Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

An npn transistor in a common emitter mode is used as a simple voltage – amplifier with a collector current of 4 mA. The terminal of 8 V battery is connected to the collector through a load resistance RL and base through load resistance  RB. The collector emitter voltage VCE=4V , base emitter voltage VBE=0.6V and current amplification factor   β= 100. Find RB (in ohm).

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

18.50×104

b

19.50×104

c

19.50×106

d

18.50×106

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Ic=4mA=4×10−3A, Vcc=8V, VCE=4V, VBE=0.6V, β=100, RB=?Using KVL in outer loop ,  Vcc-VBE -IbRB = 0, ⇒RB=Vcc−VBEIb,Also , β=IcIb, ⇒Ib=Icβ,∴RB=Vcc−VBEIcβ=βVCC−VBEIc=100(8−0.64×10−3)=25(7.410−3)=25×7.4×103=25×74×102=1850×102=18.50×104Ω
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring