JEE Mains · Physics · STD 12 - Semicondutor electronics
- JEE Mains 2021
- Hard
- Numerical
A trransistor is connected in common emitter circuit configuration, the collector supply voltage is \(10 \,{V}\) and the voltage drop across a resistor of \(1000\, \Omega\) in the collector circuit is \(0.6\, {V}\). If the current gain factor \((\beta)\) is \(24\) , then the base currect is \(....\,\mu\) \(A.\) (Round off to the Nearest Integer)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(25\)
Step-by-step Solution
Detailed explanation
\(\beta=\frac{ I _{ C }}{ I _{ B }}=24\)
\(R _{ C }=1000\)
\(\Delta V =0.6\)
\(I _{ C }=\frac{0.6}{1000}\)
\(I _{ C }=6 \times 10^{-4}\)
\(I _{ B }=\frac{ I _{ C }}{\beta}=\frac{6 \times 10^{-4}}{24}=25 \mu A\)
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