CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A PNP in common emitter configuration used as an amplifier, its current gain is 50. If input resistance is 1 kΩ and input voltage is 5 V , then output current will be-

A
250 mA
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
30 mA
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
50 mA
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
100 mA
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 250 mA
Given,

RB=1 kΩ ; VB=5 V ; β=50

From Ohm's law, V=IR

IB=VBRB=5×103 A (or)

IB=5 mA

IC=β IB=50×5×103=250 mA

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (A) is the correct answer.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Transistor
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon