3
\$\begingroup\$

Suppose we have this circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

->

->

No to find the apparent power of R1 we multiply V*I

->

The active power is <12W.

However the total power dissipated on the circuit is:

enter image description here

and because the only thing in the circuit which consumes power is the resistor the total power dissipated on the circuit is the power dissipated on the resistor. Why am I getting 2 different results?My question is how is it possible to find different values of true power using different methods?

\$\endgroup\$
3
  • \$\begingroup\$ Can you show how you calculated your value for \$Z_T\$? \$\endgroup\$ Dec 20, 2021 at 22:02
  • \$\begingroup\$ @ElliotAlderson Updated \$\endgroup\$
    – Jun Seo-He
    Dec 20, 2021 at 22:16
  • \$\begingroup\$ Check whether the inductive and capacitive reactance cancel out each other? \$\endgroup\$
    – Alex
    Dec 27, 2021 at 7:51

3 Answers 3

0
\$\begingroup\$

The issue is that the correct calculation for apparent power uses the conjugate of current (S=VI*). If you are unfamiliar with the conjugate operation it essentially reverses the angle of a complex number (or equivalently reverses the sign of the imaginary component). So in your calculations the exponential terms for voltage and current will cancel leaving 12W exactly. The other aspect of this to tip you off to where the mistake is, is that resistors should only ever dissipate real power, in your calculations the fact that you are getting a complex number for the resistors apparent power should be a red flag that something is wrong.

\$\endgroup\$
1
\$\begingroup\$

Well, the real or active power provided by the source is defined as:

$$\text{P}_\text{s}:=\text{V}_{\text{s}\space\text{|}\space\text{rms}}\cdot\text{I}_{\text{s}\space\text{|}\space\text{rms}}\cdot\cos\left(\varphi_\text{s}\right)\tag1$$

Which we can take a look at individually:

  • \$\text{V}_{\text{s}\space\text{|}\space\text{rms}}\$ is given by: $$\text{V}_{\text{s}\space\text{|}\space\text{rms}}=\frac{\displaystyle\hat{\text{u}}_\text{i}}{\displaystyle\sqrt{2}}\tag2$$ Where \$\hat{\text{u}}_\text{i}\$ is the amplitude of the input voltage.
  • \$\text{I}_{\text{s}\space\text{|}\space\text{rms}}\$ is given by: \begin{equation} \begin{split} \text{I}_{\text{s}\space\text{|}\space\text{rms}}&=\frac{\displaystyle1}{\displaystyle\sqrt{2}}\cdot\left|\frac{\displaystyle\underline{\text{V}}_{\space\text{s}}}{\displaystyle\underline{\text{Z}}_{\space\text{i}}}\right|\\ \\ &=\frac{\displaystyle1}{\displaystyle\sqrt{2}}\cdot\frac{\displaystyle\left|\underline{\text{V}}_{\space\text{s}}\right|}{\displaystyle\left|\underline{\text{Z}}_{\space\text{i}}\right|}\\ \\ &=\frac{\displaystyle1}{\displaystyle\sqrt{2}}\cdot\frac{\displaystyle\hat{\text{u}}_\text{i}}{\displaystyle\left|\text{R}+\left(\text{j}\omega\text{L}\space\text{||}\space\frac{1}{\text{j}\omega\text{C}}\right)\right|}\\ \\ &=\frac{\displaystyle\hat{\text{u}}_\text{i}}{\displaystyle\sqrt{2}}\cdot\left(\sqrt{\text{R}^2+\left(\frac{\displaystyle\text{L}\omega}{\displaystyle1-\text{CL}\omega^2}\right)^2}\right)^{-1} \end{split}\tag3 \end{equation}

Where \$\displaystyle\underline{\text{Z}}_{\space\text{i}}\$ is the input impedance of the circuit and \$\displaystyle\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$.

  • \$\displaystyle\varphi_\text{s}\$ is given by: $$\varphi_\text{s}=\arg\left(\underline{\text{Z}}_{\space\text{i}}\right)= \begin{cases} 0&\space\text{when}\space\omega=0\\ \\ \arctan\left(\frac{\displaystyle1}{\displaystyle\text{R}}\cdot\frac{\displaystyle\text{L}\omega}{\displaystyle1-\text{CL}\omega^2}\right)&\space\text{when}\space1-\text{CL}\omega^2>0\\ \\ \frac{\displaystyle3\pi}{\displaystyle2}+\arctan\left(\text{R}\cdot\frac{\displaystyle\text{CL}\omega^2-1}{\displaystyle\text{L}\omega}\right)&\space\text{when}\space1-\text{CL}\omega^2<0 \end{cases} \tag4$$ Where \$\displaystyle\arg\left(\cdot\right)\$ is the principal value of the argument.

So, we end up with:

\begin{equation} \begin{split} \text{P}_\text{s}&:=\text{V}_{\text{s}\space\text{|}\space\text{rms}}\cdot\text{I}_{\text{s}\space\text{|}\space\text{rms}}\cdot\cos\left(\varphi_\text{s}\right)\\ \\ &=\frac{\displaystyle\hat{\text{u}}_\text{i}}{\displaystyle\sqrt{2}}\cdot\frac{\displaystyle\hat{\text{u}}_\text{i}}{\displaystyle\sqrt{2}}\cdot\left(\sqrt{\text{R}^2+\left(\frac{\displaystyle\text{L}\omega}{\displaystyle1-\text{CL}\omega^2}\right)^2}\right)^{-1}\cdot\cos\left(\varphi_\text{s}\right)\\ \\ &=\frac{\displaystyle\hat{\text{u}}_\text{i}^2}{\displaystyle2}\cdot\left(\sqrt{\text{R}^2+\left(\frac{\displaystyle\text{L}\omega}{\displaystyle1-\text{CL}\omega^2}\right)^2}\right)^{-1}\cdot\cos\left(\varphi_\text{s}\right) \end{split}\tag5 \end{equation}

