MHT CET · Physics · Semiconductors
The current amplification factor of a transistor is 50. The input resistance when used in common emitter mode is \(1 \mathrm{k} \Omega\). The peak value for an a.c. input voltage of 0.01 V peak is
- A \(100 \mu \mathrm{~A}\)
- B 0.01 mA
- C 0.25 mA
- D \(500 \mu \mathrm{~A}\)
Answer & Solution
Correct Answer
(D) \(500 \mu \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Given: } \beta=50, R_i=1 \mathrm{k} \Omega=10^3 \Omega, \\ & V_i=0.01 \mathrm{~V} \\ & \beta=\frac{I_C}{I_B}=50 \Rightarrow I_C=50 \times I_B \\ & V_i=I_B \times R_i \\ \therefore \quad & I_B=\frac{V_i}{R_i}=\frac{0.01}{10^3}=10^{-5} \\ \therefore \quad & I_C=50 \times 10^{-5}=500 \times 10^{-6} \mathrm{~A}=500 \mu \mathrm{~A}\end{aligned}\)
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