JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
An \(npn\) transistor operates as a common emitter amplifier with a power gain of \(10^{6} .\) The input circuit resistance is \(100\, \Omega\) and the output load resistance is \(10\, K \Omega\). The common emitter current gain ' \(\beta\) ' will be............. (Round off to the Nearest Integer)
- A \(400\)
- B \(100\)
- C \(121\)
- D \(169\)
Answer & Solution
Correct Answer
(B) \(100\)
Step-by-step Solution
Detailed explanation
\(10^{6}=\beta^{2} \times \frac{ R _{0}}{ R _{ i }}\) \(10^{6}=\beta^{2} \times \frac{10^{4}}{10^{2}}\) \(\beta^{2}=10^{4} \Rightarrow \beta=100\)
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