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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)

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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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