Q168 / 200 · MHT CET - Medical 2026 (24 April Shift 1) · Physics
For a common emitter amplifier, the audio signal voltage across the collector resistance of $2\ \text{k}\Omega$ is 4 V. If the current amplification factor of the transistor is 100 and the base resistance is $1\ \text{k}\Omega$, then the input voltage is
- A$10\ \text{mV}$
- B$15\ \text{mV}$
- Correct answerC$20\ \text{mV}$
- D$25\ \text{mV}$
Solution
Output voltage $V_o = 4\ \text{V}$
Collector resistance $R_c = 2\ \text{k}\Omega = 2000\ \Omega$
Collector current $I_c = \dfrac{V_o}{R_c} = \dfrac{4}{2000} = 2 \times 10^{-3}\ \text{A}$
Current amplification factor $\beta = 100$
Base current $I_b = \dfrac{I_c}{\beta} = \dfrac{2 \times 10^{-3}}{100} = 2 \times 10^{-5}\ \text{A}$
Base resistance $R_b = 1\ \text{k}\Omega = 1000\ \Omega$
Input voltage $V_i = I_b R_b = (2 \times 10^{-5}) \times 1000 = 2 \times 10^{-2}\ \text{V} = 20\ \text{mV}$
Answer: $20\ \text{mV}$
Collector resistance $R_c = 2\ \text{k}\Omega = 2000\ \Omega$
Collector current $I_c = \dfrac{V_o}{R_c} = \dfrac{4}{2000} = 2 \times 10^{-3}\ \text{A}$
Current amplification factor $\beta = 100$
Base current $I_b = \dfrac{I_c}{\beta} = \dfrac{2 \times 10^{-3}}{100} = 2 \times 10^{-5}\ \text{A}$
Base resistance $R_b = 1\ \text{k}\Omega = 1000\ \Omega$
Input voltage $V_i = I_b R_b = (2 \times 10^{-5}) \times 1000 = 2 \times 10^{-2}\ \text{V} = 20\ \text{mV}$
Answer: $20\ \text{mV}$
Hence, the correct answer is option: 'C'.