Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

In the circuit shown in the figure, the input voltage Vi is 20 V, VBE = 0 and VCE = 0. The values of IB, lC and \beta  are given by

  • Option 1)

    I_{B} = 20\mu A , I_{C} = 5mA , \beta = 250

  • Option 2)

    I_{B} = 25\mu A , I_{C} = 5mA , \beta = 200

  • Option 3)

    I_{B} = 40\mu A , I_{C} = 10mA , \beta = 250

  • Option 4)

    I_{B} = 40\mu A , I_{C} = 5mA , \beta = 125

 

Answers (1)

best_answer

As we have learned

Relation between ? and ? -

eta = frac{alpha }{1-alpha }
 

- wherein

alpha = frac{I_{C}}{I_{E}}

eta = frac{I_{C}}{I_{B}} (current gain )

 

 I_{B}= \frac{V_{i}-V_{B}}{R_{B}}= 20-0/500k\Omega = 40\mu A

I_{C}= \frac{20-0}{4k\Omega } = 5mA

\beta= \frac{I_{C}}{I_{B}}=125

 

 

 

 


Option 1)

I_{B} = 20\mu A , I_{C} = 5mA , \beta = 250

This is incorrect

Option 2)

I_{B} = 25\mu A , I_{C} = 5mA , \beta = 200

This is incorrect

Option 3)

I_{B} = 40\mu A , I_{C} = 10mA , \beta = 250

This is incorrect

Option 4)

I_{B} = 40\mu A , I_{C} = 5mA , \beta = 125

This is correct

Posted by

prateek

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

NEET UG 2026 correction facility opens at neet.nta.nic.in; exam on May 3
anam-khan

NEET 2026 LIVE: NTA application correction window closes today; admit card, exam dates, pattern
anam-khan

NEET 2026 registration deadline extended to March 11 for MBBS, BDS; exam on May 3

Latest Question