Question

In a common-emitter amplifier, using output resistance of $5000\text{ohm}$ and input resistance of $2000\text{ohm}$, if the peak value of input signal voltage is $10\text{mV}$ and $\beta =50$, then the power gain is

• A. $625\text{W}$
• B. $25\text{W}$
• C. $6250\text{W}$
• D. $125\text{W}$

Answer 1

The correct option is C $6250\text{W}$

Power gain $\left(P_g\right)=\beta \times V_A$
where, $\beta =$ current gain
$V_A=$ Voltage gain
here, $V_A=\beta \times \frac{R_0}{R_I}$
so, $\left(P_g\right)={\beta }^2\times \frac{R_0}{R_I}$

from the given data,
$P_g={50}^2\times \frac{5000}{2000}=6250W$