Can superposition principle be applied on capacitor? - Page 1
Forum section: electronics / beginners

author: sevenofnine33
author_rank: Contributor
timestamp: February 09, 2019, 10:13:23 am
content: I saw someone solve a DC circuit by applying superposition to capacitor and voltage source. The circuit is given in the picture. When shoring voltage source he treated capacitor as a voltage source, and when 12V was input, capacitor was an open circuit. Capacitor has some initial voltage.My solution was to find current trough the capacitor, meaning that same current flows trough 20Ohm resistor. Then finding voltage for that resistor. Writing KVL to find voltage at middle node, and then dividing that voltage with resistance value to get the current trough 100Ohm resistor.Which one is correct, since we get different values?

author: soldar
author_rank: Super Contributor
timestamp: February 09, 2019, 10:49:28 am
content: I am not sure I understand your question. Witha DC voltage the steady state current through the capacitor is zero. From connection time the capacitor will charge exponentially to 12/5 volts. Is this what you are looking for?From the perspective of the capacitor that circuit behaves the same as the one I attach.

author: Benta
author_rank: Super Contributor
timestamp: February 09, 2019, 10:57:39 am
content: Generally, you can not use superposition when calculating a circuit with capacitors or inductors, you'll need to set up the differential equation. In special cases, superposition might be possible, but you'll need to define the limitations precisely.

author: sevenofnine33
author_rank: Contributor
timestamp: February 09, 2019, 10:59:05 am
content: This was his solution. I'm wandering if this is correct?

author: sevenofnine33
author_rank: Contributor
timestamp: February 09, 2019, 10:59:56 am
content: And this is the question:

author: sevenofnine33
author_rank: Contributor
timestamp: February 09, 2019, 11:00:43 am
content: I was thinking the same thing. That's why I used the other solution, the one I wrote in my question.

author: sevenofnine33
author_rank: Contributor
timestamp: February 09, 2019, 11:02:40 am
content: I don't think this is the steady state. The task is a little vague, it doesn't say if the capacitor is charging or discharging, but it's one of those cases, so there is some voltage across it and some current.

author: soldar
author_rank: Super Contributor
timestamp: February 09, 2019, 11:33:17 am
content: The capacitor voltage equation does not correspond to the circuit shown so... either you go by the equation given or by the circuit.Unless I am mistaken, for the circuit as shown the equation would be Vc = 2.4 (1 - e^(-t/RC) ) = 2.4 * (1 - e^(-5*t))The current going into the capacitor will be I = (2.4 - Vc) / 100

author: RoGeorge
author_rank: Super Contributor
timestamp: February 09, 2019, 11:44:45 am
content: The short answer isno, superposition can not be applied, because superposition can only be applied to linear systems.YET, if you manage to describe whatever you are after as a linear system (even thought that system might have some non-linear components), then you can safely apply the superposition principle.https://en.wikipedia.org/wiki/Superposition_principle

author: Doctorandus_P
author_rank: Super Contributor
timestamp: February 09, 2019, 12:16:40 pm
content: Simple superposition can be used here, you just have to turn it a bit around.Do not start with trying to find the current through the capacitor.Start by taking an snapshot in time (This circumvents all the differential equation stuff).On any instant the voltage over the capacitor is a constant.You can then replace the capacitor with a voltage source.Then you have a schematic with 2 voltage sources and 3 resistors, and you can use superposition on that to calculate the currents.From the current through the capacitor you can then calculate the rate the voltage is changing at that instant.For calculating voltages and currents after the whole circuit has been stabilized you simply replace the capacitor with an open connection, because there will be no current through the capacitor it will not influence the circuit during a steady state.(Inductors are replaced with a short during steady state calculations. Because the current through the inductor does not change, there is no voltage over the inductor, thus it is a short).

author: orin
author_rank: Frequent Contributor
timestamp: February 10, 2019, 03:09:58 am
content: It's a sneaky problem. There is an initial condition of charge on the capacitor, i.e. at t = 0, v(t) isn't 0.At t = 0, from the v(t) given, there is 8V on the capacitor. How is that going to be taken into account?I'd just replace the 100, 400 resistors and 12V voltage source with their Thevenin equivalent (2.4V and 80 ohms). Now you have a simple voltage divider with the 20 ohm resistor and the 80 ohm Thevenin equivalent. There is v(t) - 2.4 across the divider, so you have 0.8(v(t) - 2.4) + 2.4 across the 100 ohm resistor. Substitute in v(t) and divide by 100 to get the current through the 100 resistor.

author: soldar
author_rank: Super Contributor
timestamp: February 10, 2019, 10:11:28 am
content: Quote from: orin on February 10, 2019, 03:09:58 amYeah, sneaky in the sense that it is badly formulated. And, yeah, Thevenin is the way to go in any case.

author: emece67
author_rank: Frequent Contributor
timestamp: February 10, 2019, 05:37:18 pm
content: .