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ELECTRIC CURRENT THROUGH GASES CHAPTER 41 1.

Let the two particles have charge ‘q’ Mass of electron ma = 9.1  10–31 kg Mass of proton mp = 1.67  10–27 kg Electric field be E Force experienced by Electron = qE accln. = qE/me For time dt

1 qE   dt 2 2 me

Se =

…(1)

For the positive ion, accln. =

qE 4  mp

1 qE   dt 2 2 4  mp

Sp =

…(2)

Se 4mp = 7340.6  Sp me 2.

3

–6

E = 5 Kv/m = 5  10 v/m ; t = 1 s = 1  10 F = qE = 1.6  10 a=

–9

 5  10

s

3

qE 1.6  5  10 16  m 9.1 10 31

a) S = distance travelled =

1 2 at = 439.56 m = 440 m 2

b) d = 1 mm = 1  10 1  10

–3

2

t = 3.

=

–3

m

1 1.6  5 5 2  10  t 2 9.1

9.1 –9  1018  t = 1.508  10 sec  1.5 ns. 0.8  5

Let the mean free path be ‘L’ and pressure be ‘P’ L  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 = 0.02/2 = 0.01 m of Hg. 4.

V = f(Pd) vs = Ps ds vL = Pl dl 

Vs Ps ds 100 10 1mm      Vl Pl dl 100 20 x

 x = 1 mm / 2 = 0.5 mm 5.

i = ne or n = i/e ‘e’ is same in all cases. We know, 41.1

A

C


Electric current through gases 2

i = AST e

 / RT

 = 4.52 eV, K = 1.38  10

n(1000) = As  (1000)  e 4.521.610 2

19

/ 1.3810

–23

23

J/k

1000

–17

 1.7396  10 a) T = 300 K

19

n(T) AS  (300)2  e 4.521.610 / 1.3810  n(1000K) AS  1.7396  10 17

23

300

= 7.05  10

–55

b) T = 2000 K 19

n(T) AS  (2000)2  e4.521.610 / 1.3810  n(1000K) AS  1.7396  10 17

23

 2000

= 9.59  10

11

c) T = 3000 K 19

 6.

n(T) AS  (3000)2  e4.521.610 / 1.3810  n(1000K) AS  1.7396  10 17

23

3000

16

= 1.340  10 

i = AST e  / KT 2

i1 = i

i2 = 100 mA

A1 = 60  10

4

A2 = 3  10

S1 = S

4

S2 = S

T1 = 2000

T2 = 2000

1 = 4.5 eV

2 = 2.6 eV

K = 1.38  10

–23

J/k 4.51.61019

4

i = (60  10 ) (S)  (2000)

2

23 e1.3810 2000

2.61.61019 4

100 = (3  10 ) (S)  (2000)

2

23 e1.3810 2000

Dividing the equation  4.51.610 2.61.610  ( ) 1.382 1.3820 

i   e 100 

i i  20  e11.014   20  0.000016 100 100

 i = 20  0.0016 = 0.0329 mA = 33 A 7.

Pure tungsten

Thoriated tungsten

 = 4.5 eV

 = 2.6 eV 4

2

A = 60  10 A/m – k

2

4

i = AST e / KT 2

iThoriated Tungsten = 5000 iTungsten 4.51.610 19 4

2

So, 5000  S  60  10  T  e 1.38T10

23

2.651.61019 4

2

 S  3  10  T  e

1.38T 1023

4.51.610 19 8

 3  10  e

2

A = 3  10 A/m – k

1.38T10 23

2.651.61019

= e

1.38T 1023

 3  10

4

Taking ‘ln’  9.21 T = 220.29  T = 22029 / 9.21 = 2391.856 K 41.2

2


Electric current through gases 8.

2

i = AST e

 / KT

e / KT

12

i = AST

i T 2 e / KT  12  / KT i T e 2

2

i  T   / KT KT  T  KT  / KT   e   e i  T    T  2

i  2000  =   e i  2010   9.

4.51.61019 1.3810 23

1   1    = 0.8690  2010 2000 

i 1   1.1495 = 1.14 i 0.8699 4

2

A = 60  10 A/m – k

2 –8

 = 4.5 eV S = 2  10

 = 6  10 –5

m

2

2

/m – k

K = 1.38  10

–23

4

J/K

H = 24  The Cathode acts as a black body, i.e. emissivity = 1 4

 E =  A T (A is area) 4

T =

E 24   2  1013 K  20  1012 K 8 A 6  10  2  10 5 3

 T = 2.1147  10 = 2114.7 K = AST e / KT 2

Now, i

4.51.610 19 5

–5

= 6  10  2  10 = 1.03456  10 10. ip =

–3

2

 (2114.7)  e 1.38T10

23

A = 1 mA

CVp3 / 2

…(1)

 dip = C 3/2 Vp(3 / 2)1dv p 

dip dv p

3 CVp1/ 2 2

…(2)

Dividing (2) and (1)

3 / 2CVp1/ 2 i dip  ip dv p CVp3 / 2  

1 dip 3  ip dv p 2V dv p dip

R=

2V 3ip

2V 2  60   4  103  4k 3ip 3  10  10 3

11. 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. 12. P = 1 W, p = ? Vp = 36 V, Vp = 49 V, P = IpVp 41.3


