Hc Verma II for Class 11 Science Physics Chapter 41 - Electric Current Through Gases

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Page No 351:

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 free-path 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.

Page No 351:

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.

Page No 351:

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.

Page No 351:

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.

Page No 351:

Answer:

We will prefer a material with low work-function to be used as a cathode in a diode, so that electron emission can occur using a small amount of energy.

Page No 351:

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.

Page No 351:

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.

Page No 351:

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 non-conductors.

Page No 351:

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.

Page No 351:

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.

Page No 351:

Answer:

(a) electrons

Cathode rays consist of a stream of fast moving electrons.

Page No 351:

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.

Page No 352:

Answer:

(d)

Since the thermionic current is directly proportional to the square of the temperature of the surface emitting thermions, the graph is parabolic.

Page No 352:

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.

Page No 352:

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.

Page No 352:

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.

Page No 352:

Answer:

(c) remain almost the same

Dynamic resistance (r_P) is given by,
r_P = $\frac{∆ V_{P}}{∆ i_{P}}$
The 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.

Page No 352:

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.

Page No 352:

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.

Page No 352:

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.

Page No 352:

Answer:

(a) It travels in a straight line.
(b) It emits an X-ray when it strikes a metal.

A cathode ray travels in a straight line. When cathode rays strike a solid metal, X-rays 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.

Page No 352:

Answer:

(a) the plate current decreases

A diode valve consists of a negatively-charged region between the cathode and anode, called the space charge region. This negatively-charged 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.

Page No 352:

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.

Page No 352:

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 (bi-directional) into direct current (uni-directional), 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.

Page No 352:

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.

Page No 352:

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:

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 = $\frac{q E}{m_{e}}$
Distance travelled by the electron in time t,
$S_{e} = \frac{1}{2} \times \frac{q E}{m_{e}} \times t^{2} . . . (1)$
Acceleration of the positive ion $= \frac{q E}{4 \times m_{p}}$
Distance travelled by the helium ion,
$S_{He} = \frac{1}{2} \times \frac{q E}{4 \times m_{p}} \times t^{2} . . . (2)$

The ratio of distance travelled by the given particles is given by
$\frac{S_{e}}{S_{He}} = \frac{4 m_{p}}{m_{e}} \frac{S_{e}}{S_{He}} = \frac{4 \times 1.67 \times 10^{- 27}}{9.1 \times 10^{- 31}} \frac{S_{e}}{S_{He}} = 0.73406 \times 10^{4} = 7340.6 \Rightarrow \frac{S_{e}}{S_{He}} = 7340$

Page No 352:

Answer:

Given:
Electric field inside discharge tube, E = 5.0 kV/m
$t = 1 μs = 1 \times 10^{- 6} S F = q E = (1.6 \times 10^{- 19}) \times 5 \times 10^{3} N a = \frac{q E}{m} = \frac{8 \times 10^{- 16}}{9.1 \times 10^{- 31}}$

(a)
Let S be distance travelled by the free electron in 1 µs.
$S = \frac{1}{2} a t^{2} S = \frac{1}{2} \times \frac{8}{9.1} \times 10^{- 16 + 31} \times (10^{- 6})^{2} S = 0.43956 \times 10^{3} m \approx 440 m$

(b)
Let the time of transit of the free electron between successive collisions be t.
$S = 1 mm = 1 \times 10^{- 3} m Using, S = \frac{1}{2} \times (\frac{q E}{m}) \times t^{2}, we get : 1 \times 10^{- 3} = \frac{1}{2} \times \frac{1.6 \times 5}{9.1} \times 10^{16} \times (t)^{2} \Rightarrow t^{2} = \frac{2 \times 9.1 \times 10^{- 3}}{1.6 \times 5 \times 10^{16}} \Rightarrow t^{2} = \frac{9.1}{0.8 \times 5} \times 10^{- 19} \Rightarrow t = 0.4802 \times 10^{- 9} s = 0.5 ns$

Page No 353:

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:
$L \propto \frac{1}{P}$
Here, 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,
$P' = \frac{0.02}{2} = 0.01 mm of Hg$

Page No 353:

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.
$V = f (P d)$
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
$P_{s} d_{s} = P_{l} d_{l} \frac{P_{s}}{P_{l}} = \frac{d_{s}}{d_{l}} \frac{10}{20} = \frac{1 mm}{x} \Rightarrow \frac{20}{10} = \frac{1 mm}{x} \Rightarrow x = \frac{1 mm}{2} = 0.5 mm$
Hence, the pressure in the longer tube is 0.5 mm of Hg.

