Question:
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 $10 \mathrm{~K} \Omega$. The common emitter current gain 'lbeta' will be (Round off to the Nearest Integer)
Solution:
$(100)$
$10^{6}=\beta^{2} \times \frac{\mathrm{R}_{0}}{\mathrm{R}_{\mathrm{i}}}$
$10^{6}=\beta^{2} \times \frac{10^{4}}{10^{2}}$
$\beta^{2}=10^{4} \Rightarrow \beta=100$
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