A transistor connected in a common base configuration has the following readings I= 2 mA and IB = 20 μA. Find the current gain α.

This question was previously asked in
SSC JE EE Previous Paper 12 (Held on: 24 March 2021 Evening)
View all SSC JE EE Papers >
  1. 0.95
  2. 1.98
  3. 0.99
  4. 0.98

Answer (Detailed Solution Below)

Option 3 : 0.99
Free
Electrical Machine for All AE/JE EE Exams Mock Test
7.4 K Users
20 Questions 20 Marks 20 Mins

Detailed Solution

Download Solution PDF

Current amplification factor: It is defined as the ratio of the output current to the input current. In the common-base configuration, the output current is emitter current IC, whereas the input current is base current IE.

Thus, the ratio of change in collector current to the change in the emitter current is known as the current amplification factor. It is expressed by the α.

\(\alpha = \frac{{{\rm{\Delta }}{I_C}}}{{{\rm{\Delta }}{I_E}}}\)

Where, IE = IC + IB

Calculation:

Given,

IE = 2 mA

IB = 20 μA = 0.02 mA

From above concept,

IC = 2 mA - 0.02 mA = 1.98 mA

Current amplification factor is given as,

\(\alpha=\frac{I_C}{I_E}=\frac{1.98}{2}=0.99\)

Latest SSC JE EE Updates

Last updated on May 29, 2025

-> SSC JE Electrical 2025 Notification will be released on June 30 for the post of Junior Engineer Electrical/ Electrical & Mechanical.

-> Applicants can fill out the SSC JE application form 2025 for Electrical Engineering from June 30 to July 21.

-> SSC JE EE 2025 paper 1 exam will be conducted from October 27 to 31. 

-> Candidates with a degree/diploma in engineering are eligible for this post.

-> The selection process includes Paper I and Paper II online exams, followed by document verification.

-> Prepare for the exam using SSC JE EE Previous Year Papers.

Get Free Access Now
Hot Links: all teen patti master teen patti royal - 3 patti teen patti real cash withdrawal