# Calculating the output power from an Amplifier

This is not based on any particular hardware. I am trying to understand the theory behind it by doing tutorial questions from my uni, I have completed the question but don't understand the approach taken by the worked solutions.

simulate this circuit – Schematic created using CircuitLab

Above schematic is my representation of the circuit, bascially $i_1(t) = 0.1\cos(2\pi f)$, f= 10 Mhz and the Amp has a dB gain of 25 (50Ω matched) and the load is 50Ω.

we are asked to calculate the output power, so i simply worked out the RMS current 0.071 and times it by R1 (50Ω) to work out the input power (0.25 W) and using the gain (G=Pout/Pinput) $P_{out} = P_{input} \times G$ (316.2 ratio form) making Pout = 79.7 W. However in the answer sheet they do the same steps except for $P_{input}$ they use $P_{input} = (I_{rms} \times R_1)/4$.

I don't know where this four has come from and why $P_{input}$ needs to be divided by four???

Nb: $P_{out}$= power out; $P_{input}$ = power in

• Maximum power transfer theorem?? – nidhin Jun 18 '14 at 5:44

The current $i_1$ is split equally across $R_1$ and $R_2$. Therefore current to the input of amplifier (flowing through $R_2$) is $i_1/2$. Therefore power transferred from source to amplifier input is $(i_1/2)^2\times R_2 = i_1^2R_2/4 = i_1^2R_1/4$.
Max power transfer theorem also says that when the loads are matched, the power transferred is $I_{rms}^2R_L/4$. Here source is $i_1$ (with $R_1$ in parallel) and load is amplifier (input). Hence the answer.