Past exam of the physics 2 paper of the University of Oxford / 2013 / 8 / iii / Solution

With $R_f=R_i$, the output and input voltage magnitudes are equal. The signal power supplied at the input is $P_i=V_{i,\mathrm{rms}}^2/R_i$, while the load receives $P_L=V_{o,\mathrm{rms}}^2/R_L$. HenceThis is $10$ dB of power gain into the specified load. Output energy comes from the amplifier's supply rails; unity voltage-gain magnitude does not imply unity power gain. Dissipation in the feedback resistor is additional to the specified load power.