Question

What is the power gain in a CE amplifier, where input resistance is $3{\text{ k}}\Omega $ and load resistance is $24\,{\text{k}}\Omega $ given $\beta = 6$?

Answer 1

Hint: In the given question, we are given the value of input resistance and output resistance. We have to find out the value of power gain. We are also provided with current gain here. We have to formula the general formula where we have to make use of current gain and hence find the answer.

Complete step by step answer:
Power gain of an A.C. circuit can be generalized as a formula.
${\text{Power Gain}} = \dfrac{{{\text{Output power}}}}{{{\text{Input power}}}}$
Let the input current in the CE amplifier be ${I_i}$ and the output current in the CE amplifier be ${I_o}$. Given input resistance ${R_i}$ of the CE amplifier is $3{\text{ k}}\Omega $ and the output resistance ${R_o}$ of the CE amplifier is $24\,{\text{k}}\Omega $.We know,
$P = {I^2}R$
where,$P$ corresponds to power, $R$ corresponds to the resistance and $I$ as the current.

Now we can write,
${\text{Power Gain}} = \dfrac{{I_o^2{R_o}}}{{I_i^2{R_i}}}$
Again, current gain $\beta = \dfrac{{{I_o}}}{{{I_i}}} = 6$
Thus we get,
${\text{Power Gain}} = {\left( {\dfrac{{{I_o}}}{{{I_i}}}} \right)^2} \times \dfrac{{{R_o}}}{{{R_i}}}$
Substituting the values of current gain, load resistance (output resistance) and input resistance we get,
${\text{Power Gain}} = {\left( 6 \right)^2} \times \dfrac{{24000}}{{3000}} \\
\therefore {\text{Power Gain}}= 288$

Hence, we find the power gain of the CE amplifier as $288$.

Note: It must be noted that the power gain of a circuit is a dimensionless quantity a it the ratio of two similar dimensional quantities. Current gain$\left( \beta \right)$ is also defined as the ratio of collector current of the circuit to the base current of the circuit. There is another current gain $\left( \alpha \right)$ which is the ratio of collector current of a circuit to the emitter current of the circuit.