I've the following circuit:


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:

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


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

  • \$\begingroup\$ 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. \$\endgroup\$
    – jonk
    Jan 30, 2018 at 17:35

1 Answer 1


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.

  • \$\begingroup\$ Yes sure my mistake, I see now. \$\endgroup\$
    – hjk6
    Jan 30, 2018 at 17:28
  • \$\begingroup\$ But when I measure current and voltage I will meaure the RMS, so why isn't that the case when I measure power? \$\endgroup\$
    – hjk6
    Jan 30, 2018 at 17:30
  • \$\begingroup\$ 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. \$\endgroup\$
    – Andy aka
    Jan 30, 2018 at 17:33
  • \$\begingroup\$ Look at my answer here for clarification \$\endgroup\$
    – Andy aka
    Jan 30, 2018 at 17:38

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