
Chapter 41 – Electric Current through Gases
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Question 1:
Why is conducting 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:
When electrons move through a gas, they collide with the gaseous particles and lose energy. This increases the resistance and, hence, reduces the current. But at low pressure, as the gas particle are widely spread, there are fewer collisions. So, the electrons can pass easily and with less collisions.
If the pressure is reduced to nearly zero, the current through the gas will decrease. This is because the mean freepath of the electrons (distance that the electrons travel between collisions) is longer and they can, therefore, be accelerated to higher speeds before collision with an atom and they have more chance of causing ionisation.
Question 2:
An AC source is connected to a diode and a resistor in series. Is the current thorough the resistor AC or DC?
Answer:
If the diode and resistor are in series, for the positive half cycle of AC, the current through the resistor will be DC. But for the next half cycle the current through the resistor will be zero. As a diode is a device that converts AC into DC, current through the resistor will be DC.
Question 3:
How will the thermionic current vary if the filament current is increased?
Answer:
If the filament current is increased, it will increase the temperature of the cathode (metal) and the cathode will emit more electrons. This will lead to an increase in the number of thermions emitted per unit time. As a result, the thermionic current will increase.
Question 4:
Would you prefer a material with a high melting point or a low melting point to be used as a cathode in a diode?
Answer:
We will prefer a material with high melting point to be used as the cathode in a diode. The material of the cathode of a diode should be resistant to high temperature, have high melting point and be electrically conductive because thermionic emission occurs at high temperature.
Question 5:
Would you prefer a material with a high workfunction or a low workfunction to be used as a cathode in a diode?
Answer:
We will prefer a material with low workfunction to be used as a cathode in a diode, so that electron emission can occur using a small amount of energy.
Question 6:
An isolated metal sphere is heated to a high temperature. Will it become positively charged due to thermionic emission?
Answer:
Yes, it will become positively charged due to thermionic emission. When the metal sphere is heated, average kinetic energy of the electrons will increases, due to which, the free electrons of the metal sphere will be able to escape. As a result, they will leave a positive charge on the isolated metal sphere.
Question 7:
A diode value 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 become more and more positively charged?
Answer:
No, the cathode does not become more and more positively charged. In the diode circuit, the cathode of the diode valve is always connected to the negative terminal of the battery. So, the battery will supply the electrons to the cathode and the cathode will continuously emit electrons without becoming more and more positively charged.
Question 8:
Why does thermionic emission not take place in nonconductors?
Answer:
For thermionic emission, material should have low work function and large number of free electrons. But nonconductor does not have free electrons and they have higher work function. So, thermionic emission does not takes place in nonconductors.
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 place current decrease, increase or remain the same?
Answer:
As the rate of emission of thermions is directly proportional to the surface area of the surface emitting it, the plate current doubles if the cathode of a diode valve is replaced by another cathode of double the surface area.Hence plate current increases.
Question 10:
Why is the linear portion of the triode characteristic chosen to operate the triode as an amplifier?
Answer:
When the operating point lies on the linear portion of the characteristics curve, change in voltage across the load resistance follows the pattern of the input signal, but the amplitude is much larger. This can be done by choosing the linear portion of triode characteristic.
Question 1:
Cathode rays constitute a stream of
(a) electrons
(b) protons
(c) positive ions
(d) negative ions
Answer:
(a) electrons
Cathode rays consist of a stream of fast moving electrons.
Question 2:
Cathode rays are passing through a discharge tube. In the 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:
(c) an electric as well as a magnetic field
Cathode rays consist of beams of electrons that constitute current and, hence, magnetic field. We know electric field is produced by a charge, whether it is stationary or moving, So, electric field will also be present inside the tube.
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Question 3:
Let i_{0} be the thermionic current from a metal surface when the absolute temperature of the surface is T_{0}. 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/i_{0}) against (T/T_{0})?
Answer:
(d)
Since the thermionic current is directly proportional to the square of the temperature of the surface emitting thermions, the graph is parabolic.
