 Solution

ID: 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, rms}^{2} / R_{i}

, while the load receives

P_{L} = V_{o, rms}^{2} / R_{L}

. Hence

\frac{P _{L}}{P _{i}} = (\frac{R _{f}}{R _{i}})^{2} \frac{R _{i}}{R _{L}} = \frac{1000}{100} = 10.

This 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.

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