 # Electric Current through Gases

## Concept Of Physics

### H C Verma

1   A discharge tube contains helium at r, 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.

##### Solution :

Let the two particles have charge $‘q’$ $\\$ Mass of electron $m_a = 9.1 \times 10^{–31}$ kg $\\$ Mass of proton $m_p = 1.67 \times 10^{–27}$ kg $\\$ Electric field be $E$ $\\$ Force experienced by Electron $= qE$ $\\$ accln. $= \frac{qE}{m_e}$ $\\$ For time dt, $\\$ $S_e = \frac{1}{2} \times \frac{qE}{m_e} \times dt^2 \qquad ....(1)$ $\\$ For the positive ion,$\\$ accln $= \frac{qE}{4 \times m_p}$ $\\$ $S_p = \frac{1}{2} \times \frac{qE}{4 \times m_p} \times dt^2 \qquad ......(2)$ $\\$ $\frac{S_e}{S_p} = \frac{4m_p}{m_e} = 7340.6$

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 exists in the vicinity of the event, (a) Find the distance travelled by the free electron in 1 fis 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.

##### Solution :

$E=5Kv/m = 5 \times 10^3 v/m ; t =1 \mu s = 1 \times 10^{-6} s$ $\\$ $F = qE = 1.6 \times 10^{-9} \times 5 \times 10^3$ $\\$ $a =\frac{qE}{m} =\frac{1.6 \times 5 \times 10^{-16}}{9.1 \times 10^{-31}}$ $\\$ (a) S = distance travelled $= \frac{1}{2} at^2 = 439.56 = 440 m$ $\\$ (b) $d = 1 mm \quad = 1 \times 10^{-3} m$ $\\$ $1 \times 10^{-3} = \frac{1}{2} \times \frac{1.6 \times 5}{9.1} 10^5 \times t^2$ $\\$ $\Rightarrow t^2 = \frac{9.1}{0.8 \times 5 } \times 10^{-18} \quad \Rightarrow t=1.508 \times 10^{-9} sec \quad \Rightarrow 1.5ns$ $\\$

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.

##### Solution :

Let the mean free path be ‘L’ and pressure be ‘P’ $\\$ $L \propto \frac{1}{p}$ for L = half of the tube length, P = 0.02 mm of Hg $\\$ As ‘P’ becomes half, ‘L’ doubles, that is the whole tube is filled with Crook’s dark space. $\\$ Hence the required pressure $= \frac{0.02}{2} = 0.01$m of Hg.

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 ?

##### Solution :

$V = f(Pd)$ $\\$ $V_s = P_s d_s$ $\\$ $V_L = P_I d_I$ $\\$ $\Rightarrow \frac{V_s}{V_I} = \frac{P_s}{P_I} \times \frac{d_s}{d_I} \Rightarrow \frac{100}{100} = \frac{10}{20} \times \frac{1mm}{x}$ $\\$ $\Rightarrow x = \frac{1 mm }{2} = 0. 5 mm$

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 $4.52 eV$.

##### Solution :

$i = ne \quad or \quad n = \frac{i}{e}$ $'e'$ is same in all cases. $\\$ We know, $\\$ $i = AST^2 e^{- \phi /RT} \qquad \phi = 4.52eV, K = 1.38 \times 10^{-23}J/K$ $\\$ $n(1000) = As \times (1000)^2 \times e^{-4.52 \times 6 \times 10^{-19}/1.38 \times 10^{-23} \times 1000}$ $\\$ $\Rightarrow 1.7396 \times 10^{-17}$ $\\$ (a) $T = 300K$ $\\$ $\frac{n(T)}{n(1000K)} = \frac{AS \times (300)^2 \times e^{-4.52 \times 6 \times 10^{-19}/1.38 \times 10^{-23} \times 300}}{AS \times 1.7396 \times 10^{-17}} = 7.05 \times 10^{-56}$ $\\$ (b) $T = 2000K$ $\\$ $\frac{n(T)}{n(1000K)} = \frac{AS \times (2000)^2 \times e^{-4.52 \times 6 \times 10^{-19}/1.38 \times 10^{-23} \times 2000}}{AS \times 1.7396 \times 10^{-17}} = 9.59 \times 10^{11}$ $\\$ (c) $T = 3000K$ $\\$ $\frac{n(T)}{n(1000K)} = \frac{AS \times (3000)^2 \times e^{-4.52 \times 6 \times 10^{-19}/1.38 \times 10^{-23} \times 3000}}{AS \times 1.7396 \times 10^{-17}} = 1.340 \times 10^{16}$