Question 4:
When the diode shows saturated current, dynamic place resistance is
(a) zero
(b) infinity
(c) indeterminate
(d) different for different diodes
Answer:
(b) infinity
Saturated current is the maximum current beyond which, there is no effect of plate voltage (V_{P})
on the plate current (I_{P}). At this stage, dynamic resistance is infinite.
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:
(a) No appreciable current will pass through the diode.
If the anode is given a negative potential relative to the cathode, the electrons are pushed back to the cathode. Hence, no current will flow through the 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:
(b) 50
The number of current pulses is equal to the frequency of the AC source because one current pulse passes through the diode for one oscillation of the AC source.
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:
(c) remain almost the same
Dynamic resistance (r_{P}) is given by,
r_{P} =
∆VP∆iPThe triode is operating in the linear region. Therefore,
Change in the value of voltage = Change in the value of current
So, (r_{P}) will remain almost the same.
Question 8:
The plate current in a triode value is maximum when the potential of the grid is
(a) positive
(b) zero
(c) negative
(d) nonpositive
Answer:
(a) positive
If the grid voltage is made positive, it will help the electrons move towards the anode, which will help in increasing the current. Thus, the plate current in the triode value is maximum when the potential of the grid is 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:
(d) the separations of the grid from the cathode and the anode
When the triode operating in the linear region, value of plate voltage, grid voltage and plate current is already specified (or fixed). If the grid is near the cathode, then we assume that grid lie in the region where the space charge density is more. There is a chance that grid will pick up electrons from the space charge region of cathode and effect the plate current and hence amplification factor (as we know that amplification factor will depend on the plate current). If the grid is far from the cathode, it will reduce the grid voltage and hence the amplification factor also get affected.
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:
(a) positive ions
(b) negative ions
(c) electrons
Some ions are always present in gases due to cosmic rays and other factors. When we apply potential difference across a discharge tube, the ions get accelerated due to the electric field. If the potential difference is large, then the ions get enough energy to ionise the molecules on collision.
Thus, a large number of ions are produced and conduction starts. Generally, an electron gets detached from a molecule to make the molecule a positive ion. At low pressure, this electron moves through a large distance and gets attached to another molecule and forms a negative ion.
Thus, electric conduction takes place in a discharge tube due to the movement of positive ions, negative ions and electrons.
Question 2:
Which of the following is true for a cathode ray?
(a) It travels in a straight line.
(b) It emits an Xray when it strikes a metal.
(c) It is an electromagnetic wave.
(d) It is not deflected by a magnetic field.
Answer:
(a) It travels in a straight line.
(b) It emits an Xray when it strikes a metal.
A cathode ray travels in a straight line. When cathode rays strike a solid metal, Xrays are emitted from the metal. When a cathode ray tube is brought in a magnetic field, the path of cathode rays gets deflected. Cathode rays are the matter waves that consists of electrons thus they are not electromagnetic waves.
Question 3:
Due to 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:
(a) the plate current decreases
A diode valve consists of a negativelycharged region between the cathode and anode, called the space charge region. This negativelycharged region repels the electrons coming from the cathode and, thus, it reduces the plate current. On the other hand, this region has no effect on plate voltage, rate of emission of thermions and the saturation current.
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:
(a) the grid voltage
Since the triode value consists of a space charge region near the cathode, this space charge region is affected by the grid voltage. If the grid voltage is made negative, it repels the electrons coming from the cathode and reduces current. If this voltage is made positive, it helps the electrons move towards the anode. Hence, this voltage is more effective in changing the plate current. Since, the saturated current in the triode is the maximum value of plate current, the saturated current in the triode valve is effectively controlled by the grid voltage. More negative is the grid voltage, less is the value of saturated 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 as an amplifier.
(d) A triode valve can be used as an amplifier.
Answer:
(a) A diode valve can be used as a rectifier.
(b) A triode valve can be used as a rectifier.
(d) A triode valve can be used as an amplifier.