Electric current through gases

P 1  Ip =  Vp 36 Ip  (Vp)

3/2

Ip  (Vp) 

Ip

Ip

3/2

(Vp )3 / 2 Vp 3/2

1/ 36  36    Ip  49 

1 36 6    Ip  0.4411 36 Ip 49 7

P = Vp Ip = 49  0.4411 = 2.1613 W = 2.2 W 13. Amplification factor for triode value

Charge in Plate Voltage Vp  Charge in Grid Voltage Vg

== =

250  225 25   12.5 2.5  0.5 2

[ Vp = 250 – 225, Vg = 2.5 – 0.5]

3

14. rp = 2 K = 2  10  gm = 2 milli mho = 2  10

–3

3

mho

 = rp  gm = 2  10  2  10

–3

= 4 Amplification factor is 4.

15. Dynamic Plate Resistance rp = 10 K = 104  Ip = ? Vp = 220 – 220 = 20 V Ip =Vp / rp) / Vg = constant. 4

= 20/10 = 0.002 A = 2 mA

 Vp 16. rp =   Ip

  at constant Vg 

Consider the two points on Vg = –6 line rp =

(240  160)V (13  3)  10 3 A

80  103   8K 10

 Ip  gm =  = constant  Vg  v p   Considering the points on 200 V line, gm =

(13  3)  10 3 10  10 3  2.5 milli mho A [( 4)  ( 8)] 4

 = rp  gm = 8  103   2.5  10–3 –1 = 8  1.5 = 20 17. a) rp = 8 K = 8000  Vp = 48 V

Ip = ?

Ip = Vp / rp) / Vg = constant. So, Ip = 48 / 8000 = 0.006 A = 6 mA b) Now, Vp is constant. Ip = 6 mA = 0.006 A 41.4


Electric current through gases gm = 0.0025 mho Vg = Ip / gm) / Vp = constant. =

0.006 = 2.4 V 0.0025 3

18. rp = 10 K = 10  10   = 20

Vp = 250 V

Vg = –7.5 V

Ip = 10 mA

 Ip  a) gm =   Vg  Vp = constant    Vg =

Ip

gm

=

15  103  10  10 3  / rp

5  10 3 20 /10  10

3

5  2.5 2

rg = +2.5 – 7.5 = –5 V

 Vp b) rp =   Ip

  Vg = constnant  Vp

4

 10 =

(15  10

3

 10  10 3 )

4

–3

 Vp = 10  5  10

= 50 V

Vp – Vp = 50  Vp = –50 + Vp = 200 V 19. Vp = 250 V, Vg = –20 V a) ip = 41(Vp + 7Vg)

1.41

 41(250 – 140) b) ip = 41(Vp + 7Vg)

1.41

1.41

= 41  (110)

= 30984 A = 30 mA

1.41

Differentiating, dip = 41  1.41  (Vp + 7Vg)

dVp

Now rp = or

dVp dip

dip

0.41

 (dVp + 7dVg)

Vg = constant.

1 106

6

41 1.41 1100.41

= 10  2.51  10

–3

3

 2.5  10  = 2.5 K

c) From above, dIp = 41  1.41  6.87  7 d Vg gm =

dIp dVg

= 41  1.41  6.87  7  mho

= 2780  mho = 2.78 milli mho. d) Amplification factor 3

 = rp  gm = 2.5  10  2.78  10 20. ip = K(Vg + Vp/)

3/2

–3

= 6.95 = 7

…(1)

Diff. the equation : dip = K 3/2 (Vg + Vp/)1/2 d Vg 

dip dVg

1/ 2

V  3  K  Vg  0    2 

41.5


Electric current through gases  gm = 3/2 K (Vg + Vp/)

1/2

…(2)

From (1) ip = [3/2 K (Vg + Vp/)1/2]3  8/K2 27 3

 ip = k (gm)  gm  3 ip 21. rp = 20 K = Plate Resistance Mutual conductance = gm = 2.0 milli mho = 2  10–3 mho Amplification factor  = 30 Load Resistance = RL = ? We know A=

 A=

 rp 1 RL

where A = voltage amplification factor

rp  gm rp 1 RL

 30 =

where  = rp  gm

4RL 20  103  2  10 3 3= 20000 RL  20000 1 RL

 3RL + 60000 = 4 RL  RL = 60000  = 60 K  22. Voltage gain =

 rp 1 RL

When A = 10, RL = 4 K 10 =

1

 rp

  4  103

 10 

4  10

4  103  rp

3

3

3

 40  10  10rp = 4  10 

…(1)

when A = 12, RL = 8 K 12 =

1

 rp

 12 

8  103

  8  103 8  103  rp

3

3

 96  10 + 12 rp = 8  10 

…(2)

Multiplying (2) in equation (1) and equating with equation (2) 3

2(40  10 + 10 rp) = 96  10+3 + 12rp 3

 rp = 2  10  = 2 K Putting the value in equation (1) 3

3

3

40  10 + 10(2  10 ) = 4  10  3

3

3

 40  10 + 20  10 ) = 4  10   = 60/4 = 15 



41.6


41.electric current through gases