Home
Class 12
PHYSICS
In the above problem the current amplifi...

In the above problem the current amplification factor `beta` is

A

`83`

B

`100`

C

`133`

D

`203`

Text Solution

Verified by Experts

The correct Answer is:
C

`B=(I_(C))/(I_(B))=(3.33 xx10^(-3))/(25xx10^(-6))=1.33 xx10^(2)=133`.
Promotional Banner

Topper's Solved these Questions

  • SEMI CONDUCTOR DEVICES

    NARAYNA|Exercise Level-I (H.W)|26 Videos
  • SEMI CONDUCTOR DEVICES

    NARAYNA|Exercise Level-II (H.W)|36 Videos
  • SEMI CONDUCTOR DEVICES

    NARAYNA|Exercise Level-III (C.W)|20 Videos
  • RAY OPTICS AND OPTICAL INSTRAUMENTS

    NARAYNA|Exercise EXERCISE- 4 One or more than one correct answer type|13 Videos
  • SEMICONDUCTOR ELECTRONICS

    NARAYNA|Exercise ADDITIONAL EXERCISE (ASSERTION AND REASON TYPE QUESTIONS :)|19 Videos

Similar Questions

Explore conceptually related problems

In a common base mode of transistor, collector current is 5.488 mA for an emitter current of 5.60mA . The value of the base current amplification factor (beta) will be :

In a common base mode of a transition , the collector current is 5.488 mA for an emitter currect of 5.60 mA . The value of the base current amplification factor (beta) will be

The current amplification factor alpha of a common base transistor and the current amplification factor beta of a common emitter transistor are not related by

In a n-p-n transistor 10^(10) electrons enter the emitter in 10^(-6)s . 2% of the elecrons are lost in the base. The current transfer ratio and the current amplification factor will be

In a transistor if collector current is 25 mA and base current is 1 mA, then current amplification factor alpha is

In NPN transistor, 10^(10) electrons enters in emitter region in 10^(-6) sc. If 2% electrons are lost in base region then collector current and current amplification factor (beta) respectively are