A diode valve and a triode valve allow current to flow only in one direction. Since a rectifier is a device that converts alternating current (bidirectional) into direct current (unidirectional), a diode valve and a triode valve can be used as rectifiers.
A triode valve can control its output in proportion to the input signal. That is, it can act as an amplifier, whereas a diode valve cannot control its output in proportion to the input signal. So, it cannot be use as an amplifier.
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
The plate current varies directly with the plate voltage. Therefore, if the plate voltage is zero, the plate current is also zero. Due to the same reason, if the plate voltage is negative, the plate current will be zero. Now, if the plate is slightly positive, then it may be the reason that the plate voltage is not able to reduce the effect of space charge. Hence, the current may be zero. Now, as the temperature of the filament is low, it will not be able to emit electrons and the resulting plate current will be zero. Hence, all the options are possible.
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Question 7:
The plate current in a triode valve is zero. The temperature of the filament is high. It is possible that
(a) V_{g} > 0, V_{p} > 0
(b) V_{g} > 0, V_{p} < 0
(c) V_{g} < 0, V_{p} > 0
(d) V_{g} < 0, V_{p} < 0
Answer:
(b) V_{g} > 0, V_{p} < 0
(c) V_{g} < 0, V_{p} > 0
(d) V_{g} < 0, V_{p} < 0
If the temperature of the filament is high, then it will emit electrons. Other conditions for the plate current in the diode are:
(1) Positive grid voltage, V_{g}
(2) Positive plate voltage, V_{P}
If any of the above conditions is not satisfied, then plate current in a triode will be zero.
Page No 352:
Question 1:
A discharge tube contains helium at low pressure. A large potential difference is applied across the tube. Consider a helium atom that has just been ionised 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 ionisation.
Answer:
Charge on the electron,
q = −1.6 × 10^{−19} C
Charge on the helium ion = +q = +1.6 × 10^{−19} C
Let the electric field in the tube be E.
Mass of electron, m_{e} = 9.1 × 10^{−31} kg
Mass of proton, m_{p} = 1.67 × 10^{−27} kg
Mass of helium ion, = 4m_{p} = 4×(1.67 × 10^{−27} ) kg
Magnitude of the force experienced by the electron,
F = qE
Magnitude of acceleration of the electron,
a =
qEmeDistance travelled by the electron in time t,
Se=12×qEme×t2 …(1)
Acceleration of the positive ion
=qE4×mpDistance travelled by the helium ion,
SHe=12×qE4×mp×t2 …(2)The ratio of distance travelled by the given particles is given by
SeSHe=4mpmeSeSHe=4×1.67×10279.1×1031SeSHe=0.73406×104=7340.6⇒SeSHe=7340
Question 2:
A molecule of a gas, filled in a discharge tube, gets ionised 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 there’s 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:
Given:
Electric field inside discharge tube, E = 5.0 kV/m
t=1 μs=1×106SF=qE=(1.6×1019)×5×103 Na=qEm=8×10169.1×1031(a)
Let S be distance travelled by the free electron in 1 µs.
S=12at2S=12×89.1×1016+31×(106)2S=0.43956×103 m≈440 m(b)
Let the time of transit of the free electron between successive collisions be t.
S=1 mm=1×103 mUsing, S=12×qEm×t2, we get:1×103=12×1.6×59.1×1016×(t)2⇒t2=2×9.1×1031.6×5×1016⇒t2=9.10.8×5×1019⇒t=0.4802×109 s=0.5 ns
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Question 3:
The mean free path of electrons in a 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 of the electrons be L and pressure inside the tube be P.
It is given that the mean free path of electrons and the pressure inside the tube are related as:
L∝1PHere, L = half of tube’s length
Pressure inside the tube, P = 0.02 mm of Hg
As pressure ‘P‘ becomes half, mean free path of electrons, L, doubles. So, the whole tube is filled with Crookes dark space.
