Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
In the following common emitter circuit, if β=100VCE=7 VRC=2 kΩ and VBE is negligible, then IB is

PhysicsSemiconductorsNEET
Options:
  • A 0.01mA
  • B 0.04mA
  • C 0.02mA
  • D 0.03mA
Solution:
1031 Upvotes Verified Answer
The correct answer is: 0.04mA
V=VCE+ICRC

15=7+IC×2×103      Ic=mA

β=ICIB        IB=4100=0.04 mA

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.