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I've the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

I know that the relation between the input voltage and the input current is given by:

$$\begin{cases} \text{V}_{\space\text{in}}\left(t\right)=\text{I}_{\space\text{in}}\left(t\right)\cdot\text{R}_1+\text{I}_{\space\text{in}}'\left(t\right)\cdot\text{L}_1+\text{V}_{\space\text{R}_2\space||\space\text{L}_2}\left(t\right)\\ \\ \text{I}_{\space\text{in}}'\left(t\right)=\text{V}_{\space\text{R}_2\space||\space\text{L}_2}'\left(t\right)\cdot\frac{1}{\text{R}_2}+\text{V}_{\space\text{R}_2\space||\space\text{L}_2}\left(t\right)\cdot\frac{1}{\text{L}_2} \end{cases}\tag1 $$

Now, when I measure the real input power, I will get an the RMS value of the real input power, and my question is: can I state the following:

$$\text{P}_{\space\text{in RMS}}=\text{P}_{\space\text{R}_1\space\text{RMS}}+\text{P}_{\space\text{R}_2\space\text{RMS}}=\text{I}_{\space\text{in RMS}}^2\cdot\left(\text{R}_1+\text{R}_2\right)\tag2$$

Where:

$$\text{I}_{\space\text{in RMS}}^2=\lim_{\text{n}\to\infty}\sqrt{\frac{1}{\text{n}}\int_0^\text{n}\left(\text{I}_{\space\text{in}}^2\left(t\right)\right)^2\space\text{d}t}=\lim_{\text{n}\to\infty}\sqrt{\frac{1}{\text{n}}\int_0^\text{n}\text{I}_{\space\text{in}}^4\left(t\right)\space\text{d}t}\tag3$$

hjk6
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  • Power is either instantaneous or an average. RMS makes no sense for power (well, unless you define a new meaning for it that I don't know about.) Also, I see flaws in your thinking about (2) and (3). Hopefully, someone will help you sort this. For me, I don't have the energy right now. – jonk Jan 30 '18 at 17:35

1 Answers1

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No you can't state what you have stated because \$I_{in}\$ is not exclusively shared by both resistors R1 and R2. Some of \$I_{in}\$ flows through L2.

when I measure the real input power, I will get an the RMS value of the real input power

When you use a wattmeter you measure average power. RMS power is a confusion.

Andy aka
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  • Yes sure my mistake, I see now. – hjk6 Jan 30 '18 at 17:28
  • But when I measure current and voltage I will meaure the RMS, so why isn't that the case when I measure power? – hjk6 Jan 30 '18 at 17:30
  • No, RMS current and RMS voltage (if in phase and sinewaves) when multiplied together produce average power. In other words there is a waveform produced for power (at twice the frequency of voltage and current) but its average value will be true power. – Andy aka Jan 30 '18 at 17:33
  • [Look at my answer here for clarification](https://electronics.stackexchange.com/questions/202667/whats-the-most-economical-way-to-digitally-measure-240v-mains-voltage-current) – Andy aka Jan 30 '18 at 17:38