Page No 353:

Answer:

According to Richardson-Dushman equation, the number of thermions (n) emitted by a surface, in a given time (t), is given by
$i = n e = A S T^{2} e^{- ϕ / k T} A' = \frac{A}{e} \Rightarrow n = A' S T^{2} e^{- ϕ / k T}$
Here,
$ϕ = 4.52 e . V = 4.52 \times (1.6 \times 10^{- 19}) J k = 1.38 \times 10^{- 23} J / K n (1000) = A' S \times (1000)^{2} \times e^{(- 4.52 \times 1.6 \times 10^{- 19}) / (1.38 \times 10^{- 23} \times 1000)} n (1000) = A' S \times 10^{6} \times 1.7396 \times 10^{- 23} n (1000) = A' S \times 1.7396 \times 10^{- 17}$
$\frac{n (300 K)}{n (1000 K)} = \frac{A' S \times (300)^{2} \times e^{(- 4.52 \times 1.6 \times 10^{- 19}) / (1.38 \times 10^{- 23} \times 300)}}{A' S \times 1.7396 \times 10^{- 17}} \frac{n (300 K)}{n (1000 K)} = \frac{9 \times 10^{4} \times 1.364 \times 10^{- 76}}{1.7396 \times 10^{- 17}} \frac{n (300 K)}{n (1000 K)} = 7.056 \times 10^{- 55}$
$\frac{n (2000 K)}{n (1000 K)} = \frac{A' S \times (2000)^{2} \times e^{(- 4.52 \times 1.6 \times 10^{- 19}) / (1.38 \times 10^{- 23} \times 2000)}}{A' S \times 1.7396 \times 10^{- 17}} \frac{n (2000 K)}{n (1000 K)} = \frac{4 \times 10^{6} \times (4.1712 \times 10^{- 12})}{(1.7396 \times 10^{- 17})} \frac{n (2000 K)}{n (1000 K)} = 9.73 \times 10^{11}$

$\frac{n (3000 K)}{n (1000 K)} = \frac{A' S \times (3000)^{2} \times e^{(- 4.52 \times 1.6 \times 10^{- 19}) / (1.38 \times 10^{- 23} \times 3000)}}{A' S \times 1.7396 \times 10^{- 17}} \frac{n (3000 K)}{n (1000 K)} = \frac{(9 \times 10^{6}) \times (2.5913 \times 10^{- 8})}{(1.7396 \times 10^{- 17})} \frac{n (3000 K)}{n (1000 K)} = 1.34 \times 10^{16}$

Page No 353:

Answer:

According to Richardson-Dushman equation, the current of thermions is given by
$i = A S T^{2} e^{- ϕ / K T}$
For the thoriated tungsten cathode:
Saturation current, i = 100 mA
Temperature, T = 2000 K
A = 3.0 × 10⁴ Am⁻²K⁻²
ϕ = 2.6 eV.

For the pure tungsten cathode:
Let the saturation current be i.
Temperature, T = 2000 K
A = 60 × 10⁴ Am ⁻² K⁻²
ϕ = 2.6 eV

k = 1.38× 10⁻²³ J/K

Substituting the values of the quantities in the Richard-Dushman equation, we get:
$i = (60 \times 10^{4}) (S) \times (2000)^{2} e^{\frac{- 4.5 \times 1.6 \times 10^{- 19}}{1.38 \times 10^{- 23} \times 2000}} . . . (1) 100 \times 10^{- 3} = (3 \times 10^{4}) (S) (2000)^{2} e^{\frac{- 2.6 \times 1.6 \times 10^{- 19}}{1.38 \times 10^{- 23} \times 2000}} . . . (2)$ ^{Dividing equation (1) by (2)}

