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So I've been solving some basic instantaneous power/ average power problems and I've been asked to "draw" $$p(t)$$ which is the instantaneous power function of the circuit, I guess there are two ways to accomplish this task:

first, you basically sketch the voltage waveform and the current wave form and keep picking points on both traces and multiplying them (I don't think i'm going to do this)

second, and this is what I actually did, found $$p(t) = ((VmIm)/2)(cos(2wt+ϴv+ϴi) + avgpower$$

and sketched the graph manually.

ok my question is if it's possible to check my answer using any simulation program -multisim maybe- by constructing the circuit and plotting the power waveform just like the oscilloscope plots voltage waveform?

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  • \$\begingroup\$ Sorry didn't get what you mean by the voltage sources allowing angular offset what does this has to do with the power graph? \$\endgroup\$ – Seraj Nov 12 '17 at 23:04
  • \$\begingroup\$ I did a little searching on Google, and it does appear that you can plot something on Multisim. However, you will have to use custom formulas. Check this out.... by the way, I deleted my last comment :P \$\endgroup\$ – KingDuken Nov 12 '17 at 23:12
  • \$\begingroup\$ @KingDuken seems legit, still going to give scilab a try because I'm a "cheap sucker" :D \$\endgroup\$ – Seraj Nov 12 '17 at 23:14
  • \$\begingroup\$ LOL okay man, good luck :) \$\endgroup\$ – KingDuken Nov 12 '17 at 23:15
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Well, for the average power in a series RLC-circuit we can write:

$$\overline{\text{P}}_{\space\text{in}}:=\lim_{\text{T}\to\infty}\frac{1}{\text{T}}\int_0^\text{T}\text{P}_{\space\text{in}}\left(t\right)\space\text{d}t=\lim_{\text{T}\to\infty}\frac{1}{\text{T}}\int_0^\text{T}\text{U}_{\space\text{in}}\left(t\right)\cdot\text{I}_{\space\text{in}}\left(t\right)\space\text{d}t\tag1$$

Let us assume a few things to make life easier:

  1. $$\text{R}=\text{C}=\text{L}=1\tag2$$
  2. The input voltage looks like: $$\text{V}_{\space\text{in}}\left(t\right)=\text{A}\cdot\sin\left(2\pi\cdot\text{f}\cdot t+\varphi\right)\tag3$$ Where \$\text{A}=\text{f}=1\$ and \$\varphi=0\$.
  3. For the initial conditons we have: \$\text{I}_{\space\text{in}}\left(0\right)=0\$ and \$\text{I}_{\space\text{in}}'\left(0\right)=0\$.

So, we need to solve:

$$ \begin{cases} \frac{\partial}{\partial t}\left(\text{A}\cdot\sin\left(2\pi\cdot\text{f}\cdot t+\varphi\right)\right)=\text{I}_{\space\text{in}}'\left(t\right)\cdot\text{R}+\text{I}_{\space\text{in}}''\left(t\right)\cdot\text{L}+\text{I}_{\space\text{in}}\left(t\right)\cdot\frac{1}{\text{C}}\\ \\ \text{A}=\text{f}=\text{R}=\text{C}=\text{L}=1\\ \\ \varphi=0\\ \\ \text{I}_{\space\text{in}}\left(0\right)=\text{I}_{\space\text{in}}'\left(0\right)=0 \end{cases}\tag4 $$

And when we solve that and using equation \$\left(1\right)\$, we find:

$$\overline{\text{P}}_{\space\text{in}}=\frac{2\pi^2}{1-4\pi^2+16\pi^4}\approx0.0129857\tag5$$

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