Thus, the pressure required for filling the whole tube with Crookes dark space,
P’=0.022=0.01 mm of Hg
Question 4:
Two discharge tubes have identical material structures 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:
According to Paschen’s equation, sparking potential (V) is a function of the product of pressure (P) inside the tube and the length (d) of the tube.
V=f(Pd)As the sparking potential is same in both the tubes, the product of P and d must be equal for both the tubes.
Length of the smaller tube, d_{s} = 10 cm
Pressure inside the smaller tube, P_{s} = 1.0 mm of mercury
Let the pressure inside the longer tube, P_{l,,} be x mm of mercury.
Length of the longer tube, d_{l} = 20 cm
Psds=PldlPsPl=dsdl1020=1 mmx⇒2010=1 mmx⇒ x=1 mm2=0.5 mmHence, the pressure in the longer tube is 0.5 mm of Hg.
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 RichardsonDushman equation, the number of thermions (n) emitted by a surface, in a given time (t), is given by
i=ne=AST2eϕ/kTA’=Ae⇒n=A’ST2eϕ/kTHere,
ϕ=4.52 e.V =4.52×(1.6×1019) Jk=1.38×1023J/Kn(1000)=A’S×(1000)2×e(4.52×1.6×1019)/(1.38×1023×1000)n(1000)=A’S×106×1.7396×1023n(1000)=A’S×1.7396×1017
n(300K)n(1000K)=A’S×(300)2×e(4.52×1.6×1019)/(1.38×1023×300)A’S×1.7396×1017n(300K)n(1000K)=9×104×1.364×10761.7396×1017 n(300K)n(1000K)=7.056×1055
n(2000K)n(1000K)=A’S×(2000)2×e(4.52×1.6×1019)/(1.38×1023×2000)A’S×1.7396×1017n(2000K)n(1000K)=4×106×(4.1712×1012)(1.7396×1017) n(2000K)n(1000K)=9.73×1011n(3000K)n(1000K)=A’S×(3000)2×e(4.52×1.6×1019)/(1.38×1023×3000)A’S×1.7396×1017n(3000K)n(1000K)=(9×106)×(2.5913×108)(1.7396×1017)n(3000K)n(1000K)=1.34×1016
Question 6:
The saturation current from a thoriatedtungsten cathode at 2000 K is 100 mA. What will be the saturation current for a puretungsten cathode of the same surface area operating at the same temperature? The constant A in the Richardson−Dushman equation is 60 × 10^{4} Am ^{−2} K^{−2} for pure tungsten and 3.0 × 10^{4} Am ^{−2} k^{−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 RichardsonDushman equation, the current of thermions is given by
i=AST2eϕ/KTFor the thoriated tungsten cathode:
Saturation current, i = 100 mA
Temperature, T = 2000 K
A = 3.0 × 10^{4} Am^{−2}K^{−2}
Ï• = 2.6 eV.
For the pure tungsten cathode:
Let the saturation current be i.
Temperature, T = 2000 K
A = 60 × 10^{4} Am ^{−2} K^{−2}
Ï• = 2.6 eV
k = 1.38× 10^{−2}^{3} J/K
Substituting the values of the quantities in the RichardDushman equation, we get:
i=(60×104)(S)×(2000)2e4.5×1.6×10191.38×1023×2000 …(1)100×103=(3×104)(S)(2000)2e2.6×1.6×10191.38×1023×2000 …(2)
Dividing equation (1) by (2)
i100×103=20×e4.5×1.6×10191.38×1023×2000–2.6×1.6×10191.38×1023×2000i100×103=20×e(2.64.5)×1.6×102×1.38i100×103=20×e8×(1.9)1.38i100×103=20×0.000016⇒i=32.9×106 A ≈ 33 μA
Question 7:
A tungsten cathode and a thoriatedtungsten cathode have the same geometric dimensions and are operated at the same temperature. The thoriatedtungsten cathode gives 5000 times more current than the other cathode. Find the operating temperature. Take relevant data from the previous problem.
