A Colpitts oscillator is designed as a radio frequency oscillator. Which of the following statements is INCORRECT?

This question was previously asked in
RRB JE ECE 22 Apr 2025 Shift 2 CBT 2 Official Paper
View all RRB JE Papers >
  1. It operates on the principle of parallel resonance.
  2. In a Colpitts oscillator, two capacitors and an inductor form the feedback network.
  3. The frequency of oscillation is \(\omega = \frac{1}{\sqrt{L\left(\frac{C_1 + C_2}{C_1 C_2}\right)}}\)
  4. An LC network is used in the design of Colpitts oscillators.

Answer (Detailed Solution Below)

Option 3 : The frequency of oscillation is \(\omega = \frac{1}{\sqrt{L\left(\frac{C_1 + C_2}{C_1 C_2}\right)}}\)
Free
General Science for All Railway Exams Mock Test
2.1 Lakh Users
20 Questions 20 Marks 15 Mins

Detailed Solution

Download Solution PDF

Concept:

The Colpitts oscillator is a type of LC oscillator used to generate high-frequency sinusoidal oscillations, especially in RF applications.

It works based on the principle of LC parallel resonance and uses a combination of capacitors and inductors to determine its frequency of oscillation.

Explanation:

The formula provided in statement 3 is incorrect in the way it's expressed.

The correct frequency of oscillation is given by:

\( \omega = \frac{1}{\sqrt{L \cdot C_{eq}}} \), where \( C_{eq} = \frac{C_1 C_2}{C_1 + C_2} \)

This is the equivalent capacitance of two capacitors in series.

Conclusion:

Statement 3 is incorrect due to the incorrect formula for the oscillation frequency in a Colpitts oscillator.

Latest RRB JE Updates

Last updated on Jun 7, 2025

-> RRB JE CBT 2 answer key 2025 for June 4 exam has been released at the official website.

-> Check Your Marks via RRB JE CBT 2 Rank Calculator 2025

-> RRB JE CBT 2 admit card 2025 has been released. 

-> RRB JE CBT 2 city intimation slip 2025 for June 4 exam has been released at the official website.

-> RRB JE CBT 2 Cancelled Shift Exam 2025 will be conducted on June 4, 2025 in offline mode. 

-> RRB JE CBT 2 Exam Analysis 2025 is Out, Candidates analysis their exam according to Shift 1 and 2 Questions and Answers.

-> The RRB JE Notification 2024 was released for 7951 vacancies for various posts of Junior Engineer, Depot Material Superintendent, Chemical & Metallurgical Assistant, Chemical Supervisor (Research) and Metallurgical Supervisor (Research). 

-> The selection process includes CBT 1, CBT 2, and Document Verification & Medical Test.

-> The candidates who will be selected will get an approximate salary range between Rs. 13,500 to Rs. 38,425.

-> Attempt RRB JE Free Current Affairs Mock Test here

-> Enhance your preparation with the RRB JE Previous Year Papers

Get Free Access Now
Hot Links: teen patti king lotus teen patti teen patti casino download