In a transistor, the change produced in collector current is 3 mA, when the base current is changed by 60 μA. This transistor is used as a common emitter amplifier with a load resistance of 5 KΩ. The input resistance of this silicon transistor is 50 Ω. The voltage gain of the amplifier is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 5000
The voltage gain of a common emitter amplifier is the ratio of the change in output voltage to the change in input voltage. The output voltage change is \( \Delta I_C R_L \) and the input voltage change is \( \Delta I_B R_{\mathrm{in}} \).
\( \displaystyle A_v = \frac{\Delta I_C\,R_L}{\Delta I_B\,R_{\mathrm{in}}} = \beta_{\mathrm{ac}}\,\frac{R_L}{R_{\mathrm{in}}} \)
\( \displaystyle \beta_{\mathrm{ac}} = \frac{\Delta I_C}{\Delta I_B} = \frac{3\times10^{-3}\,\mathrm{A}}{60\times10^{-6}\,\mathrm{A}} = 50 \)
\( \displaystyle A_v = 50\times\frac{5000\,\Omega}{50\,\Omega} = 5000 \)
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Transistor as a Device – “… We can express the voltage gain AV in terms of the resistors in the circuit and the current gain of the transistor as follows. … So, the voltage gain of this CE amplifier (Fig. 14.32) is given by
\( \displaystyle A_V = -\beta_{ac}\left(R_C / R_B\right) \qquad (14.14) \)”