Answer:
For the pure tungsten cathode,
work function, Ï• = 4.5 eV
A = 60×10^{4} A/m^{2}K^{2}
For the thoriatedtungsten cathode
work function, Ï• = 2.6 eV
A = 3×10^{4} A/m^{2}K^{2}
Saturation current,
i=AST2eϕ/KTFor thoriated tungsten,
iThorited Tungsten=5000iTungstenSo, ⇒S×3×104×T2×e2.6×1.6×10191.38×1023×T=5000×60×104×S×T2×e4.5×1.6×10191.38×T×1023⇒e2.5×1.6×10191.38×1023×T=105×e4.5×1.6×10191.38×1023×TTaking ‘ln’ of both sides, we get:
2.89×104T=11.51+5.22×104T⇒11.51T=2.33×104⇒T=2024 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:
Given:
Work function of tungsten = 4.5 eV
Initial temperature of tungsten filament, T = 2000 K
Final temperature of tungsten filament, T‘ = 2010 K
Let the emission current at (T = 2000 K) be i.
Let the emission current at (T’ = 2010 K) be i‘.
The emission currents are given by
i=AST2eϕ/kT
i’=AST’2ee/kT’Dividing i by i’, we get:
ii’=T2T’2eϕ/kT’eϕ/kT’ii’=TT’2eϕ/kT+ϕ/kT’=TT’2eϕk1T’1Tii’=200020102e4.5×1.6×10191.38×10231201012000.=40000(201)2e4.5×1.61.38(0.0248)=0.8786(201)2×40000=0.8699ii’=10.8699=1.1495=1.14
Question 9:
The constant A in the Richardson−Dushman equation for tungsten is 60 × 10^{4} A m^{−2} K^{−2}. The work function of tungsten is 4.5 eV. A tungsten cathode with a surface area 2.0 × 10^{−5} m^{2} is heated by a 24 W electric heater. In steady state, the heat radiated by the heater and the cathode 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’s Constant = 6 × 10^{−8} W m^{−2} K^{−1}. Assume that the thermions take only a small fraction of the heat supplied.
Answer:
Given:
A = 60 × 10^{4} A m^{−2} K^{−2}
Work function of tungsten, Ï• = 4.5 eV
Stefan’s Constant, σ = 6 × 10^{−8} W m^{−2} K^{−1}
Surface area of tungsten cathode, S = 2.0 × 10^{−5} m^{2}
Boltzmann’s Constant, k = 1.38× 10^{−23} J/K
Heat supplied by the heater, H = 24 W
The cathode acts as a black body; thus, its emissivity is 1.
According Stefan’s Law:
The power (P) radiated by a blackbody with surface area (A) and temperature (T),
P=σAT4 ⇒T4=PσA⇒T4=24(6×108)×(2×105)⇒T4=2×1013 K=20×1012 K⇒ T=2.1147×103=2114.7 KAccording to the RichardDushmann equation, emission current,
i=AST2eϕ/KTi=6×105×2×105×(2114.7)2×e4.5×1.6×10191.38×1023×2114.7i=1.03456×103 A≈1.0 mA
Question 10:
A plate current of 10 mA is obtained when 60 volts are applied across a diode tube. Assuming the LangmuirChild relation
ip∞Vp3/2to hold, find the dynamic resistance r_{p} in this operating condition.
Answer:
According to LamgmuirChild Law,
the relation between plate current (i_{p}) and the plate voltage (V_{p}) is given by
ip=CVp3/2 …(1)Differentiating equation (1) with respect V_{p}, we get:
dipdVp=32CVp1/2 …(2)Dividing (2) and (1), we get:
1ipdipdvp=3/2CVp1/2CVp3/2⇒1ip.dipdvp=32VpThe dynamic resistance is given by:
dvpdip=2Vp3iprp=2Vp3iprp=2×603×10×103rp=4×103=4 kΩ
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 70 V?
Answer:
For 50 V or 60 V, the plate current is 20 mA. That means 20 mA is the saturation current.
