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JEE Main Physics Class 12 Electronic Devices Part 2 Questions
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© examsnet.com
Question : 80
Total: 111
In a given circuit diagram, a
5
V
zener diode along with a series resistance is connected across a
50
V
power supply. The minimum value of the resistance required, if the maximum zener current is
90
m
A
will be _____
Ω
.
[22 Jul 2021 Shift 2]
Your Answer:
Validate
Solution:
Voltage across
R
L
=
5
V
⇒
i
2
=
5
R
L
Also voltage across
R
=
50
−
5
=
45
volt
B
y
v
=
i
R
⇒
R
=
v
i
=
45
i
i
+
i
2
R
=
45
90
m
A
+
5
R
L
Current in zener diode is maximum when
R
L
→
∞
(
i
2
→
0
and
i
i
=
i
)
So
R
=
45
90
m
A
=
500
Ω
© examsnet.com
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