Question:

In a common emitter configuration, a transistor has $\beta = 50$ and input resistance $1\, k\Omega$. If the peak value of a.c. input is $0.01\, V$ then the peak value of collector current is

Updated On: Apr 27, 2024
  • $0.01\, \mu A$
  • $0.25\, \mu A$
  • $100\, \mu A$
  • $500\, \mu A$
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The Correct Option is D

Solution and Explanation

Given that the current gain of the transistor is $\beta=50$
Input resistance $=1 k \Omega$, input voltage $=0.01\, V$
Hence, $50=\frac{\Delta i_{C}}{\Delta i_{B}} $
$\Rightarrow \Delta i_{C}=50 \Delta i_{B}$
Also the change in base current is
$\Delta i_{B}=\frac{\text { input voltage }}{\text { input resistance }}=\frac{0.01}{1 \times 10^{3}}=10^{-5}\, A$
so, $\Delta i_{C}=50 \times 10^{-5}=500 \,\mu A$
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