For a Common-Emitter amplifier, the audio signal voltage across the collector resistance of \( 2\,\mathrm{k\Omega} \) is 2 V. Suppose the current amplification factor of the transistor is 100. The base resistance is \( 1\,\mathrm{k\Omega} \). The input signal voltage will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.01 V
In a common-emitter amplifier the collector current is \( \beta \) times the base current, where \( \beta \) is the current amplification factor. The output voltage across \( R_C \) gives the collector current, and the input voltage across \( R_B \) is \( I_B R_B \).
\( V_{o} = 2\,\mathrm{V} \)
\( R_{C} = 2\,\mathrm{k\Omega} = 2000\,\Omega \)
\( R_{B} = 1\,\mathrm{k\Omega} = 1000\,\Omega \)
\( \beta = 100 \)
\( \displaystyle I_{C} = \frac{V_{o}}{R_{C}} = \frac{2}{2000} = 10^{-3}\,\mathrm{A} \)
\( \displaystyle I_{B} = \frac{I_{C}}{\beta} = \frac{10^{-3}}{100} = 10^{-5}\,\mathrm{A} \)
\( \displaystyle V_{i} = I_{B}\,R_{B} = 10^{-5} \times 1000 = 0.01\,\mathrm{V} \)
Shortcut: The voltage gain is \( A_{V} = \beta\,R_{C}/R_{B} = 100 \times 2/1 = 200 \), so \( V_{i} = V_{o}/A_{V} = 2/200 = 0.01\,\mathrm{V} \).
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):
Exercises – …
