Question:
In a common emitter configuration with suitable bias, it is given that \(R_L\) is the load resistance and \(R_{BE}\) is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by :
\(\beta\) is current gain, \(I_B\), \(I_C\) and \(I_E\) are respectively base, collector and emitter currents.
In a common emitter configuration with suitable bias, it is given that \(R_L\) is the load resistance and \(R_{BE}\) is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by :
\(\beta\) is current gain, \(I_B\), \(I_C\) and \(I_E\) are respectively base, collector and emitter currents.
\(\beta\) is current gain, \(I_B\), \(I_C\) and \(I_E\) are respectively base, collector and emitter currents.
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Voltage gain = current gain times the resistance ratio; power gain is the product of voltage gain and current gain.
Updated On: Nov 11, 2026
- \(\beta\frac{R_L}{R_{BE}},\ \frac{\Delta I_C}{\Delta I_B},\ \beta^2\frac{R_L}{R_{BE}}\)
- \(\beta\frac{R_L}{R_{BE}},\ \frac{\Delta I_E}{\Delta I_B},\ \beta^2\frac{R_L}{R_{BE}}\)
- \(\beta^2\frac{R_L}{R_{BE}},\ \frac{\Delta I_C}{\Delta I_E},\ \beta^2\frac{R_L}{R_{BE}}\)
- \(\beta^2\frac{R_L}{R_{BE}},\ \frac{\Delta I_C}{\Delta I_B},\ \beta\frac{R_L}{R_{BE}}\)
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The Correct Option is A
Solution and Explanation
In the common emitter circuit the input current is \(I_B\) and the output current is \(I_C\).
Current gain: \(\beta=\frac{\Delta I_C}{\Delta I_B}\).
Voltage gain: \(A_V=\frac{\Delta V_{out}}{\Delta V_{in}}=\frac{\Delta I_C R_L}{\Delta I_B R_{BE}}=\beta\frac{R_L}{R_{BE}}\).
Power gain: \(A_P=A_V\times\beta=\beta^2\frac{R_L}{R_{BE}}\).
This matches option (1). In options (2) and (3) the current gain is wrongly written with \(\Delta I_E\), and in (4) the voltage and power gains are interchanged.
Answer: (1) \(\beta\frac{R_L}{R_{BE}},\ \frac{\Delta I_C}{\Delta I_B},\ \beta^2\frac{R_L}{R_{BE}}\)
Current gain: \(\beta=\frac{\Delta I_C}{\Delta I_B}\).
Voltage gain: \(A_V=\frac{\Delta V_{out}}{\Delta V_{in}}=\frac{\Delta I_C R_L}{\Delta I_B R_{BE}}=\beta\frac{R_L}{R_{BE}}\).
Power gain: \(A_P=A_V\times\beta=\beta^2\frac{R_L}{R_{BE}}\).
This matches option (1). In options (2) and (3) the current gain is wrongly written with \(\Delta I_E\), and in (4) the voltage and power gains are interchanged.
Answer: (1) \(\beta\frac{R_L}{R_{BE}},\ \frac{\Delta I_C}{\Delta I_B},\ \beta^2\frac{R_L}{R_{BE}}\)
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