Electronic – RC circuit resistor power dissipation

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How do you calculate the actual power dissipation across the resistor in an RC circuit?

For instance, 12 V source, 10 Ohm resistor in series with 100 uF capacitor, no other load connected.

12/10 = 1.2A x 12V = 12 Watts (This calculator says 14.4).

But only for 4 milliseconds.

I tried a 0.5 Watt 10 Ohm resistor with 300 uF and 12V supply and the resistor didn't even get warm, so I assume the power dissipation is very small. But I'd like to calculate for other source voltages, resistors, capacitors.

Is there a simple way to calculate the RMS power dissipation per this partial answer: RC Network resistor power rating with DC supply

Best Answer

Well, according to Ohm's law we know that:

$$\text{V}_\text{R}\left(t\right)=\text{I}_\text{R}\left(t\right)\cdot\text{R}\tag1$$

And the power in a resistor is given by:

$$\text{P}_\text{R}\left(t\right)=\text{V}_\text{R}\left(t\right)\cdot\text{I}_\text{R}\left(t\right)\tag2$$

Combining both equations we can see that:

$$\text{P}_\text{R}\left(t\right)=\text{I}_\text{R}^2\left(t\right)\cdot\text{R}\tag3$$


In a series RC-circuit we know that the input current is given by:

$$\text{I}_\text{in}\left(t\right)=\mathscr{L}_\text{s}^{-1}\left[\frac{\hat{\text{u}}_\text{i}}{\text{s}}\cdot\frac{1}{\text{R}+\frac{1}{\text{sC}}}\right]_{\left(t\right)}\tag4$$

Where \$\mathscr{L}_\text{s}^{-1}\left[\cdot\right]_{\left(t\right)}\$ is the inverse Laplace transform.

Using the table of Laplace transforms we can see that:

$$\text{I}_\text{in}\left(t\right)=\frac{\hat{\text{u}}_\text{i}}{\text{R}}\cdot\exp\left(-\frac{t}{\text{CR}}\right)\tag5$$

Because the circuit is in series we know that the current in the resistor and the capacitor are equal, so we get:

$$\text{P}_\text{R}\left(t\right)=\left(\frac{\hat{\text{u}}_\text{i}}{\text{R}}\cdot\exp\left(-\frac{t}{\text{CR}}\right)\right)^2\cdot\text{R}=\frac{\hat{\text{u}}_\text{i}^2}{\text{R}}\cdot\exp\left(-\frac{2t}{\text{CR}}\right)\tag6$$

Using your values, we get:

$$\text{P}_\text{R}\left(t\right)=\frac{12^2}{10}\cdot\exp\left(-\frac{2t}{100\cdot10^{-6}\cdot10}\right)=\frac{72}{5}\cdot\exp\left(-2000t\right)\tag7$$

And the value found by the calculator is \$\frac{72}{5}=14.4\space\text{W}\$, because at \$t=0\$ we can see that \$\text{P}_\text{R}\left(0\right)=\frac{72}{5}=14.4\space\text{W}\$.