$\frac{i}{100 \times 10^{- 3}} = 20 \times e (^{\frac{- 4.5 \times 1.6 \times 10^{- 19}}{1.38 \times 10^{- 23} \times 2000} - \frac{- 2.6 \times 1.6 \times 10^{- 19}}{1.38 \times 10^{- 23} \times 2000}}) \frac{i}{100 \times 10^{- 3}} = 20 \times e^{\frac{(2.6 - 4.5) \times 1.6 \times 10}{2 \times 1.38}} \frac{i}{100 \times 10^{- 3}} = 20 \times e^{\frac{8 \times (- 1.9)}{1.38}} \frac{i}{100 \times 10^{- 3}} = 20 \times 0.000016 \Rightarrow i = 32.9 \times 10^{- 6} A \approx 33 μA$

Page No 353:

Answer:

For the pure tungsten cathode,
work function, ϕ = 4.5 eV
A = 60×10⁴ A/m²-K²
For the thoriated-tungsten cathode
work function, ϕ = 2.6 eV
A = 3×10⁴ A/m²-K²

Saturation current,
$i = A S T^{2} e^{- ϕ / K T}$
For thoriated tungsten,
$i_{Thorited Tungsten} = 5000 i_{Tungsten} So, \Rightarrow S \times 3 \times 10^{4} \times T^{2} \times e^{\frac{- 2.6 \times 1.6 \times 10^{- 19}}{1.38 \times 10^{- 23} \times T}} = 5000 \times 60 \times 10^{4} \times S \times T^{2} \times e^{\frac{- 4.5 \times 1.6 \times 10^{- 19}}{1.38 \times T \times 10^{- 23}}} \Rightarrow e^{\frac{- 2.5 \times 1.6 \times 10^{- 19}}{1.38 \times 10^{- 23} \times T}} = 10^{5} \times e^{\frac{- 4.5 \times 1.6 \times 10^{- 19}}{1.38 \times 10^{- 23} \times T}}$

Taking 'ln' of both sides, we get:

$\frac{- 2.89 \times 10^{4}}{T} = 11.51 + \frac{- 5.22 \times 10^{4}}{T} \Rightarrow 11.51 T = 2.33 \times 10^{4} \Rightarrow T = 2024 K$

Page No 353:

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
$i = A S T^{2} e^{- ϕ / k T}$
$i' = A S T'^{2} e^{- e / k T'}$
Dividing i by i', we get:
$\frac{i}{i'} = \frac{T^{2}}{T'^{2}} \frac{e^{- ϕ / k T'}}{e^{- ϕ / k T'}} \frac{i}{i'} = {(\frac{T}{T'})}^{2} e^{- ϕ / k T + ϕ / k T'} = {(\frac{T}{T'})}^{2} e^{\frac{ϕ}{k} (\frac{1}{T'} - \frac{1}{T})} \frac{i}{i'} = {(\frac{2000}{2010})}^{2} e^{\frac{4.5 \times 1.6 \times 10^{- 19}}{1.38 \times 10^{- 23}} (\frac{1}{2010} - \frac{1}{2000}) .} = \frac{40000}{(201)^{2}} e^{\frac{4.5 \times 1.6}{1.38} (- 0.0248)} = \frac{0.8786}{(201)^{2}} \times 40000 = 0.8699 \frac{i}{i'} = \frac{1}{0.8699} = 1.1495 = 1.14$

Page No 353:

Answer:

Given:
A = 60 × 10⁴ A m⁻² K⁻²
Work function of tungsten, ϕ = 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 m²
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),
$P = σ A T^{4} \Rightarrow T^{4} = \frac{P}{σ A} \Rightarrow T^{4} = \frac{24}{(6 \times 10^{- 8}) \times (2 \times 10^{- 5})} \Rightarrow T^{4} = 2 \times 10^{13} K = 20 \times 10^{12} K \Rightarrow T = 2.1147 \times 10^{3} = 2114.7 K$

According to the Richard-Dushmann equation, emission current,
$i = A S T^{2} e^{- ϕ / K T} i = 6 \times 10^{5} \times 2 \times 10^{- 5} \times (2114.7)^{2} \times e^{\frac{- 4.5 \times 1.6 \times 10^{- 19}}{1.38 \times 10^{- 23} \times 2114.7}} i = 1.03456 \times 10^{- 3} A \approx 1.0 mA$

