Question:
An NPN transistor is used in common emitter configuration as an amplifier with \(1\ \mathrm{k}\Omega\) load resistance. Signal voltage of 10 mV is applied across the base-emitter. This produces a 3 mA change in the collector current and \(15\mu\mathrm{A}\) change in the base current of the amplifier. The input resistance and voltage gain are :
An NPN transistor is used in common emitter configuration as an amplifier with \(1\ \mathrm{k}\Omega\) load resistance. Signal voltage of 10 mV is applied across the base-emitter. This produces a 3 mA change in the collector current and \(15\mu\mathrm{A}\) change in the base current of the amplifier. The input resistance and voltage gain are :
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Input resistance \(=\Delta V_{BE}/\Delta I_B\) and voltage gain \(=\Delta I_C R_L/\Delta V_{BE}\).
Updated On: Nov 5, 2026
- 0.67 k\(\Omega\), 200
- 0.33 k\(\Omega\), 1.5
- 0.67 k\(\Omega\), 300
- 0.33 k\(\Omega\), 300
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The Correct Option is C
Solution and Explanation
Input resistance: \(R_{in}=\frac{\Delta V_{BE}}{\Delta I_B}=\frac{10\times10^{-3}}{15\times10^{-6}}=666.7\ \Omega\approx0.67\ \mathrm{k}\Omega\).
Output voltage change: \(\Delta V_{out}=\Delta I_C R_L=3\times10^{-3}\times10^3=3\) V.
Voltage gain \(=\frac{3}{10\times10^{-3}}=300\).
Answer: (3) 0.67 k\(\Omega\), 300
Output voltage change: \(\Delta V_{out}=\Delta I_C R_L=3\times10^{-3}\times10^3=3\) V.
Voltage gain \(=\frac{3}{10\times10^{-3}}=300\).
Answer: (3) 0.67 k\(\Omega\), 300
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