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Class 12 NEET Physics Semiconductor Electronics Part 2
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© examsnet.com
Question : 28
Total: 43
In the circuit shown in the figure, the input voltage
V
1
is 20V,
V
B
E
=
0
and
V
C
E
=
0
.The values of
I
B
,
I
C
and β are given by
[NEET 2018]
I
B
=
40
µ
A
,
I
C
=
10
m
A
,
β
=
250
I
B
=
25
µ
A
,
I
C
=
5
m
A
,
β
=
200
I
B
=
40
µ
A
,
I
C
=
5
m
A
,
β
=
125
I
B
=
20
µ
A
,
I
C
=
5
m
A
,
β
=
250
Validate
Solution:
👈: Video Solution
V
BE
=
0
V
CE
=
0
V
b
=
0
I
C
=
(
20
−
0
)
4
×
10
3
I
C
=
5
×
10
−
3
=
5
m
A
V
i
=
V
BE
+
I
B
R
B
V
i
=
0
+
I
B
R
B
20
=
I
B
×
500
×
10
3
I
B
=
20
500
×
10
3
=
40
µ
A
β
=
I
C
I
b
=
25
×
10
−
3
40
×
10
−
6
=
125
© examsnet.com
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