Page No 353:

Answer:

According to Lamgmuir-Child Law,
the relation between plate current (i_p) and the plate voltage (V_p) is given by
$i_{p} = C {V_{p}}^{3 / 2} . . . (1)$

Differentiating equation (1) with respect V_p, we get:
$\frac{d i_{p}}{d V_{p}} = \frac{3}{2} C {V_{p}}^{1 / 2} . . . (2)$

Dividing (2) and (1), we get:
$\frac{1}{i_{p}} \frac{d i_{p}}{d v_{p}} = \frac{3 / 2 C {V_{p}}^{1 / 2}}{C {V_{p}}^{3 / 2}} \Rightarrow \frac{1}{i_{p}} . \frac{d i_{p}}{d v_{p}} = \frac{3}{2 V_{p}}$

The dynamic resistance is given by:
$\frac{d v_{p}}{d i_{p}} = \frac{2 V_{p}}{3 i_{p}} r_{p} = \frac{2 V_{p}}{3 i_{p}} r_{p} = \frac{2 \times 60}{3 \times 10 \times 10^{- 3}} r_{p} = 4 \times 10^{3} = 4 kΩ$

Page No 353:

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.

Page No 353:

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

$P = I_{p} V_{p} \Rightarrow I_{p} = \frac{P}{V_{P}} = \frac{1}{36}$
According to Langmuir-Child equation,
$I_{p} \propto (V_{p})^{3 / 2}, I_{p}' \propto (V_{p}')^{3 / 2}, \Rightarrow \frac{I_{p}}{I'_{p}} = \frac{(V_{p})^{3 / 2}}{(V_{p}')^{3 / 2}} \Rightarrow \frac{1 / 36}{I_{p}'} = {(\frac{36}{49})}^{3 / 2} \Rightarrow I_{p} = 0.04411$

Thus, power delivered when the plate voltage is increased to 49 V,
$P' = V_{p}' \times I_{p}' P' = 49 \times 0.04411 W P' = 2.1613 W = 2.2 W$

Page No 353:

Answer:

Amplification factor for a triode valve,
$μ = \frac{Change in Plate Voltage}{Change in Grid Voltage}$
At constant plate current,
$μ = (\frac{δ V_{P}}{δ V_{G}})_{i_{P} = constant} μ = \frac{250 - 225}{2.5 - 0.5} μ = \frac{25}{2} = 12.5$

This implies that the amplification factor of the triode is 12.5 .

Page No 353:

Answer:

Given:
Plate resistance, $r_{p} = 2 kΩ$

Transconductance of plate, $g_{m} = 2 mili mho = 2 \times 10^{- 3} mho$

Therefore, the amplification factor is given as:
$μ = r_{P} \times g_{m} μ = 2 \times 10^{3} \times 2 \times 10^{- 3} = 4$
That is, amplification factor is 4.

Page No 353:

Answer:

Given:
Plate resistance, $r_{p} = 10 KΩ = 10^{4} Ω$
Change in plate voltage, $δ V_{p} = 220 - 200 = 20 V$
Plate resistance at constant grid voltage is given as:

$r_{P} = (\frac{δ V_{P}}{δ I_{P}})_{V_{G} = C onstant} \Rightarrow δ I_{P} = \frac{δ V_{P}}{r_{P}} δ I_{P} = \frac{δ V_{p}}{r_{P}} δ I_{P} = \frac{20}{10^{4}} = 0.002 A = 2 mA$

Page No 353:

Answer:

Dynamic plate resistance,
$r_{p} = (\frac{{δV}_{p}}{{δI}_{p}})$ , 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:
$r_{p} = \frac{(240 - 160) V}{(13 - 3) \times 10^{- 3} A} r_{p} = \frac{80}{10} \times 10^{3} Ω r_{p} = 8 KΩ g_{m} = {(\frac{δ I_{p}}{δ V_{g}})}_{V_{P} = constant (200 V)}$

Consider two points on the 200 V line:
$g_{m} = \frac{(13 - 3) \times 10^{- 3}}{[(- 4) - (- 8)]} A g_{m} = \frac{10 \times 10^{- 3}}{4} = 2.5 mili mho$
Amplification factor, $μ = - {(\frac{∆ V_{P}}{∆ V_{G}})}_{i_{P} = constant}$
$μ = - \frac{100 - 180}{- 6 - (- 10)} μ = \frac{80}{4} = 20$

