# Calculating Real Power Dissipation in a Series RC Circuit

Given circuit:

simulate this circuit – Schematic created using CircuitLab

Question:

What is the real power dissipated in the resistor of the given circuit?

My Solution:

Note: I am calculating the real power analytically without inserting specific numerical values.

Using complex numbers:

• $$\\overline{V}=\hat{V}\$$
• $$\\overline{Z}=R-jX_C\$$

Then I calculated the current:

• $$\\overline{I}=\frac{\overline{V}}{\overline{Z}} = \frac{\hat{V}}{R-jX_C} = \cdots = \frac{\hat{V}\cdot R}{R^2+X_C^2}+j\frac{\hat{V}\cdot X_C}{R^2+X_C^2}\$$

Next, I calculated the apparent power:

• $$\\overline{S}=\overline{V} \cdot \overline{I}^*=\hat{V}\cdot \left( \frac{\hat{V}\cdot R}{R^2+X_C^2}-j\frac{\hat{V}\cdot X_C}{R^2+X_C^2} \right)\$$

Finally, the real power is the real part of the apparent power:

• $$\P=\text{Re}\{\overline{S}\}= \frac{\hat{V}^2\cdot R}{R^2+X_C^2} \quad \text{with} \ X_C = \frac{1}{2\cdot \pi\cdot f \cdot C} \$$

Is my solution correct? Or does anyone have any better ideas or improvements?

Example with values:

• $$\\hat{V}=3.2V\$$ (amplitude)
• $$\f=100kHz\$$
• $$\R=50m\Omega\$$
• $$\C=10\mu F\$$

$$\P=\text{Re}\{\overline{S}\}= \frac{3.2V^2\cdot 50\cdot10^{-3}\Omega}{(50\cdot10^{-3}\Omega)^2+\left(\frac{1}{2\cdot \pi \cdot 100 \cdot 10^3 s^{-1} \cdot 10 \cdot 10^{-6} s\ \Omega^{-1}}\right)^2} \approx \underline{\underline{18.397W}} \$$

$$\P_{average}=\frac{P}{2}=\frac{18.397W}{2}\approx9.199W\$$

• Just a quick skim and I think you are okay. If I suppose 100 kHz and 10 V peak value and your values above then I find that your final power equation arrives at the real part of 89.8301624 + j285.938288 given (10/sqrt(2))*(10/sqrt(2))/(50m+10uF). So I think you are good. I'll spend more time if you have more questions. But right now I think you are okay. You also correctly used the conjugate of the current. So it really does look like you took the right path to me. Commented Jun 2 at 7:25
• With 3.2 V peak I get (3.2/sqrt(2))*(3.2/sqrt(2))/(50m+10uF) = 9.19860863 + j29.2800806 or 30.6910007 ∠ 72.5594055. Power is an average over one or more cycles (an integer number of them) and not a maximum. For instantaneous values you need to use Euler's. And that's a whole other question. Commented Jun 2 at 10:25
• Oki If so, to get the average (real) power you need to divide your result by 2. Thus, P = 18.397W/2 = 9.198W
– G36
Commented Jun 2 at 11:57
• Yes, 18.397W is the peak real powe.
– G36
Commented Jun 2 at 12:14
• @MarcoMoldenhauer Everything seems settled now. Best wishes! Commented Jun 2 at 17:36

Well, the real power in the circuit is given by:

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

So, let's find the three parts:

1. $$\text{V}_\text{rms}=\frac{\displaystyle\hat{\text{u}}_\text{i}}{\displaystyle\sqrt{2}}\tag2$$
2. $$\text{V}_\text{rms}=\frac{\displaystyle1}{\displaystyle\sqrt{2}}\cdot\frac{\displaystyle\hat{\text{u}}_\text{i}}{\displaystyle\left|\text{Z}_\text{i}\right|}=\frac{\displaystyle1}{\displaystyle\sqrt{2}}\cdot\frac{\displaystyle\hat{\text{u}}_\text{i}}{\displaystyle\left|\text{R}+\frac{\displaystyle1}{\displaystyle\text{j}\omega\text{C}}\right|}=\frac{\displaystyle1}{\displaystyle\sqrt{2}}\cdot\frac{\displaystyle\hat{\text{u}}_\text{i}}{\displaystyle\sqrt{\text{R}^2+\left(\frac{\displaystyle1}{\displaystyle\omega\text{C}}\right)^2}}\tag3$$
3. $$\varphi=\arg\left(\text{Z}_\text{i}\right)=\arg\left(\text{R}+\frac{\displaystyle1}{\displaystyle\text{j}\omega\text{C}}\right)=\arg\left(\text{R}-\frac{\displaystyle\text{j}}{\displaystyle\omega\text{C}}\right)=\frac{\displaystyle3\pi}{\displaystyle2}+\arctan\left(\omega\text{CR}\right)\tag4$$

So, we end up with:

$$\text{P}=\frac{\displaystyle\hat{\text{u}}_\text{i}}{\displaystyle\sqrt{2}}\cdot\frac{\displaystyle1}{\displaystyle\sqrt{2}}\cdot\frac{\displaystyle\hat{\text{u}}_\text{i}}{\displaystyle\sqrt{\text{R}^2+\left(\frac{\displaystyle1}{\displaystyle\omega\text{C}}\right)^2}}\cdot\cos\left(\frac{\displaystyle3\pi}{\displaystyle2}+\arctan\left(\omega\text{CR}\right)\right)\tag5$$

Which simplifies to:

$$\text{P}=\frac{\displaystyle\text{R}\left(\omega\text{C}\hat{\text{u}}_\text{i}\right)^2}{\displaystyle2\left(1+\left(\omega\text{CR}\right)^2\right)}\tag6$$

$$\text{P}=\frac{\displaystyle50\cdot10^{-3}\left(2\pi\cdot100\cdot10^3\cdot10\cdot10^{-6}\cdot\frac{\displaystyle16}{\displaystyle5}\right)^2}{\displaystyle2\left(1+\left(2\pi\cdot100\cdot10^3\cdot10\cdot10^{-6}\cdot50\cdot10^{-3}\right)^2\right)}=\frac{\displaystyle512\pi^2}{\displaystyle5\left(100+\pi^2\right)}\approx9.19861\space\text{W}\tag7$$