# Simple complex power question

#### Lifelongstudent

Joined Jul 30, 2019
8
so I have been out of the game for a while with hardware
Mostly been a software person for a while.

I used to have lots of experience with 3 phase

It’s been a long while since I messed with DC power.

Question:
If you have say a simple RC circuit that is a complex impedance
And you put DC through on it, is there any reactive power? Is it all watts?

I am trying to recall the rules
Now logic tells me that at steady state, the cap is going to have infinite impedance

So is it consuming any power?

How can we prove this?

#### Papabravo

Joined Feb 24, 2006
16,810
Rule #1 You can't put DC current through a capacitor. It's blocked. You can however charge it and discharge it to a voltage. Charging and discharging don't cause any real power dissipation in the capacitor
Rule #2 You can put DC current through a real inductor because it has some small dc resistance in the coil. That resistance will dissipate real power.

Read the article on RC and RL circuits
https://en.wikipedia.org/wiki/RC_circuit
https://en.wikipedia.org/wiki/RL_circuit

#### Lifelongstudent

Joined Jul 30, 2019
8
Rule #1 You can't put DC current through a capacitor. It's blocked. You can however charge it and discharge it to a voltage. Charging and discharging don't cause any real power dissipation in the capacitor
Rule #2 You can put DC current through a real inductor because it has some small dc resistance in the coil. That resistance will dissipate real power.

Read the article on RC and RL circuits
https://en.wikipedia.org/wiki/RC_circuit
https://en.wikipedia.org/wiki/RL_circuit
Yes technically true I suppose

My question to be more specific

How would you calculate the power consumed in that circuit at steady state because p=Iv seems to simple lol

#### crutschow

Joined Mar 14, 2008
27,734
How would you calculate the power consumed in that circuit at steady state
You replace all the capacitors with an open circuit, all the inductors with their internal resistance, and then calculate the power as you would any DC circuit using P = I*V.

#### Lifelongstudent

Joined Jul 30, 2019
8
You replace all the capacitors with an open circuit, all the inductors with their internal resistance, and then calculate the power as you would any DC circuit using P = I*V.
So there are no reactive vars, the power factor is 1

#### WBahn

Joined Mar 31, 2012
26,398
Rule #1 You can't put DC current through a capacitor. It's blocked. You can however charge it and discharge it to a voltage. Charging and discharging don't cause any real power dissipation in the capacitor
Rule #2 You can put DC current through a real inductor because it has some small dc resistance in the coil. That resistance will dissipate real power.

Read the article on RC and RL circuits
https://en.wikipedia.org/wiki/RC_circuit
https://en.wikipedia.org/wiki/RL_circuit
You seem to be saying that the reason that you can put DC current through a real inductor is that has some small DC resistance in the coil, implying that if it didn't have some small DC resistance that you couldn't put DC current through it. That makes no sense and you can most definitely put a DC current through a superconducting coil, which is both a real and inductor and one that does not have some small DC resistance in it.

#### WBahn

Joined Mar 31, 2012
26,398
so I have been out of the game for a while with hardware
Mostly been a software person for a while.

I used to have lots of experience with 3 phase

It’s been a long while since I messed with DC power.

Question:
If you have say a simple RC circuit that is a complex impedance
And you put DC through on it, is there any reactive power? Is it all watts?

I am trying to recall the rules
Now logic tells me that at steady state, the cap is going to have infinite impedance

So is it consuming any power?

How can we prove this?
Once you reach DC steady state, then none of your voltages and currents can be changing (else it isn't in steady state). Since the current in a capacitor is proportional to the rate of change of the voltage across it and since the voltage across it can't be changing, then the current has to be identically zero. Since the power being absorbed by it (real or reactive) is the product of the voltage across it and the current through it and since the current through it is zero, there is no power flowing into or out of it. A similar situation for inductors -- the voltage across them is proportional to the rate of change of the current through them, which is identically zero in steady state, so once again they can't be either absorbing or delivering any power at all.

Most real components, particularly inductors, have parasitics that can and do dissipate real power even in DC steady state. These appear as a (usually small) resistance in series with an ideal inductor or a (usually large) resistance in parallel with an ideal capacitor. But the reactive part of the component does not absorb or deliver any power at all, real or reactive, in DC steady state.