Page No 353:

Answer:

Given:
Plate resistance, $r_{p} = 8 k Ω = 8000 k Ω$
Change in plate voltage, $δ V_{p} = 48 V$
Formula for plate resistance:
$r_{P} = {(\frac{δ V_{P}}{δ I_{P}})}_{V_{G} = constant} \Rightarrow δ I_{p} = \frac{δ V_{p}}{r_{p}}_{V_{G = constant}} \Rightarrow δ I_{p} = \frac{48}{8000} = 0.006 A = 6 mA$

(b) Now, V_p is kept constant.
Change in plate current,
$δ I_{p} = 6 mA = 0.006 A$
Transconductance,
$g_{m} = 0.0025 mho δ V_{G} = \frac{δ I_{p}}{g_{m}} = \frac{0.006}{0.0025} δ V_{G} = 2.4 V, at constant plate voltage$

Page No 353:

Answer:

Given:
Plate resistance, $r_{p} = 10 KΩ$
Amplification factor, $μ = 20$
Plate voltage, $V_{p} = 240 V$
Grid voltage, $V_{G} = 7.5 V$
$I_{p} = 10 mA$

(a) Transconductance,
$δ V_{G} = \frac{δ I_{p}}{g_{m}} δ V_{G} = \frac{(15 \times 10^{- 3} - 10 \times 10^{- 3})}{μ / r_{p}} δ V_{G} = \frac{5 \times 10^{- 3}}{20 / 10 \times 10^{- 3}} δ V_{G} = \frac{5 \times 10^{- 3}}{2 \times 10^{- 3}} = \frac{5}{2} = 2.5 V_{G} = + 2.5 - 7.5 = - 5.0 V$
$r_{p} = {(\frac{δ V_{p}}{δ I_{p}})}_{V_{g} = Constant} \Rightarrow 10^{4} = \frac{δ V_{p}}{(10 \times 10^{- 3} - 15 \times 10^{- 3})} \Rightarrow δ V_{p} = - 10^{4} \times 5 \times 10^{- 3} = - 50 V ∴ V_{p}' - V_{p} = - 50 \Rightarrow V_{p}' = - 50 + V_{p} = 200 V$

Page No 353:

Answer:

Given:
Plate voltage, V_P = 250 V
Grid voltage, V_G = $-$ 20 V

(a) As given in the question, plate current varies as,
$i_{p} = 41 (V_{p} + 7 V_{g})^{1.41} i_{p} = 41 (250 - 140)^{1.41} i_{p} = 41 \times (110)^{1.41} i_{p} = 30984.71 μA = 30.98 mA$

(b) $i_{p} = 41 (V_{p} + 7 V_{G})^{1.41}$
Differentiating this equation, we get:
$d i_{P} = 41 \times 1.41 \times (V_{p} + 7 V_{g})^{0.41} \times (d V_{p} + 7 d V_{g}) (1) P late resistance is defined as : r_{p} = {(\frac{d V_{p}}{d i_{p}})}_{V_{g} = Constant} From equation (1), \frac{d V_{p}}{d i_{p}} = \frac{1 \times 10^{6}}{41 \times 1.41 \times 110^{0.41}} \frac{d V_{p}}{d i_{p}} = 10^{6} \times 2.51 \times 10^{- 3} \frac{d V_{p}}{d i_{p}} = 2.5 \times 10^{3} Ω = 2.5 KΩ$

(c) From above:
$As d I_{_{P}} = 41 \times 1.41 \times (250 + 7 \times (- 20))^{0.41} \times 7 d V_{g,} g_{m} = (\frac{d I_{p}}{d V_{g}})_{V_{P} = Constant} From equation (1), \frac{1}{7} {(\frac{d I_{p}}{d V_{g}})}_{V_{P} = Constant} = 41 \times 1.41 \times (250 + 7 \times (- 20))^{0.41} {(\frac{d I_{p}}{d V_{g}})}_{V_{P} = Constant} = 41 \times 1.41 \times (110)^{0.41} \times 7 mho {(\frac{d I_{p}}{d V_{g}})}_{V_{P} = Constant} = 41 \times 1.41 \times 6.87 \times 7 mho {(\frac{d I_{p}}{d V_{g}})}_{V_{P} = Constant} = 2.78 milli mho$
(d) Amplification factor,
$μ = r_{p} \times g_{m} = 2.5 \times 10^{3} \times 2.78 \times 10^{- 3} = 6.95 \approx 7$