At the given temperature, the plate current is 20 mA for all other values of voltages.
Thus, the current at 70 V will be 20 mA.
Question 12:
The power delivered in the plate circular 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 LangmuirChild equation to hold.
Answer:
Given:
When plate voltage, V_{p}_{, }is 36 V, power delivered in the plate circular of a diode, P, is 1.0 W.
Let the plate current be I_{p}.
Let the power delivered be P‘ and plate current be I_{p}‘ when plate voltage, V_{p}, is increased to 49 V
P=IpVp⇒ Ip=PVP=136According to LangmuirChild equation,
Ip ∝ (Vp)3/2,Ip’ ∝ (Vp’)3/2,⇒ IpI’p=(Vp)3/2(Vp’)3/2⇒1/36Ip’=36493/2⇒ Ip=0.04411Thus, power delivered when the plate voltage is increased to 49 V,
P’=Vp’×Ip’P’=49×0.04411 WP’=2.1613 W=2.2 W
Question 13:
A triode value operates at V_{p} = 225 V and V_{g} = −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:
Amplification factor for a triode valve,
μ=Change in Plate VoltageChange in Grid VoltageAt constant plate current,
μ= (δVPδVG)iP =constantμ=2502252.50.5μ=252=12.5This implies that the amplification factor of the triode is 12.5 .
Question 14:
Calculate the amplification factor of a triode valve that has plate resistance of 2 kâ„¦ and transconductance of 2 millimho.
Answer:
Given:
Plate resistance,
rp = 2 kΩTransconductance of plate,
gm=2 mili mho=2×103 mhoTherefore, the amplification factor is given as:
μ=rP×gmμ=2×103×2×103=4That is, amplification factor is 4.
Question 15:
The dynamic plate resistance of a triode value is 10 kâ„¦. Find the change in the plate current if the plate voltage is changed from 200 V to 220 V.
Answer:
Given:
Plate resistance,
rp = 10 KΩ = 104ΩChange in plate voltage,
δVp = 220200 = 20VPlate resistance at constant grid voltage is given as:
rP =(δVPδIP)VG = Constant⇒δIP = δVPrPδIP =δVprPδIP =20104=0.002 A = 2 mA
Question 16:
Find the values of r_{p}, µ and g_{m} of a triode operating at plate voltage 200 V and grid voltage −6. The plate characteristics are shown in the figure.
Figure
Answer:
Dynamic plate resistance,
rp=δVpδIp, at constant grid voltage
We need to find the slope of the graph for a particular value of grid voltage, i.e. V_{g} = −6 V.
Consider two points for the plot of V_{g} = −6 V:
rp = (240160) V(133)×103 Arp= 8010×103 Ωrp=8 KΩgm= δIpδVgVP=constant (200 V)Consider two points on the 200 V line:
gm = (133)×103[(4)(8)]Agm= 10×1034=2.5 mili mho
Amplification factor,
μ = ∆VP∆VGiP=constant
μ=1001806(10)μ=804=20
Question 17:
The plate resistance of a triode is 8 kâ„¦ and the transconductance is 2.5 millimho. (a) If the plate voltage is 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, by how much should the grid voltage be decreased in order to bring the plate current back to its initial value?
Answer:
Given:
Plate resistance,
rp=8 kΩ=8000 kΩChange in plate voltage,
δVp=48 VFormula for plate resistance:
rP =δVPδIPVG = constant ⇒δIp=δVprp VG =constant ⇒δIp=488000=0.006 A=6 mA(b) Now, V_{p} is kept constant.
