एक स्थिर अभिनति परिपथ में सिलिकॉन NPN ट्रांजिस्टर, उभयनिष्ट उत्सर्जक विन्यास जिसमें β = 50 का उपयोग किया जाता है। जब RB = 106 Ω, RC = 5 kΩ और VCC = 10V हो, तो स्थिर बिंदु पर VCE की गणना कीजिए।

This question was previously asked in
SSC JE Electrical 06 Jun 2024 Shift 2 Official Paper - 1
View all SSC JE EE Papers >
  1. 7.67 V
  2. 8.50 V
  3. 7.50 V
  4. 6.67 V

Answer (Detailed Solution Below)

Option 1 : 7.67 V
Free
RRB JE CBT I Full Test - 23
100 Qs. 100 Marks 90 Mins

Detailed Solution

Download Solution PDF

सिद्धांत

स्थिर अभिनति परिपथ का परिपथ नीचे दिखाया गया है:

संग्राहक से उत्सर्जक वोल्टेज का मान निम्न द्वारा दिया गया है:

जहां, IC का मान निम्न द्वारा दिया गया है:

गणना

दिया गया है, β = 50

RB = 106 Ω, RC = 5 kΩ और VCC = 10V

mA

VCE = 7.67 V

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.

More Bipolar Junction Transistors Questions

Hot Links: teen patti master gold teen patti 50 bonus teen patti joy official teen patti master app teen patti real cash game