I cannot understand the solution to the following problem: enter image description here

The specific maths I can't understand are outlined in the following figure, along with what I thought the answer was. Is \$P=\frac{V^2}{R}\$ as the power dissipated over a resistor correct? So why isn't it the case in the solutions?

If someone could meticulously explain the second bubble, that would be great as well, since I don't get it based off of my understanding. Should it just be \$\frac{(45.3mV)^2}{1}\$?

enter image description here

  • \$\begingroup\$ Yeah, that formula. That was my mistake in the post, but that is what I did. I applied that formula. \$\endgroup\$ – Carl Sep 20 '16 at 22:42

There's a difference between the amplitude of a signal and the RMS wich is used for power calculations. The amplitude is \$\hat{V}\$.

$$P = \frac{V_{eff}^2}{R}$$

$$V_{eff} = \frac{\hat{V}}{\sqrt{2}} \rightarrow V_{eff}^2 = \frac{\hat{V}^2}{2}$$

This is where the 2 is coming from.

  • \$\begingroup\$ What about the second bubble? What about the 2's under both fractions? \$\endgroup\$ – Carl Sep 21 '16 at 1:06
  • \$\begingroup\$ this is where the 2's in both bubbles are coming from. \$V_o\$ is the amplitude of the signal, so you have to divide it by \$\sqrt{2}\$ to get the "DC-equivalent". If you square \$V_o/\sqrt{2}\$ you get the two in the denominator. \$\endgroup\$ – Felix S Sep 21 '16 at 8:28

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