Page No 353:

Answer:

Given: The plate current varies with plate and grid voltage as
$i_{p} = K {(V_{g} + \frac{V_{p}}{μ})}^{3 / 2} . . . (1)$
Differentiating the equation w.r.t $V_{G}$ ,we get:
$d i_{p} = K \frac{3}{2} {(V_{g} + \frac{V_{p}}{μ})}^{1 / 2} d V_{g} \Rightarrow g_{m} = \frac{d i_{p}}{d V_{g}} = \frac{3}{2} K {(V_{g} + \frac{V_{p}}{μ})}^{1 / 2}$
From (1), plate current can be written in terms of transconductance as:
$i_{p} = {[\frac{3}{2} K {(V_{g} + \frac{V_{p}}{μ})}^{1 / 2}]}^{3} \times K' H ere, K^{'} is a constant = (\frac{2}{3})^{3} \times \frac{1}{K^{2}} i_{p} = K' (g_{m})^{3} \Rightarrow g_{m} \propto \sqrt[3]{i_{p}}$

Page No 353:

Answer:

Plate resistance, r_p = 20 kΩ
Mutual Conductance, g_m= 2.0 milli mho
g_m= 2 × 10⁻³ mho

Amplification factor, µ = 30
Load Resistance = R_L = ?
We know: $A = \frac{μ}{1 + \frac{r_{p}}{R_{L}}}$ ,
where A is the voltage amplification factor.
Also, amplification factor, $μ = r_{P} \times g_{m}$
$A = \frac{r_{p} \times g_{m}}{1 + \frac{r_{p}}{R_{L}}} On substitution of all the given values, we get : 30 = \frac{20 \times 10^{3} \times 2 \times 10^{- 3}}{1 + \frac{20000}{R_{L}}} \Rightarrow 1 + \frac{20000}{R_{L}} = \frac{40}{30} \frac{20000}{R_{L}} = \frac{1}{3} \Rightarrow R_{L} = 60000 Ω = 60 kΩ$

Page No 354:

Answer:

We know:
Voltage gain = $\frac{μ}{1 + \frac{r_{p}}{R_{L}}}$ ...(1)
When voltage amplification factor, A = 10,
R_L = 4 kΩ
$10 = \frac{μ}{1 + \frac{r_{p}}{4 \times 10^{3}}} \Rightarrow 10 = \frac{μ \times 4 \times 10^{3}}{4 \times 10^{3} + r_{p}} \Rightarrow 4 \times 10^{4} + 10 r_{P} = 4000 μ . . . (2)$
Now, increased gain, A = 12
Substituting this value in (1) ,we get:
$12 = \frac{μ}{1 + \frac{r_{P}}{R_{L}}} 12 = \frac{μ}{1 + \frac{r_{p}}{8 \times 10^{3}}} \Rightarrow 12 = \frac{μ \times 8000}{8000 + r_{P}} \Rightarrow 96000 + 12 r_{P} = 8000 μ . . . (3)$
On solving equations (2) and (3), we get:
$μ = 15 r_{P} = 2000 Ω = 2 kΩ$

Page No 354:

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
$R_{p} = \frac{r_{P_{1}} \times r_{P_{2}}}{r_{P_{1}} + r_{P_{2}}} if both the triodes have equal plate resistance i . e . r_{P_{1}} = r_{P_{2}} = r_{P} R_{P} = \frac{r_{P}}{2}$

To 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
$G_{m} = 2 g_{m} if the 2 triodes have equal trens conductance i . e g_{m_{1}} = g_{m_{2}} = g_{m}$
Now the equivalent amplification factor is,
$μ_{equivalent} = G_{m} \times R_{P} μ_{equivalent} = 2 g_{m} \times \frac{r_{P}}{2} μ_{equivalent} = μ$

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