Change in plate current,
δIp = 6 mA = 0.006 ATransconductance,
gm = 0.0025 mhoδVG = δIpgm = 0.0060.0025δVG= 2.4 V, at constant plate voltage
Question 18:
The plate resistance and amplification factor of a triode are 10kâ„¦ 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:
Plate resistance,
rp=10 KΩAmplification factor,
μ=20Plate voltage,
Vp=240 VGrid voltage,
VG=7.5 V
Ip=10 mA(a) Transconductance,
δVG=δIpgmδVG=(15×10310×103)μ/rpδVG=5×10320/10×103δVG=5×1032×103=52=2.5VG=+2.57.5=5.0 V
rp = δVpδIpVg = Constant⇒ 104=δVp(10×10315×103)⇒ δVp = 104×5×103= 50 V∴Vp’Vp = 50⇒ Vp’=50+Vp=200 V
Question 19:
The plate current, plate voltage and grid voltage of a 6F6 triode tube are related as
i_{p} = 41 (V_{p} + 7 V_{g})^{1.41}
Here, V_{p} and V_{g} are in volts and i_{p} in microamperes.
The tube is operated at V_{p} = 250 V, V_{g} = −20 V. Calculate (a) the tube current, (b) the plate resistance, (c) the mutual conductance and (d) the amplification factor.
Answer:
Given:
Plate voltage, V_{P} = 250 V
Grid voltage, V_{G} =
20 V
(a) As given in the question, plate current varies as,
ip= 41(Vp+7 Vg)1.41ip= 41(250140)1.41ip=41×(110)1.41ip=30984.71 μA=30.98 mA(b)
ip=41(Vp+7VG)1.41Differentiating this equation, we get:
diP = 41×1.41×(Vp+7Vg)0.41×(dVp+7dVg) (1)Plate resistance is defined as: rp= dVpdip Vg = ConstantFrom equation (1), dVpdip=1×10641×1.41×1100.41dVpdip=106×2.51×103dVpdip=2.5×103 Ω=2.5 KΩ(c) From above:
As dIP=41×1.41×(250+7×(20))0.41×7dVg,gm = (dIpdVg)VP = ConstantFrom equation (1), 17dIpdVgVP = Constant =41×1.41×(250+7×(20))0.41dIpdVgVP = Constant= 41×1.41×(110)0.41×7 mhodIpdVgVP = Constant= 41×1.41×6.87×7 mhodIpdVgVP = Constant=2.78 milli mho(d) Amplification factor,
μ = rp×gm= 2.5×103×2.78×103= 6.95≈7
Question 20:
The plate current in a triode can be written as
ip=kvg+Vpμ3/2Show that the mutual conductance is proportional to the cube root of the plate current.
Answer:
Given: The plate current varies with plate and grid voltage as
ip = KVg+Vpμ3/2 …(1)Differentiating the equation w.r.t
VG,we get:
dip = K32Vg+Vpμ1/2dVg⇒ gm = dipdVg = 32KVg+Vpμ1/2From (1), plate current can be written in terms of transconductance as:
ip = 32KVg+Vpμ1/23×K’Here, K’ is a constant =(23)3×1K2ip = K'(gm)3⇒ gm ∝ ip3
Question 21:
A triode has mutual conductance of 2.0 millimho and plate resistance of 20 kâ„¦. It is desired to amplify a signal by a factor of 30. What load resistance should be added in the circuit?
Answer:
Plate resistance, r_{p} = 20 kâ„¦
Mutual Conductance, g_{m }= 2.0 milli mho
g_{m }= 2 × 10^{−3} mho
Amplification factor, µ = 30
Load Resistance = R_{L} = ?
We know:
A=μ1+rpRL,
where A is the voltage amplification factor.
Also, amplification factor,
μ = rP×gm
A=rp×gm1+rpRLOn substitution of all the given values, we get: 30=20×103×2×1031+20000RL⇒1+20000RL = 403020000RL = 13⇒RL=60000 Ω=60 kΩ
Page No 354:
Question 22:
The gain factor of an amplifier in increased from 10 to 12 as the load resistance is changed from 4 kâ„¦ to 8 kâ„¦. Calculate (a) the amplification factor and (b) the plate resistance.
