JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
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)
- A \(5\)
- B \(10\)
- C \(25\)
- D \(30\)
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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