In a common emitter amplifier the audio signal voltage across the collector resistance of 2 kΩ is 2 V. The current amplification factor is 50. The value of base current is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 20 μA
The collector current follows from Ohm’s law across the collector resistance, and the current amplification factor of a common emitter amplifier is \( \beta = I_C/I_B \).
\( \displaystyle I_C = \frac{V_C}{R_C} = \frac{2\,\mathrm{V}}{2000\,\Omega} = 1\times10^{-3}\,\mathrm{A} = 1\,\mathrm{mA} \)
\( \displaystyle I_B = \frac{I_C}{\beta} = \frac{1\,\mathrm{mA}}{50} = 0.02\,\mathrm{mA} = 20\,\mu\mathrm{A} \)
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 an Amplifier (CE-Configuration) – “… Example 14.10 For a CE transistor amplifier, the audio signal voltage across the collector resistance of 2.0 kΩ is 2.0 V. Suppose the current amplification factor of the transistor is 100, What should be the value of RB in series with VBB supply of 2.0 V if the dc base current has to be 10 times the signal current. … Solution The output ac voltage is 2.0 V. So, the ac collector current iC = 2.0/2000 = 1.0 mA. The signal current through the base is, therefore given by iB = iC/β = 1.0 mA/100 = 0.010 mA.”