Test Index

CBSE Class 12 Physics 2017 Delhi Set 1 Paper

© examsnet.com
Question : 21 of 26
Marks: +1, -0
For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 k Ω2\ \text{k}\ \Omega is 2V. Given the current amplification factor of the transistor is 100 , find the input signal voltage and base current, if the base resistance is 1 k Ω1\ \text{k}\ \Omega.
Solution:  
Calculation of collector current IeI_e base current IBI_B and input signal voltage ViV_i
Given Re  =2 k ΩR_e\;=2\ \text{k}\ \Omega
  =2×103 Ω\;=2 \times 10^3\ \Omega
VCE  =ICRCV_{C E}\;=I_C R_C
IC  =  VCERC  =  22×103 AI_C\;=\;\frac{V_{C E}}{R_C}\;=\;\frac{2}{2 \times 10^3}\ \text{A}
  =10−3 A\;=10^{-3}\ \text{A}
  =1 mA\;=1\ \text{mA}
    Current gain    =  ICIB\;\;\text{Current gain}\;\;=\;\frac{I_C}{I_B}
  ∴  100  =  10−3IB\;\therefore\;100\;=\;\frac{10^{-3}}{I_B}
  ∴  IB  =10−5 A\;\therefore\;I_B\;=10^{-5}\ \text{A}
Input signal voltage
Vi=IBRBV_i=I_B R_B
=1×10−5×103 Ω  =1 \times 10^{-5} \times 10^3\ \Omega\;
=10−2 V=10^{-2}\ \text{V}
[Note : Give full credit if student calculates the required quantities by any other alternative method]
© examsnet.com
Go to Question:
Prev Question
Next Question