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Before posing the question I would like to write the assumption I make for the power concept in passive circuits. As far as I know the power dissipated in an RLC circuit is the active power which is actually only dissipated through the resistor R. Power through L and C components are associated with reactive power which means they in average do not dissipate power, they absorb and give back the energy to the circuit continuously.

If the above is true, then I assume a simulation should indicate the average power through an inductor or a capacitor as zero.

Here is a simulation for inductor case where the input is a periodic pulse (plot shows the power through the inductor):

enter image description here

It seems the integral of the above plot with time is zero.

But below in capacitor case I don't get the expected plot:

enter image description here

In this case the integral of the above power plot with time is non-zero. And it is increasing.

Why in the capacitor case the average power is non-zero unlike in inductor case in this LTspice simulation?

EDIT: Below is the input signal asked as request:

enter image description here

Here are the settings:

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Edit 2: Here in case of sine input:

enter image description here

Edit 3:

Capacitor power from beginning to steady state (source is pulse again):

enter image description here

And a zoom section from the steady-state:

enter image description here

And here below the capacitor power in mW(blue line) and its integral(green line) for the first 500 sec:

enter image description here

As you see the integral goes to a constant which indicates the capacitor does not suck energy anymore from the circuit after certain amount of seconds.

Is the energy at the beginning called as dissipated or stored? In other words is the power at the beginning active or reactive?

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    \$\begingroup\$ What is your input signal? Does this have a different result using an input waveform with no DC content (sine wave, \$\pm1\$V square wave)? \$\endgroup\$ – user2943160 Jun 7 '16 at 23:45
  • \$\begingroup\$ No the input signal is pulse PULSE(0 1 0 0.00001 0.00001 0.001 0.002) i.e. 500Hz with %50 duty cycle 1V in ON time, 0V in OFF time. \$\endgroup\$ – user16307 Jun 7 '16 at 23:48
  • \$\begingroup\$ How are you calculating 'power' \$\endgroup\$ – Chu Jun 7 '16 at 23:48
  • \$\begingroup\$ @Chu When you hover the mouse pointer on the component in LTspice with AltGr it plots I*V (current X voltage accross the component). I plot that way. \$\endgroup\$ – user16307 Jun 7 '16 at 23:50
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    \$\begingroup\$ The pulse train is always positive so it's charging the capacitor, ie the capacitor is storing energy. Energy is the integral of power. \$\endgroup\$ – Chu Jun 8 '16 at 0:27
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What I think you're seeing is the start of an exponential where the capacitor is charging. The time constant is 22s and the pulse duration is very small compared with this, so you're only seeing the early stages of the exponential.

In charging a capacitor there is a flow of power in, but the resultant energy is stored and not dissipated.

I don't know why there is no discharge curve; it may be worth decreasing the pulse frequency significantly so that we can see what's happening in the time frame of the time constant. The integration sampling time may also be a factor.

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  • \$\begingroup\$ I think you are absolutely right. I edited my question. Please see Edit 3. So after the circuit current reaches steady state, the capacitor's average power becomes always zero. But at the beginning the things are not periodic and not steady. During transition it charges exponentially and absorbs energy from the circuit. That's what we see as non zero power. Actually this power in the form of stored charges is not dissipated as heat like in the resistor. It is still in the part of the circuit's total energy. Should we consider that power part as dissipated active power or reactive power? \$\endgroup\$ – user16307 Jun 8 '16 at 8:52
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Generator output may be splitted onto two components: DC 0.5 V and square wave with no DC. So, we see superposition of two processes: slow capacitor charging (to 0.5 V DC voltage) and fast oscillations. Hence, positive mean power (active power) corresponds to that charging. If at some time generator will stop at 0 V, we will see negative power consumption corresponding to capacitor discharge.

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