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 1018 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 4.521.610 2
19
/ 1.3810
–23
23
J/k
1000
–17
1.7396 10 a) T = 300 K
19
n(T) AS (300)2 e 4.521.610 / 1.3810 n(1000K) AS 1.7396 10 17
23
300
= 7.05 10
–55
b) T = 2000 K 19
n(T) AS (2000)2 e4.521.610 / 1.3810 n(1000K) AS 1.7396 10 17
23
2000
= 9.59 10
11
c) T = 3000 K 19
6.
n(T) AS (3000)2 e4.521.610 / 1.3810 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 4.51.61019
4
i = (60 10 ) (S) (2000)
2
23 e1.3810 2000
2.61.61019 4
100 = (3 10 ) (S) (2000)
2
23 e1.3810 2000
Dividing the equation 4.51.610 2.61.610 ( ) 1.382 1.3820
i e 100
i i 20 e11.014 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 4.51.610 19 4
2
So, 5000 S 60 10 T e 1.38T10
23
2.651.61019 4
2
S 3 10 T e
1.38T 1023
4.51.610 19 8
3 10 e
2
A = 3 10 A/m – k
1.38T10 23
2.651.61019
= e
1.38T 1023
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.
4.51.61019 1.3810 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
4.51.610 19 5
–5
= 6 10 2 10 = 1.03456 10 10. ip =
–3
2
(2114.7) e 1.38T10
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
Ip (Vp)
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 = Vp Ip = 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
=
15 103 10 10 3 / rp
5 10 3 20 /10 10
3
5 2.5 2
rg = +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
Vp – Vp = 50 Vp = –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