Class 12-science H C VERMA Solutions Physics Chapter 19 - Electric Current through Gases
Electric Current through Gases Exercise 352
Solution 1
Force is given as:
and 
For time period 'dt'
Also,
(for positive ion)
Taking ratio of (1) and (2), we get
Solution 2
We know,
and 
(a) 
m
(b) 
t=1.5ns
Electric Current through Gases Exercise 353
Solution 3
We know,
If P becomes half then L will double.
Pressure required=
m of Hg.
Solution 4
We know,
mm
Solution 5
And
(a) If T=300K then,
(b) If T=2000K then,
(c) If T=3000K then,
Solution 6
We know,
For 
For 
Dividing eq (1) and (2), we get
A
Solution 7
We know,
Also,
Taking log and solving we get,
T=2391.85K
Solution 8
and 
Dividing both equations above we get,
Solution 9
We know,
[AS emissivity is 1]
T=2114.7K
Also,
mA
Solution 10
We know,
Differentiating we get
Now, dividing above equations we get,
[
]
kΩ
Solution 11
Plate have voltage of 50V or 60V, so its acts as saturation current.
Hence, for other value of voltages current is 20mA.
Solution 12
We know,
and 
Also,
Now,
W
Solution 13
Amplification factor is:
Solution 14
We know,
Solution 15
We know,
mA
Solution 16
We know,
and
Also, we know
Solution 17
(a) 
mA
(b) 
V
Solution 18
(a) 
[As 
]
And thus, grid voltage should be charged by,
(b)
V
or 
V
Solution 19
(a) 
mA
(b) 
Differentiating above equation we get
Also, 
(c) 
(d) 
Solution 20
Differentiating we get,
Comparing (1) and (2) we get
or 
Solution 21
We know,
RL = 60kΩ
Electric Current through Gases Exercise 354
Solution 22
Voltage gain, 
Case 1: 
Case 2:
Solving equation (1) and (2)
We get: 
And 
Solution 23
Resistances are connected in parallel, thus equivalent resistance is given as:
Also, equivalent conductance is given as:
Now,
Equivalent amplification is given as:
Which is same as individual mutual conductance.
