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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

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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}\$.

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    \$\begingroup\$ Excellent answer, Jan. Since the 14.4W dissipation is so very brief, are watts a bit misleading for sizing a resistor for an RC circuit? Would Joules be better? Since we're talking 4-5 milliseconds to charge, would the average Joules be 14.4W*0.005 = 0.072 Joules? Is that why even a 1/4 watt 10 Ohm resistor doesn't even get warm with 12V and 300 uF? \$\endgroup\$ May 3 at 21:41
  • \$\begingroup\$ @BeachInMexico Yes you can calculate the total energy dissipated in the resistor. You will get: $$\text{E}_\text{total}=\int_\mathbb{R}\text{P}_\text{R}\left(t\right)\space\text{d}t=\frac{\text{C}\hat{\text{u}}_\text{i}^2}{2}$$ \$\endgroup\$
    – Jan
    May 3 at 21:44
  • \$\begingroup\$ Thank you! For my example of 12V, 100 Ohm, 300 uF, is that Etotal = (Capacitance * volts^2) / 2, or (0.0003*(12*12))/2 = 0.0216J? That's such a tiny amount it begins to explain why even a 1/4 Watt resistor wouldn't get warm. \$\endgroup\$ May 3 at 22:17
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The energy stored in the cap = \$\frac{1}{2}CV^2\$ = 100e-6/2*144= 7.2mJ

  • with a half power time constant of \$0.5P_{max}=\dfrac{\tau}{2\sqrt{2}}=~0.35\tau\$

  • for RC = 10 x 1e-4= 1ms

  • This means a start current of 1.2A, 14.4W and after 350us it is 7.2W

  • in 1ms when the Vcap=64% Vin leaving 36% Voltage across the resistor and the square of that yields 13% Pmax at t=RC

  • But assume that the thermal mass requires X seconds to reach 1/2 it’s rated 100’C max temp rise above 25’C the power in 1ms will result in a case temperature rise of 0.1% of X temperature rise.

    • But what if the part is only rated for 1/8W or a peak power of 115 times the parts rated power ?

    • it might be expected to rise 11 deg C but convection and lead conduction cooling would reduce that to barely warm.

    • follow absolute max specs with a 50% safety margin

  • semiconductors have a SOA Power vs pulse width curve that spans a wide range of Pk/avg powers

  • Yet 5mm LEDs do not as the micron size gold whisker wire bond can fail(fuse open) at some level after a peak current that is usually 30mA for a 20mA thermal rating.

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The power dissipation being large in the beginning won't necessarily make it perceptibly hot in a single charging, you have to take in account its thermal capacity in account.

In a RC circuit, charging the capacitor from 0 to the source voltage always dissipates half of the energy given by the source in the resistance and the other half is stored in the capacitor, which also means these energies are equal.

Now, consider this charging was very fast and that no heat was lost. So the resistor would have a temperature change ΔT which will depend solely on its thermal capacity (c) and the energy it dissipated (Er):

If you assume its total thermal capacity is equal to that of 1 gram of water (4.2 J/K), with a 12 V DC source, a 10 ohm resistor and a 100 uF capacitor, you get a ΔT of approximately 0.0017 Celsius (or Kelvin).

You'd never feel that. And although it was an "worst case scenario", as this charging will be very fast, then it's probably close to the real temperature variation.

Of course that a tiny resistor can have a thermal capacity well lesser than that of 1 gram of water (because it has way less material) but still, it's hard to heat it like that, the ΔT could be a thousand times larger and you wouldn't feel it warming yet.

Now, if instead of just charging it once you're constantly charging and discharging, then you need to bother about your average power.

More specifically the average power in a time window that's coherent to the "thermal inertia" of the resistor, and not any arbitrary time window (if it's too large then you could have intervals inside it where the power rating would be exceeded for way too long, even if the average over the whole time window is small).

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  • \$\begingroup\$ Excellent answer. Thank you! I wonder how many amps I could run through a 1/4 watt resistor in this same circuit without a problem, for instance 12V source, 300 uF cap, and 1 Ohm resistor, so 12 Amps and 144 watt power dissipation. Your delta T equation is very useful. \$\endgroup\$ May 5 at 2:11
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You don't say whether the 12 V source is AC or DC. If it is DC, then your math is incorrect and the online calculator is correct. 1.2 x 12 = 14.4.

But that is only the instantaneous power at the moment of voltage application. As the capacitor charges up, the voltage across it increases so the voltage across the resistor (and hence its power dissipation) decreases - rapidly. The voltage decreases exponentially, and the power decreases as the square of that (Watt's Law).

In your case, the R-C time constant (R x C) is 0.001 s. Here is a piece-wise approximation of the power dissipation in the resistor:

time 0 s -- 14.400 W

time 1 ms -- 1.949 W

time 2 ms -- 0.264 W

time 3 ms -- 0.036 W

At 3 time constants, the voltage across the capacitor is at 95% of the source voltage, leaving only 5% of the source voltage across the resistor. this is a reasonable place to stop the approximating.

As you can see, the instantaneous power dissipation drops to under 1/20 W in under 3 milliseconds. This is why a 0.5 W resistor's temperature increase is barely noticeable. The total energy dissipated in the resistor over the first 3 milliseconds is approx. 0.025 watt-seconds, about 5% of the resistor's continuous power dissipation rating.

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  • \$\begingroup\$ Good catch. DC source. How low do you think the resistor Ohms could be before this would even be noticeable. Same circuit with 1 Ohm? So 12 A and 144 W in the first millisecond. \$\endgroup\$ May 5 at 2:13

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