


T2=2010 K , Φ =4.5 eV


Now,



Required factor is




Question 9.

The constant A in the Richardson-Dushman equation of tungsten is 60 × 104 A m-2 K-2. The work function surface area 2.0 × 10–5 m2 is heated by a 24 W electric equals the energy input by the heater and the temperature becomes constant. Assuming that the cathode radiates like a blackbody, calculate the saturation current due to thermions. Take Stefan constant =6 × 10-8 W m-2 K-4. Assume that the thermions take only a small fraction of the heat supplied.


Answer:

Given

A=60 × 104 A m-2 K-2


S=2.0 × 10–5 m2


P=24 W


=6 × 10-8 W m-2 K-4


The power radiated by Stefen’s law is given by






According to Richardson-Dushman Equation thermionic current i is



where, n is the thermions emitted, S is the surface area, T is the absolute temperature, k is the Boltzmann constant, A is the constant depend on nature of metal and ϕ is the work function






Question 10.

A plate current of 10 mA is obtained when 60 volts are applied across a diode tube. Assuming the Langmuir-Child equation to hold, find the dynamic resistance rp in this operating condition.


Answer:

Given,


where ip is the plate current and Vp is the plate voltage


V= 60 volts


I =10 mA


(1) (k is a constant)


Taking derivative on both sides with respect to ip


(2)


Dividing (2)/(1)





Question 11.

The plate current in a diode is 20 mA when the plate voltage is 50 V or 60 V. What will be the current if the plate voltage is 70V?


Answer:

Given the plate current is 20 mA for 50 V and 60 V, i.e. 20 mA is the saturation current.

∴, for a given temperature current remains same for all voltages. . i.e. for 70 V current is 20 mA.



Question 12.

The power delivered in the plate circuit of a diode is 1.0 W when the plate voltage is 36 V. Find the power delivered if the plate voltage is increased to 49 V. Assume Langmuir – Child equation to hold.


Answer:

Given P1=1 W, VP1= 36 V, VP2= 49 V

We know,


According to Langmuir-Child equation


where ip is the plate current and Vp is the plate voltage


Now,



=.04411×49=2.16


W=2.2 W



Question 13.

A triode valve operates at Vp = 225 V and Vg = –0.5 V. The plate current remains unchanged if the plate voltage is increased to 250 V and the grid voltage is decreased to –2.5 V. Calculate the amplification factor.


Answer:

Given VP1=225 V, VP2=250 V, Vg1=-0.5, Vg2=-2.5 V


where VT is the change in plate voltage and VT is the change in grid voltage





Question 14.

Calculate the amplification factor of a triode valve which has plate resistance of 2 kΩ and transconductance of 2 millimho.


Answer:

Given rP = 2 KΩ = 2× 103 (plate resistance)

gm =2 millimho = 2× 10-3 mho(transconductane)


Amplification factor for triode



Question 15.

The dynamic plate resistance of a triode valve is 10 kΩ. Find the change in the plate current if the plate voltage is changed from 200 V to 220 V.


Answer:

Given rp=10 KΩ = 104Ω, VP1=200V, VP2=220 V

The dynamic resistance of triode is defined as,



where ΔVp is the change in plate voltage and Vg is the grid voltage and ΔIp is the change in plate current




Question 16.

Find the values of rp, μ and gm of a triode operating at plate voltage 200 V and grid voltage –6 V. The plate characteristics are shown in figure.




Answer:

The dynamic resistance of triode is defined as,


where ΔVp is the change in plate voltage and Vg is the grid voltage and ΔIp is the change in plate current


From figure, considering the line Vg=-6 V, we get two value of VP i.e.


VP1=160 V , VP2=240 V , Also ip1=3 mA ip2=13mA


Substituting values in equation we get,



The mutual inductance gm of a triode valve is defined as



Considering two points in the graph where Vp=200 V we get ip1=13 mA ip2=3mA Vg1=-4 V, Vg2=-8 V


Substituting values in equation we get.





Question 17.

The plate resistance of a triode is 8 kΩ and the transconductance is 2.5 millimho.

(a) If the plate voltage in increased by 48 V, and the grid voltage is kept constant, what will be the increase in the plate current?

(b) With plate voltage kept constant at this increased value, how much should the grid voltage be decreased in order to bring the plate current back to its initial value?