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 $60 \times 10^ 4 A/m ^2 -K ^2$ for pure tungsten and $3.0 \times 10^4 A/m^2 - K^2$ for thoriated tungsten. The work function of pure tungsten is 4.5 eV and that of thoriated tungsten is 2.6 eV

##### Solution :

$i = AST^2 e^{- \phi / KT}$ $\\$ $i_1 = i \qquad i_2 = 100mA$ $\\$ $A_1 = 60 \times 10^4 \qquad A_2 = 3 \times 10^4$ $\\$ $T_1 = 2000 \qquad T_2 =2000$ $\\$ $\phi_1 = 4.5 eV \qquad \phi_2= 2.6eV$ $\\$ $K = 1.38 \times 10^{-23} J/k$ $\\$ $i = (60 \times 10^4)(S) \times (2000)^2 e^{\frac{-4.5 \times 1.6 \times 10^{-19}}{1.38 \times 10^{-23 \times 2000}}}$ $\\$ $100 = (3 \times 10^4)(S) \times (2000)^2 e^{\frac{-4.5 \times 1.6 \times 10^{-19}}{1.38 \times 10^{-23 \times 2000}}}$ $\\$ Dividing the equation $\\$ $\frac{i}{100} = e^{\Bigg[ \frac{-4.5 \times 1.6 \times 10}{1.38 \times 2} (\frac{-2.6 \times 1.6 \times 10}{1.38 \times 20}) \Bigg]}$ $\\$ $\Rightarrow \frac{i}{100} = 20 \times e^{-11.014} \qquad \Rightarrow \frac{i}{100} = 20 \times 0.000016$ $\\$ $i = 20 \times 0.0016 = 0.0329 mA = 33 \mu A$

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.

##### Solution :

Pure tungsten $\\$ $\phi = 4.5eV$ $\\$ $A = 60 \times 10^4 A/m^2-K^2$ $\\$ Thoriated tungsten $\\$ $\phi = 2.6eV$ $\\$ $A = 3 \times 10^4 A/m^2-K^2$ $\\$ $i= AST^2 e^{-\phi/KT}$ $\\$ $i_Thoriated Tungsten = 5000 i_Tungsten$ $\\$ so, $5000 \times S \times 60 \times 10^4 \times T^2 \times e^{\frac{-4.5 \times 1.6 \times 10^{-19}}{1.38 \times T \times 10^{-23}}}$ $\\$ $\Rightarrow S \times 3 \times 60 \times 10^4 \times T^2 \times e^{\frac{-4.5 \times 1.6 \times 10^{-19}}{1.38 \times T \times 10^{-23}}} = e^{\frac{-4.5 \times 1.6 \times 10^{-19}}{1.38 \times T \times 10^{-23}}} \times 3 \times 10^4$ $\\$ Taking 'In' $\\$ $\Rightarrow 9.21 T = 220.29$ $\\$ $T = 22029 / 9.21 = 2391.856 K$

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.

##### Solution :

$i = AST^2 e^{\pi /KT}$ $\\$ $i' = AST^{12} e ^{-\pi /KT'}$ $\\$ $\frac{i}{i'} = \frac{T^2 }{T^{12}} \frac{e ^{\pi /KT}}{e ^{-\pi /KT'}}$ $\\$ $\Rightarrow \frac{i}{i'} = \Bigg( \frac{T}{T'} \Bigg)^2 e^{-\pi /KT + \pi KT'} = \Bigg( \frac{T}{T'} \Bigg)^2 e^{\pi KT' - \pi / KT}$ $\\$ $= \frac{i}{i'} = \Bigg( \frac{2000}{2010} \Bigg)^2 e^{\frac{4.5 \times 1.6 \times 10^{-19}}{1.38 \times 10^{-23}}} \Bigg( \frac{1}{2010} - \frac{1}{2000} \Bigg) = 0.8690$ $\\$ $\Rightarrow \frac{i}{i'} = \frac{1}{0.8699} = 1.1495 = 1.14$

9   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 ?

##### Solution :

For plate current 20 mA, we find the voltage 50 V or 60 V. $\\$ Hence it acts as the saturation current. Therefore for the same temperature, the plate current is 20 mA for all other values of voltage.$\\$ Hence the required answer is 20 mA

10   The constant A in the Richardson-Dushman equation for tungsten is $60 \times 10^4 A/m^2 -K^2.$ The work function of tungsten is 4'5 eV. A tungsten cathode having a surface area $2.0 \times 10^{-5} m^2$ is heated by a 24 W electric heater. In steady state, the heat radiated by 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 constant $= 6 \times 10^{-8} w/m^2 - K^4$. Assume that the thermions take only a small fraction of the heat supplied.

