Correct Answer - Option 3 : 20
Concept:
Power gain is the ratio of the output power (P0) to the input power (Pin), i.e.
\({A_p} = \frac{{{P_0}}}{{{P_{in}}}}\)
In terms of dB, the power gain is expressed as:
\(P(dB) = 10\log_{10}{\frac{P_{out}}{P_{in}}}\)
Property: logx x = 1
Calculation:
With AP = 100, we can write:
\(P(dB) = 10\log_{10}{A_P}=10\log_{10}{100}\)
\(P(dB) =10\log_{10}{10^2}\)
\(P(dB) =2\times10\log_{10}{10}\)
Since log1010 = 1, the power gain in dB will be:
\(P(dB) =20 ~dB\)
Important Note:
It is the voltage gain or the current gain that is calculated as:
\(A_V=20log(\frac{V_2}{V_1})\)
\(A_I=20log(\frac{I_2}{I_1})\)