Answer:

Given, rp=8 kΩ, gm=2.5 millimho

a) CASE 1: △VP=48 V, Vg= constant


The dynamic resistance of triode is defined as,



where ΔVp is the change in plate voltage and Vg is the grid voltage and ΔIp is the change in plate current


substituting the values




b) CASE 2: gm=2.5 millimho=.0025 mho, △Ip =6 mA =.006 A


The mutual inductance gm of a triode valve is defined as






Question 18.

The plate resistance and the amplification factor of a triode are 10 kΩ and 20. The tube is operated at plate voltage 250 V and grid voltage –7.5 V. The plate current is 10 mA

(a) To what value should the grid voltage be changed so as to increase the plate current to 15 mA?

(b) To what value should the plate voltage be changed to take the plate current back to 10 mA?


Answer:

Given, rp=10kΩ, μ=20, VP1=250 V. Vg1=-7.5 V, ip1=10 mA


a) CASE 1: ip2=15 mA. VP= Constant, Vg2=? ip1=10 mA


We know,


where rp is the plate resistance and gm is the trans-conductance


mho


Now,



where ΔVg is the change in grid voltage and Vp is the plate voltage and ΔIp is the change in plate current






b)CASE 2: VP3=?,ip3=10 mA, ip2=15mA, ip3=10mA


The dynamic resistance of triode is defined as,



where ΔVp is the change in plate voltage and Vg is the grid voltage and ΔIp is the change in plate current




-


V, required voltage



Question 19.

The plate current, plate voltage and grid voltage of a 6F6 triode tube are related as

Ip = 41(Vp + 7 Vg) 1.41, Where Vp and Vg are in volts and ip in microamperes. The tube is operated at Vp = 250 V, Vg = –20 V. Calculate

(a) the tube current,

(b) the plate resistance,

(c) the mutual conductance and

(d) the amplification factor.


Answer:

a) Given, VP = 250 V, Vg = –20 V


Also



b) Differentiating the given equation with respect to ip keeping Vg constant.



The dynamic resistance of triode is defined as,



c) Differentiating the given equation with respect to Vg keeping Vp constant




d)


where rp is the plate resistance and gm is the trans-conductance





Question 20.

The plate current in a triode can be written as

Show that the mutual conductance is proportional to the cube root of the plate current.


Answer:

Given:



gm is the mutual conductance, ip is the plate current, Vp is the plate current and Vg is the grid voltage.



Taking cube on both sides,



hence proved



Question 21.

A triode has mutual conductance = 2.0 millimho and plate resistance = 20 kΩ. It is desired to amplify a signal by a factor of 30. What load resistance should be added in the circuit?


Answer:

Given gm=2millimho=2×10-3mho, rP=20 kΩ, A=30


Here,


where rp is the plate resistance and gm is the trans-conductance



We know,


where A is the voltage gain, μ is the amplification factor, RL is the load resistance and rp is the plate resistance





Question 22.

The gain factor of an amplifier is increased from 10 to 12 as the load resistance is changed from 4 kΩ to 8 KΩ. Calculate (a) the amplification factor and (b) the plate resistance.


Answer:

…(i)


where A is the voltage gain, μ is the amplification factor, RL is the load resistance and rp is the plate resistance


Given:


A1=10, Rl1 = 4kΩ


A2=12, Rl2 = 8kΩ


Putting these two values in equation (i)




We get a linear equation in 2 variables (rp and μ)



…(ii)


We will get another equation by substituting the values of A2 and Rl2 is equation (i)




… (iii)


Solving (ii) and (iii) to find the values amplification factor and the plate resistance.


&


From (ii)


…(iv)


Substituting (iv) in (iii)





Using μ =15, the value of rp is 2000Ω or 2 kΩ



Question 23.

Figure shows two identical triode tubes connected in parallel. The anodes are connected together, the grids are connected together and the cathodes are connected together. Show that the equivalent plate resistance is half to the individual plate resistance. The equivalent mutual conductance is double the individual mutual conductance and the equivalent amplification factor is the same as the individual amplification factor




Answer:

Given the anodes are connected together, the grids are connected together and the cathodes are connected together i.e. two triodes have same voltage and same current which means rP1 = rP2=r


Here the equivalent resistance,



i.e., the equivalent plate resistance is half to the individual plate resistance.


Let gm1 and gm2 be the individual conductance.


From figure, gm1 = gm1 = g. As two triodes are parallel i.e. equivalent conductance



i.e. The equivalent mutual conductance is double the individual mutual conductance


Now,


Also,



I.e. the equivalent amplification factor is the same as the individual amplification factor