# Calculating the reactive and active power

This is probably an easy one for you. In the following circuit simulate this circuit – Schematic created using CircuitLab

I want to calculate the active and reactive power in load R2-C3. I've calculated the voltage over these elements to be

$$u=\hat{U} \sin(\omega t + \phi) = 0.0465 \sin(1000 t + 3.53 \text{ rad}) V$$

Now I calculate the active power, P, to be

$$P = \frac{U_e^2}{R2}$$

and the reactive power, Q, to be

$$Q = \frac{U_e^2}{\Im (Z_{C3})}$$

Where $$Z_{C3}$$ is the impedance corresponding to C3 and $$U_e = \frac{\hat{U}}{\sqrt{2}}$$ is the effective value, or root mean square, of voltage u.

Am I doing this correctly?

• Are you assuming that the phase angle of $U_e$ is zero or does it have a reactive component? – Elliot Alderson Dec 12 '18 at 16:29
• The input power comes from I1 and not a voltage source so how would you calculate the voltage? Plus, there is no statement about the transformer's turns ratio or parasitic components. – Andy aka Dec 12 '18 at 16:33
• See my edits. I've already dealt with the transformer and know the voltage over R2-C3. So it was probably confusing, and unnecessary, to include the previous circuit in the question. – user1176517 Dec 12 '18 at 16:45
• Take the voltage as the zero phase angle reference, then calculate $U^2/R$ for the power and $U^2/X$ for the reactive VA, where $U$ is the RMS value. – Chu Dec 12 '18 at 23:17

## 2 Answers

If the capacitor were a second resistor, you would write

$$P=\frac{U_e^2}{R_{tot}}=\frac{U_e^2}{(R_2\parallel R_3)}=\frac{U_e^2}{\frac{R_2R_3}{R_2+R_3}}$$

Same applies to this complex network:

$$P=\frac{U_e^2}{Z_{tot}}=\frac{U_e^2}{(R_2\parallel Z_{C_3})}$$

So, it's now a little more complicated to get $$\\Re(P)\$$ and $$\\Im(P)\$$, but that's just math...

Well, the real power is given by:

$$\text{P}=\text{V}_\text{rms}\text{I}_\text{rms}\cos\left(\varphi\right)\tag1$$

Where $$\\varphi=\left|\arg\left(\underline{\text{V}}\right)-\arg\left(\underline{\text{I}}\right)\right|=\left|\arg\left(\underline{\text{Z}}\right)\right|\$$.

The complex power is given by:

$$\text{Q}=\text{V}_\text{rms}\text{I}_\text{rms}\sin\left(\varphi\right)\tag2$$

And the apparent power is given by:

$$\text{S}=\sqrt{\text{P}^2+\text{Q}^2}=\text{V}_\text{rms}\text{I}_\text{rms}\tag3$$

So, for your circuit:

1. Resistor: $$\text{P}=\frac{0.0465}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}\cdot\frac{0.0465}{8.5\cdot1000}\cdot\cos\left(\left|\arg\left(8.5\cdot1000\right)\right|\right)=$$ $$\frac{0.0465^2}{2\cdot8500}\cdot\cos\left(0\right)\approx0.00000013\space\text{W}\tag4$$
2. Capacitor: $$\text{Q}=\frac{0.0465}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}\cdot\frac{0.0465}{\frac{1}{2000\pi\cdot170\cdot10^{-6}}}\cdot\sin\left(\left|\arg\left(\frac{1}{\text{j}\cdot2000\pi\cdot170\cdot10^{-6}}\right)\right|\right)=$$ $$\frac{2000\pi\cdot170\cdot10^{-6}\cdot0.0465^2}{2}\cdot\sin\left(\frac{3\pi}{2}\right)\approx-0.00115479\space\text{VAR}\tag4$$