In a common emitter transistor circuit the base current is \[40\mu {\text{A}}\], then \[{V_{BE}}\] is:

A) $2V$
B) $0.2V$
C) $0.8V$
D) $Zero$
Answer
298.5k+ views
Hint: By using current formula $V = IR$, the current flowing through a voltage V and resistance R. show the figure in common emitter transistor circuit, and given values are base current is \[40\mu {\text{A}}\], voltage current is \[{V_{CC}} = 10V\]and the base resistance is $245K\Omega $. Find the base emitter voltage\[{V_{BE}}\].
Complete step by step answer:
The given values are,
Base current is\[40\mu {\text{A}}\]
Voltage current\[{I_V} = {V_{CC}} = 10V\]
Base resistance ${R_B} = 245K\Omega $
We know,
Input voltage ${I_V} = {V_{BE}} + ({I_B}) \times ({R_B})$
Where,
${I_B}$ Is the base current, and
${R_B}$ Is the resistance offered in the base region.
${V_{BE}} = {I_V} - {I_B} \times {R_B}$
\[{I_V} = {V_{CC}} = 10V\] And
${I_B} = 40\mu A$
$ \Rightarrow 40 \times {10^{ - 6}}A$
Also,${R_B} = 245K\Omega $
$ \Rightarrow 245 \times {10^3}\Omega $ [From the figure]
Putting their respective values, we get:
${V_{BE}} = 10 - (40 \times {10^{ - 6}}A)(245 \times {10^3}\Omega )$
On simplification the bracket term and we get
$ \Rightarrow (10 - 9.8)V$
On subtracting we get,
$ \Rightarrow 0.2V$
Therefore, option (B) is the correct answer.
Additional information:
Common emitter transistor:
As such the common emitter configuration may be a good all round circuit to be used in many applications. It's also worth noting at this stage that the common emitter transistor amplifier inverts the signal at the input. Therefore if a waveform that's rising enters the input of the common emitter amplifier, it'll cause the output voltage to fall.
Note: Advantages of Common Emitter amplifier:
1. A common emitter amplifier is inverting and has low I/p impedance
2. High output impedance
3. High voltage gain.
4. High current gain.
Disadvantages of Common Emitter amplifier:
1. It’s a high output resistance.
2. It responds poorly to high frequencies.
3. Its high thermal instabilities.
4. Its voltage gain is extremely unstable
Complete step by step answer:
The given values are,
Base current is\[40\mu {\text{A}}\]
Voltage current\[{I_V} = {V_{CC}} = 10V\]
Base resistance ${R_B} = 245K\Omega $
We know,
Input voltage ${I_V} = {V_{BE}} + ({I_B}) \times ({R_B})$
Where,
${I_B}$ Is the base current, and
${R_B}$ Is the resistance offered in the base region.
${V_{BE}} = {I_V} - {I_B} \times {R_B}$
\[{I_V} = {V_{CC}} = 10V\] And
${I_B} = 40\mu A$
$ \Rightarrow 40 \times {10^{ - 6}}A$
Also,${R_B} = 245K\Omega $
$ \Rightarrow 245 \times {10^3}\Omega $ [From the figure]
Putting their respective values, we get:
${V_{BE}} = 10 - (40 \times {10^{ - 6}}A)(245 \times {10^3}\Omega )$
On simplification the bracket term and we get
$ \Rightarrow (10 - 9.8)V$
On subtracting we get,
$ \Rightarrow 0.2V$
Therefore, option (B) is the correct answer.
Additional information:
Common emitter transistor:
As such the common emitter configuration may be a good all round circuit to be used in many applications. It's also worth noting at this stage that the common emitter transistor amplifier inverts the signal at the input. Therefore if a waveform that's rising enters the input of the common emitter amplifier, it'll cause the output voltage to fall.
Note: Advantages of Common Emitter amplifier:
1. A common emitter amplifier is inverting and has low I/p impedance
2. High output impedance
3. High voltage gain.
4. High current gain.
Disadvantages of Common Emitter amplifier:
1. It’s a high output resistance.
2. It responds poorly to high frequencies.
3. Its high thermal instabilities.
4. Its voltage gain is extremely unstable
Recently Updated Pages
If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

A point charge is placed at the corner of a cube The class 12 physics JEE_Main

What changes occur if the monochromatic light used class 12 physics JEE_Main

The unit of specific conductance is A Ohm B Ohmmetre class 12 physics JEE_Main

Which lens is used in magnifying glass A Concave lens class 12 physics JEE_Main

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

What Are Current and Potential Difference in Electricity?

Understanding Uniform Acceleration in Physics

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