##### Solution :

$A = 60 \times 10^4 A/m^2 - k^2$ $\\$ $\pi = 4.5 eV \qquad \sigma = 6 \times 10^{-8} \omega /m^2 -k^4$ $\\$ $S = 2 \times 10^{-5} m^2 \qquad K =1.38 \times 10^{-23} J/K$ $\\$ $H = 24 \omega'$ $\\$ The Cathode acts as a black body , i.e. emissivity = 1 $\\$ $\therefore E = \sigma A T^4$ (A is area) $\\$ $\Rightarrow T^4 = \frac{E}{\sigma A} =\frac{24}{6 \times 10^{-8} \times 2 \times 10^{-5}} = 2\times 10^{13} K = 20 \times 10^{12} K$ $\\$ $\Rightarrow T = 2.1147 \times 10^3 = 2114.7 K$ $\\$ Now, $i = AST^2 e^{-\pi /KT}$ $\\$ $= 6 \times 10^5 \times 2 \times 10^{-5} \times (2114.7)^2 \times e^{\frac{-4.5 \times 1.6 \times 10^{-19}}{1.38 \times T \times 10^{-23}}}$ $\\$ $=1.03456 \times 10^{-3} A = 1mA$ $\\$

11   A plate current of 10 mA is obtained when 60 volts are applied across a diode tube. Assuming the Langmuir-Child equation $i _p \propto V _p^{3/2}$ to hold, find the dynamic resistance $r_p$ in this operating condition.

##### Solution :

$i_p =CV _p^{3/2}\qquad \qquad ...........(1)$ $\\$ $\Rightarrow di_p = C 3/2 V_p^{(3/2) -1} dv_p$ $\\$ $\Rightarrow \frac{di_p}{dv_p} = \frac{3}{2} CV_p^{1/2} \qquad ............(2)$ $\\$ Dividing (2) and (1) $\\$ $\frac{i}{i_p} \frac{di_p}{dv_p} = \frac{3/2 CV_p^{1/2}}{CV_p^{3/2}}$ $\\$ $\Rightarrow \frac{1}{i_p} \frac{di_p}{dv_p} = \frac{3}{2V}$ $\\$ $\Rightarrow \frac{di_p}{dv_p} = \frac{2V}{3i_p}$ $\\$ $\Rightarrow R = \frac{2V}{3i_p} = \frac{2 \times 60}{3 \times 10 \times 10^{-3}} = 4 \times 10^3 = 4k \Omega$

12   e 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.

##### Solution :

$P =1W, \qquad p =?$ $\\$ $V_p = 36V, \quad V_p = 49V, \quad P =I_pV_p$ $\\$ $\Rightarrow I_p = \frac{P}{V_p} = \frac{1}{36}$ $\\$ $I_p \propto (V_p)^{3/2}$ $\\$ $I'_p \propto (V'_p)^{3/2}$ $\\$ $\Rightarrow \frac{I_p}{I'_p} = \frac{(V_p)^{3/2}}{V'_p}$ $\\$ $\Rightarrow \frac{1/36}{I'_p} = \Bigg( \frac{36}{49} \Bigg)^{3/2}$ $\\$ $\Rightarrow \frac{1}{36I'_p} = \frac{36}{49} \times \frac{6}{7} \Rightarrow I'_p = 0.4411$ $\\$ $P' = V'_pI'_p = 49 \times 0.4411 = 2.1613 W = 2.2 W$

13   A triode valve 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.

##### Solution :

Amplification factor for triode value $= \mu = \frac{Charge \quad in \quad Plate \quad Voltage}{Charge \quad in \quad Grid \quad Voltage}$ $\\$ $= \frac{250 - 225}{2.5 - 0.5} = \frac{25}{2} = 12.5 \qquad [ \therefore \delta V_p = 250- 225, \delta V_g = 2.5 -0.5 ]$ $\\$

14   Calculate the amplification factor of a triode valve which has plate resistance of $2 k \Omega$ and transconductance of 2 millimho.