And using \$(4)\$, we get:

\begin{equation} \begin{split} \cos\left(\varphi_\text{s}\right)&=\cos\left(\begin{cases} 0&\space\text{when}\space\omega=0\\ \\ \arctan\left(\frac{\displaystyle1}{\displaystyle\text{R}}\cdot\frac{\displaystyle\text{L}\omega}{\displaystyle1-\text{CL}\omega^2}\right)&\space\text{when}\space1-\text{CL}\omega^2>0\\ \\ \frac{\displaystyle3\pi}{\displaystyle2}+\arctan\left(\text{R}\cdot\frac{\displaystyle\text{CL}\omega^2-1}{\displaystyle\text{L}\omega}\right)&\space\text{when}\space1-\text{CL}\omega^2<0 \end{cases}\right)\\ \\ &=\begin{cases} 1&\space\text{when}\space\omega=0\\ \\ \frac{\displaystyle1}{\displaystyle\sqrt{1+\epsilon_1^2}}&\space\text{when}\space1-\text{CL}\omega^2>0\\ \\ \frac{\displaystyle\epsilon_2}{\displaystyle\sqrt{1+\epsilon_2^2}}&\space\text{when}\space1-\text{CL}\omega^2<0 \end{cases} \end{split}\tag6 \end{equation}

Where \$\displaystyle\epsilon_1:=\frac{\displaystyle1}{\displaystyle\text{R}}\cdot\frac{\displaystyle\text{L}\omega}{\displaystyle1-\text{CL}\omega^2}\$ and \$\displaystyle\epsilon_2:=\text{R}\cdot\frac{\displaystyle\text{CL}\omega^2-1}{\displaystyle\text{L}\omega}\$.


In your case, we have:

$$\underline{\text{Z}}_{\space\text{i}}=3+\frac{\displaystyle400\pi\text{j}}{\displaystyle625-32\pi^2}\tag7$$

So, we get:

$$\left(\sqrt{\text{R}^2+\left(\frac{\displaystyle\text{L}\omega}{\displaystyle1-\text{CL}\omega^2}\right)^2}\right)^{-1}=\frac{\displaystyle625-32\pi^2}{\displaystyle\sqrt{\pi^2\left(9216\pi^2-200000\right)+3515625}}\tag8$$

And:

$$1-\text{CL}\omega^2=1-\frac{\displaystyle32\pi^2}{\displaystyle625}\approx0.494676>0\tag9$$

And:

$$\epsilon_1=\frac{\displaystyle1}{\displaystyle\text{R}}\cdot\frac{\displaystyle\text{L}\omega}{\displaystyle1-\text{CL}\omega^2}=\frac{\displaystyle400\pi}{\displaystyle1875-96\pi^2}\tag{10}$$

So:

$$\frac{\displaystyle1}{\displaystyle\sqrt{1+\epsilon_1^2}}=\frac{\displaystyle3\left(625-32\pi^2\right)}{\displaystyle\sqrt{\pi^2\left(9216\pi^2-200000\right)+3515625}}\tag{11}$$

So, we get for the real or active power:

$$\text{P}_\text{s}=\frac{\displaystyle600\left(625-32\pi^2\right)^2}{\displaystyle\pi^2\left(18432\pi^2-400000\right)+7031250}\approx11.75534\space\text{W}\tag{12}$$

\$\endgroup\$
6
  • \$\begingroup\$ Why can't you calculate the active power using complex analysis of the circuit? \$\endgroup\$
    – Jun Seo-He
    Dec 21, 2021 at 15:05
  • \$\begingroup\$ @JunSeo-He I did use complex analysis. I only can't find where you got wrong. Can you tell me what the amplitude of the input voltage is? \$\endgroup\$ Dec 21, 2021 at 15:07
  • \$\begingroup\$ It is 10V rms. I am converting the impedances and currents in complex exponential form I find the voltage drop and the power in complex exponential form but it doesnt give me the value found by using the power factor. \$\endgroup\$
    – Jun Seo-He
    Dec 21, 2021 at 15:12
  • \$\begingroup\$ @JunSeo-He Well, I found the exact figure of the active power and it is almost 12 Watt, so I think that your analysis is right but not exact. \$\endgroup\$ Dec 21, 2021 at 15:32
  • \$\begingroup\$ @J.W.L.JanEerland By "exact" I think you mean "as approximated with double-precision floating-point arithmetic", don't you? \$\endgroup\$ Dec 21, 2021 at 15:46
0
\$\begingroup\$

This is what I know about power for sinusoidal signals: -

Real/active power: \$P=V_{RMS}I_{RMS} \cdot \cos(\phi_v-\phi_i) = \frac{V_pI_p}{2}\cdot \cos(\phi_v-\phi_i) \$

Reactive/imaginary power: \$Q= V_{RMS}I_{RMS} \cdot \sin(\phi_v-\phi_i) = \frac{V_pI_p}{2} \cdot \sin(\phi_v-\phi_i)\$

Apparent power: \$\text{apparent power} = \sqrt{P^2+Q^2} \$

Instantaneous power: \$p(t) = P\bigg(1+\cos(2\omega t) \bigg)-Q\sin(2\omega t) \$

This is your mistake: -

enter image description here

It should be, as my first formula shows,: \$P_{R1} = V_{RMS}I_{RMS}\cdot \cos\bigg((-53^\circ)+53^\circ \bigg)=V_{RMS}I_{RMS} \$

This is obvious when you realize that voltage and current are always in-phase for a resistor.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.