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For question a, why is the power absorbed by the DC source : P=Vo*Io

However, for the same exact question, the power absorbed in the resistor is P = (Irms)^2 * R

Hence when do we use RMS current/voltage in calculations and when do we use the average current/voltage in calculations?

As the RMS current includes the dc current and harmonics, would this not be used in calculating for the power absorbed by the dc source as that current is going through it (not just the dc value (i.e. average value)?

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  • \$\begingroup\$ @Mattman944 I don't think RMS always works and average sometimes. They both work only when appropriate. E.g. you have a current flowing through a resistor and a constant voltage drop (say an idealized diode drop). Then you need RMS current to work out resistor power, but you have to use average one for the constant voltage drop. You can't swap or use RMS in both cases \$\endgroup\$ – carloc May 6 at 9:30
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why is the power absorbed by the DC source : P=Vo*Io

This is making an approximation that the current is approximately constant (or we're defining \$I_0\$ to be the time average of the current). The voltage is known to be constant because it's the voltage of an ideal DC source.

power absorbed in the resistor is P = (Irms)^2 * R

The rms current is the square root of the mean of the squared current.

Therefore \$I_{rms}^2\$ is the mean of the squared current. Since the instantaneous power consumed by the resistor is \$ i^2(t)R\$, the mean of the squared current is exactly what we need to get the mean power consumption.

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Calculus has all answers to all questions:

$$\text{d}E_t=V_t\cdot I_t\:\:\text{d}t$$

This simply says that the infinitesimal change in energy is equal to the voltage at a particular time, times the current at a particular time, times the infinitesimal instant of time. \$V_t\$ and \$I_t\$ are finite values (measurable), but the instant of time is small, beyond your imagination. And therefore, two finite values times a very tiny instant of time must result in an infinitesimal of energy that is also small, beyond your imagination. That makes sense.

To find a finite value of energy, we have to integrate both sides over some specific period of time, so:

$$E_{t_0\rightarrow t_1}=\int_{t_0}^{t_1} \text{d}E_t=\int_{t_0}^{t_1} V_t\cdot I_t\:\:\text{d}t$$

But to get power, we need to divide this by the time period in question. So:

$$P_{t_0\rightarrow t_1}=\frac{1}{t_1-t_0}\int_{t_0}^{t_1} V_t\cdot I_t\:\:\text{d}t$$

Since one cycle is the same as any other cycle, it would be very convenient to set the start time as \$t_0=0\:\text{s}\$ and ending time as \$t_1=\left(\frac{1}{f}\right)\:\text{s}\$ and somehow compute the average over that cycle. (Each being the same over a cycle, this means the average over one cycle will be the same over each successive cycle and is therefore a very convenient way to look at the problem.) So, knowing that \$V_t=V_0\,\operatorname{sin}\left(2\pi\,f\: t\right)\$:

$$\begin{align*} \overline{P}&=f\int_{0}^{\frac{1}{f}} V_t\cdot I_t\:\:\text{d}t\\\\&=f\int_{0}^{\frac{1}{f}} \frac{V_0^2\,\operatorname{sin}^2\left(2\pi\,f\: t\right)}{R}\: \text{d}t\\\\&=\frac{f\,V_0^2}{R}\int_{0}^{\frac{1}{f}} \operatorname{sin}^2\left(2\pi\,f\: t\right)\: \text{d}t\\\\&=\frac{f\,V_0^2}{R}\left[\frac{t}{2}-\frac{\operatorname{sin}\left(4\pi\,f\: t\right)}{8\pi\,f}\right]_0^\frac{1}{f}\\\\ &=\frac{f\,V_0^2}{R}\left[\frac{1}{2\,f}\right] = \frac{V_0^2}{2\,R} = \frac{V_0^2}{2}\frac{1}{R} = \left[\frac{V_0}{\sqrt{2}}\right]^2\frac{1}{R} \end{align*}$$

But what we want is some voltage that we can use for the typical Ohm's Law version of \$\overline{P}=\frac{V_?^2}{R}\$. What could \$V_?\$ be?? Well, clearly it must be \$V_?=\frac{V_0}{\sqrt{2}}\$. And we call this \$V_?=V_\text{RMS}\$. So, \$V_\text{RMS}=\frac{V_0}{\sqrt{2}}\$. Then you can compute \$\overline{P}=\frac{V_\text{RMS}^{2}}{R}\$. And that's pretty useful.

The upshot of all the above is that the fact that power is proportional to voltage-squared (or current-squared) means that there is a \$\sqrt{2}\$ factor involved. At least for sine waves. (You can perform all of the above for any definition for \$V_t\$ or \$I_t\$ and get various relationships -- but for sine waves, the above holds.)

Note, though. Voltage by itself isn't power. Current by itself isn't power. You have to multiply them together to get something related to power. And while power is inherently an average (as you can see in the above equation development), the average of voltage (or the average of current) isn't in the equations and isn't usable for computing power. Voltages and currents are instantaneous values and their average doesn't help you in working out the power equations.

Also, energy storage devices, such as inductors and capacitors, can temporarily store energy. So the formulas become more complex when you add in these devices. But for a simple resistor load, the relationships are simpler and the above works fine.

Bottom line? For sine wave voltage supplies and where average power is sought, voltages and currents must be handled as root-mean-square values to keep the formulas simple and in the usual Ohm's Law form.

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