##### Solution :

$r_p = 2k \Omega = 2 \times 10^3 \Omega$ $\\$ $g_m = 2 milli \quad mho = 2 \times 10^{-3} mho$ $\\$ $\mu = r_p \times g_m = 2 \times 10^3 \times 2 \times 10^{-3} = 4$ Amplification factor is 4. $\\$

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

##### Solution :

Dynamic Plate Resistance $r_p = 10 k \Omega = 10^4 \Omega$ $\\$ $\delta I_p =?$ $\\$ $\delta V_p =220- 220 = 20 V$ $\\$ $\delta I_p = (\delta V_p / r_p) V_g = constant$ $\\$ $=20/10^4 = 0.002A = 2mA$

16   Find the values of $r_p , \mu$ and $g_m$ of a triode operating at plate voltage $200 V$ and grid voltage $- 6 V$. The plate characteristics are shown in figure $(41-E1)$.

##### Solution :

$r_p = \Bigg( \frac{\delta V_p}{\delta V_g } \Bigg)$ at a constant $V_g$ $\\$ Consier the two points on $V_g = -6 line$ $\\$ $r_p =\frac{(240 -160)V}{(13-3) \times 10^{-3}A} = \frac{80}{10} \times 10^3 \Omega = 8K \Omega$ $\\$ $g_m = \Bigg( \frac{\delta I_p}{ \delta V_g} \Bigg) v_p = constant$ $\\$ Considering the points on 200 V line, $\\$ $g_m = \frac{(13-3) \times 10^{-3}}{[(-4) + (-8)]}A = \frac{10 \times 10^{-3}}{4} = 2.5 milli \quad mho$ $\\$ $\mu = r_p \times g_m$ $\\$ $= 8 \times 10^3 \Omega \times 2.5 \times 10^{-3} \Omega ^{-1} = 8 \times 1.5 = 20$ $\\$

17   The plate resistance of a triode is $8 k\Omega$ 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, how much should the grid voltage be decreased in order to bring the plate current back to its initial value ?

##### Solution :

(a) $r_p = 8k \Omega = 8000 \Omega$ $\\$ $\delta V_p = 48 V, \qquad \delta I_p = ?$ $\\$ $\delta I_p =(\delta V_p / r_p) /V_g = constant$ $\\$ So, $\delta I_p = 48/8000 = 0.006A =6mA$ $\\$ (b) Now, $V_p$ is constant. $\\$ $\delta I_p = 6mA = 0.006A$ $\\$ $g_m = 0.0025 mho$ $\\$ $\delta V_g = (\delta I_p / g_m) /V_p = constant$ $= \frac{0.006}{0.0025} = 2.4 V$

18   The plate resistance and the amplification factor of a triode are $10 k \Omega$ 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 ?

##### Solution :

$r_p = 10 k \Omega = 10 \times 10^3 \Omega$ $\\$ $\mu = 20, \qquad V_p = 250V$ $\\$ $V_g = -7.5 V, \qquad I_p =250V$ $\\$ (a) $g_m = \Bigg(\frac{\delta I_p}{\delta V_g} \Bigg) V_p = constant$ $\\$ $\Rightarrow \delta V_g = \frac{\delta I_p}{g_m} = \frac{15 \times 10^{-3} - 10 \times 10^{-3}}{ \mu/ r_p}$ $\\$ $\frac{5 \times 10^{-3}}{20/10 \times 10^3}= \frac{5}{2} = 2.5$ $\\$ $r'_g = +2.5 - 7.5 = -5V$ $\\$ (b) $r_p = \Bigg(\frac{\delta V_p}{\delta I_p} \Bigg) V_g = constant$ $\\$ $\Rightarrow 10^4 = \frac{\delta V_p}{15 \times 10^{-3} - 10 \times 10^{-3}}$ $\\$ $\Rightarrow \delta V_p = 10^4 \times 5 \times 10^{-3} = 50V$ $\\$ $V'_p - V_p = 50 \quad \Rightarrow V'_p = -50 + V_p = 200V$ $\\$

19   The plate current, plate voltage and grid voltage of a 6F6 triode tube are related as $i_p = 41(V_p + 7 V_g)^{1.41}$ where $V_ p$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

