Question

TIn a CE amplifier β = 50, \[{R_L} = 4k\Omega .\],\[{R_1} = 500\Omega .\]Power gain of the amplifier is.
A. \[2 \times {10^4}\]
B. \[2 \times {10^2}\]
C. \[2 \times {10^3}\]
D. \[2 \times {10^2}\]

Answer 1

Hint: To calculate the required Power gain we have to find voltage gain first then after multiplying them with \[\beta \] . In order to do so we will use formulae:-
\[{\text{Power gain}} = \beta \times {\text{Voltage gain}}\]
Since,
\[{\text{Voltage gain = }}\beta \left( {\dfrac{{{R_{output}}}}{{{R_{input}}}}} \right)\], where, \[\beta {\text{ is voltage amplification factor}}\]
 \[{R_{output}}\]and \[{R_{input}}\] are the output( load)
input and output resistances respectively.
 Therefore, \[{\text{Power gain}} = {\beta ^2}\left( {\dfrac{{{R_{output}}}}{{{R_{input}}}}} \right)\]

Complete step-by-step answer:
To find the required Power gain.
The formula used: \[{\text{Power gain}} = {\beta ^2}\left( {\dfrac{{{R_{output}}}}{{{R_{input}}}}} \right)\] …………….(i)
Given:-
\[
  \beta = 60 \\
  {R_{output}} = {R_L} = 4k\Omega \\
   \Rightarrow {\text{ }}{R_{output}} = {R_L} = 4 \times {10^3}\Omega \\
  {\text{and }}{R_{input}} = {R_1} = 500\Omega \\
\]
Substituting the given values of \[\beta ,{\text{ }}{R_{output}}{\text{ and }}{R_{input}}{\text{ in eqn(i)}}\]we get
\[
  {\text{Power gain = (50}}{{\text{)}}^2} \times \left( {\dfrac{{4 \times {{10}^3}}}{{500}}} \right) \\
   \Rightarrow {\text{Power gain = 2}} \times {10^4}{\text{ }} \\
\]
Hence Option (A) is the correct answer.

Note: To solve such kinds of questions one should have to revise the short formulae in this topic with the basic algorithm behind deducing the expressions involving terms like Power gain, voltage gain, amplification factors input, and the load or output resistance. Always start with what is given in questions and what we will have to find. We should not confuse with Power gain in amplifiers with the electrical power here. Power gain is a unitless quantity while electrical power has unit watt.