#### Lifelongstudent

Joined Jul 30, 2019
8
You seem to be saying that the reason that you can put DC current through a real inductor is that has some small DC resistance in the coil, implying that if it didn't have some small DC resistance that you couldn't put DC current through it. That makes no sense and you can most definitely put a DC current through a superconducting coil, which is both a real and inductor and one that does not have some small DC resistance in it.
Any info on my original question?
I am trying to deduce if given a complex load
Say 4+5j and then are told you apply a dc source to it, what power do you have?
Is it purely real power or includes some amount of reactive vars

#### WBahn

Joined Mar 31, 2012
26,398
Any info on my original question?
Which question? I count four of them in the original post. I provided enough information for you to answer all of them, even though you didn't provide enough information for one of them for us to answer it completely.

Let's walk through the questions in your original post.

If you have say a simple RC circuit that is a complex impedance
And you put DC through on it, is there any reactive power?
If I were to tell you that, "the reactive part of the component does not absorb or deliver any power at all, real or reactive, in DC steady state," does that not provide some information regarding this question? The real part of the circuit clearly can't have any reactive power. If the reactive part can't have any power at all, real or reactive, is there any reactive power?

What other information do you need to feel confident of answering your first question?

Is it all watts?
If there's no reactive power, the wouldn't that mean that any power that there is must all be real power?

Note that you didn't say whether this simple RC circuit is a series RC circuit or a parallel RC circuit. If it a series RC circuit then you don't have any power of any kind at all in steady state DC. If it is a parallel RC circuit, then you do have some real power being dissipated by the resistor.

I am trying to recall the rules
Now logic tells me that at steady state, the cap is going to have infinite impedance

So is it consuming any power?
If told: "Since the current in a capacitor is proportional to the rate of change of the voltage across it and since the voltage across it can't be changing, then the current has to be identically zero. Since the power being absorbed by it (real or reactive) is the product of the voltage across it and the current through it and since the current through it is zero, there is no power flowing into or out of it." What else do you need in order to conclude whether or not it is consuming any power?

How can we prove this?
I walked through a mathematical proof based on the constitutive equations for an ideal inductor and an ideal capacitor and explained how the constraints imposed in DC steady state preclude ANY power, real or reactive, flowing into or out of either of them. What else is needed?

I am trying to deduce if given a complex load
Say 4+5j and then are told you apply a dc source to it, what power do you have?
The complex load is 4+5j what? Ohms? Milliohms? Kilohms? For the sake of discussion, I'll assume you mean ohms.

How are you getting a reactance of 5 Ω at DC? The reactance is the product of the inductance and the radian frequency. At DC the radian frequency is zero.

Is it purely real power or includes some amount of reactive vars
The question has no meaning since it is referring to a physically unrealizable DC complex impedance.

#### Papabravo

Joined Feb 24, 2006
16,810
You seem to be saying that the reason that you can put DC current through a real inductor is that has some small DC resistance in the coil, implying that if it didn't have some small DC resistance that you couldn't put DC current through it. That makes no sense and you can most definitely put a DC current through a superconducting coil, which is both a real and inductor and one that does not have some small DC resistance in it.
No. Only that real inductors with a small DC resistance will dissipate real power.

#### AnalogKid

Joined Aug 1, 2013
9,355
Any info on my original question?
I am trying to deduce if given a complex load, Say 4+5j and then are told you apply a dc source to it, what power do you have?
Is it purely real power or includes some amount of reactive vars
After all transients settle down, it is all real power. The transient energy used to charge up any reactive components (either capacitors or inductors) can be approximated by integrating out to five time constants. Be sure to include the power source impedance in the equivalent R-L or R-C calculation.

Or, jump to the answer with the total energy calculation. For example, the total energy in a capacitor that is charged up to some DC voltage is

P = 1/2 x C x V^2 (one-half C V squared). The result is in watt-seconds. Once the capacitor is charged up, zero energy is needed to maintain the voltage across the capacitor. That is, zero energy for a theoretically perfect capacitor with zero leakage current.

ak

#### WBahn

Joined Mar 31, 2012
26,398
No. Only that real inductors with a small DC resistance will dissipate real power.
Then I would recommend rewording your Rule #2 so that it says that. Perhaps change:

Rule #2 You can put DC current through a real inductor because it has some small dc resistance in the coil. That resistance will dissipate real power.

to

Rule #2 You can put DC current through a real inductor; but if it has some small dc resistance in the coil, that resistance will dissipate real power.

#### MrChips

Joined Oct 2, 2009
24,207
Reactance of an inductor is ωL.
Reactance = 0 when ω = 0.

Reactance of a capacitor is 1/ωC.
When ω = 0, the capacitor becomes an infinite resistance.