##### Solution :

$V_p = 250V, \qquad V_g =-20V$ $\\$ (a) $i_p = 41(V_p + 7 V_g)^{1.41}$ $\\$ $\Rightarrow 41 (250 - 140)^{1.41} = 41 \times (110)^{1.41} = 30984 \mu A = 30mA$ $\\$ (b) $i_p = 41(V_p + 7 V_g)^{1.41}$ $\\$ Differentiating, $\\$ $di_p =41 \times 1.41 \times (V_p + 7V_g)^{0.41} \times (dV_p + 7dV_g)$ $\\$ Now, $r_p = \frac{dV_p}{di_p}V_g =constant$ $\\$ or $\frac{dV_p}{di_p} = \frac{1 \times 10^6}{41 \times 1.41 \times 110^{0.41}} = 10^6 \times 2.51 \times 10^{-3} \Rightarrow 2.5 \times 10^3 \Omega = 2.5 k \Omega$ $\\$ (c) From above, $dI_p = 41 \times 1.41 \times 6.87 \times 7 d V_g$ $\\$ $g_m = \frac{d I_p}{d V_g} = 41 \times 1.41 \times 6.87 \times 7 \mu mho$ $\\$ $= 2780 \mu mho = 2.78 milli \quad mho$ $\\$ (d) Amplifier factor $\\$ $\mu = r_p \times g_m = 2.5 \times 10^3 \times 2.78 \times 10^{-3} = 6.95 = 7$

20   The plate current in a triode can be written as $\\$ $i_p = k \Bigg[ V_g + \frac{V_p}{\mu} \Bigg] ^{3/2 }$ Show that the mutual conductance is proportional to the cube root of the plate current.

##### Solution :

$i_p = K(V_g + V_p/ \mu)$ $\\$ Diff the equation : $\\$ $di_p = K 3/2 (V_g + V_p / \mu)^{1/2} dV_g$ $\\$ $\Rightarrow \frac{di_p}{dV_g} = \frac{3}{2}K\Bigg( V_g + \frac{V_0}{\mu} \Bigg)$ $\Rightarrow g_m = 3/2 K (V_g + V_p / \mu)^{1/2} \qquad .......(2)$ $\\$ From (1) $i_p = [ 3/2 K ( V_g + V_p/ \mu )^{1/2} ]^3 \times 8/K^2 27$ $\\$ $\Rightarrow i_p= k'(g_m)^3 \quad \Rightarrow g_m \propto 3 \sqrt{i_p}$

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

##### Solution :

$r_p = 20 k\Omega =$ Plate Resistance $\\$ Mutual conductance $= g_m = 2.0 milli \quad mho = 2 \times 10^{-3} mho$ $\\$ Amplification factor $\mu = 30$ $\\$ Load Resistance $= R_L = ?$ $\\$ We know, $\\$ $A = \frac{\mu}{1 + \frac{r_p}{R_L}} \qquad$ where A = voltage amplification factor $\\$ $\Rightarrow A =\frac{r_p \times g_m}{1 + \frac{r_p}{R_L}} \qquad where \quad \mu = r_p \times g_m$ $\\$ $\Rightarrow 30 = \frac{20 \times 10^3 \times 2 \times 10^{-3}}{1 + \frac{20000}{R_L}} \Rightarrow 3 = \frac{4R_L}{R_L + 20000}$ $\\$ $\Rightarrow 3R_L + 60000 = 4R_L$ $\\$ $\Rightarrow R_L = 60000 \Omega = 60 k \Omega$

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

##### Solution :

Voltage gain $= \frac{\mu }{1+ \frac{r_p}{R_L}}$ $\\$ When $A = 10, \quad R_L =4 k \Omega$ $\\$ $10 = \frac{\mu}{1 + \frac{r_p}{4 \times 10^3}} \Rightarrow 10 = \frac{\mu \times 4 \times 10^3}{4 \times 10^3 + r_p}$ $\\$ $\Rightarrow 40 \times 10^3 \times 10 r_p = 4 \times 10^3 \mu \qquad .....(1)$ $\\$ when $A =12, \quad R_L = 8k\Omega$ $\\$ $12 = \frac{\mu}{1+ \frac{r_p}{8 \times 10^3}} \Rightarrow 12 = \frac{\mu \times 8 \times 10^3}{8 \times 10^3 + r_p}$ $\\$ $\Rightarrow 96 \times 10^3 + 12r_p = 8 \times 10^3 \mu \qquad ....(2)$ $\\$ Multiplying (2) in equation (1) and equating with equation(2) $\\$ $2(40 \times 10^3 + 10r_p) = 96 \times 10+ 3 +12r_p$ $\\$ $\Rightarrow r_p = 2 \times 10^3 \Omega = 2 k \Omega$ $\\$ Putting the value in equation (1) $\\$ $40 \times 10^3 \times 10(2 \times 10^3) = 4 \times 10^3 \mu$ $\\$ $\Rightarrow 40 \times 10^3 + 20 \times 10^3 = 4 \times 10^3 \mu$ $\\$ $\Rightarrow \mu = \frac{60}{4} = 15$