Answer:
We know:
Voltage gain =
μ1+rpRL …(1)
When voltage amplification factor, A = 10,
R_{L} = 4 kâ„¦
10=μ1+rp4×103⇒ 10=μ×4×1034×103+rp⇒4×104+10rP = 4000 μ …(2)
Now, increased gain, A = 12
Substituting this value in (1) ,we get:
12 = μ1+rPRL12 = μ1+rp8×103⇒12=μ×80008000+rP⇒96000+12rP = 8000 μ …(3)On solving equations (2) and (3), we get:
μ = 15rP = 2000 Ω=2 kΩ
Question 23:
The figure shows two identical triode tubes connected in parallel. The anodes are connected together and the cathodes are connected together. Show that the equivalent plate resistance is half 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.
Figure
Answer:
From the circuit,
Cathode of both the triodes are connected to a common point and their anodes are also connected together. Thus, the 2 triodes are connected in parallel.
As plate resistance is obtained by observing the variation of plate voltage with plate current, keeping grid voltage as constant.
Hence equivalent plate resistance
Rp = rP1×rP2rP1+rP2if both the triodes have equal plate resistance i.e. rP1 = rP2 =rPRP = rP2To find the equivalent transconductance
We need to look at the variation of current with respect to grid voltage, keeping plate voltage as constant.
From the circuit
Now the 2 triodes are in series w.r.t grid voltage.
Hence equivalent transconductance will be equal to
Gm = 2gmif the 2 triodes have equal trens conductance i.e gm1 = gm2 = gmNow the equivalent amplification factor is,
μequivalent =Gm×RPμequivalent= 2gm×rP2μequivalent=μ
HC Verma Solutions for Class 11 Physics – Part 1
 Chapter 1 – Introduction to Physics
 Chapter 2 – Physics and Mathematics
 Chapter 3 – Rest and Motion: Kinematics
 Chapter 4 – The Forces
 Chapter 5 – Newton’s Laws of Motion
 Chapter 6 – Friction
 Chapter 7 – Circular Motion
 Chapter 8 – Work and Energy
 Chapter 9 – Center of Mass, Linear Momentum, Collision
 Chapter 10 – Rotational Mechanics
 Chapter 11 – Gravitation
 Chapter 12 – Simple Harmonic Motion
 Chapter 13 – Fluid Mechanics
 Chapter 14 – Some Mechanical Properties of Matter
 Chapter 15 – Wave Motion and Wave on a String
 Chapter 16 – Sound Wave
 Chapter 17 – Light Waves
 Chapter 18 – Geometrical Optics
 Chapter 19 – Optical Instruments
 Chapter 20 – Dispersion and Spectra
 Chapter 21 – Speed of Light
 Chapter 22 – Photometry
HC Verma Solutions for Class 12 Physics – Part 2
 Chapter 23 – Heat and Temperature
 Chapter 24 – Kinetic Theory of Gases
 Chapter 25 – Calorimetry
 Chapter 26 – Laws of Thermodynamics
 Chapter 27 – Specific Heat Capacities of Gases
 Chapter 28 – Heat Transfer
 Chapter 29 – Electric Field and Potential
 Chapter 30 – Gauss’s Law
 Chapter 31 – Capacitors
 Chapter 32 – Electric Current in Conductors
 Chapter 33 – Thermal and Chemical Effects of Electric Current
 Chapter 34 – Magnetic Field
 Chapter 35 – Magnetic Field due to a Current
 Chapter 36 – Permanent Magnets
 Chapter 37 – Magnetic Properties of Matter
 Chapter 38 – Electromagnetic Induction
 Chapter 39 – Alternating Current
 Chapter 40 – Electromagnetic Waves
 Chapter 41 – Electric Current through Gases
 Chapter 42 – Photoelectric Effect and Wave Particle Duality
 Chapter 43 – Bohr’s Model and Physics of the Atom
 Chapter 44 – Xrays
 Chapter 45 – Semiconductors and Semiconductor Devices
 Chapter 46 – The Nucleus
 Chapter 47 – The Special Theory of Relativity