# Multiple sources with Capacitors only

#### Human2

Joined Dec 22, 2017
51
Alright so I am studying this now and started this assigment but I dont understand it completley:

Alright so as far as I got it, I should chose a junction (in this case I chose A) and for that junction, only first closest capacitor in each branch that goes to that junction, his charge will be added into sumation with a sign thats turned to that junction,
and then I should write Voltage equations like this:
1. Choose a direction
2. Going through a loop, the sign of a voltage is the sign out of which the current from direction leaves.

So:

Q1 + Q3 = 0
E1 = Q2/C2 + Q1/C1
-E2 - E1 = - Q3/C3

And for U3 I got it correctly, its 36V.

for U2 and U1 values should be
U2 = 8 V
U1 = 4 V

but I got something else... So can someone please explain to me what mistakes am I making while writing equations either for charges
or voltages :/ ?

#### WBahn

Joined Mar 31, 2012
25,294
I don't follow what you mean by, "only first closest capacitor in each branch that goes to that junction, his charge will be added into sumation with a sign thats turned to that junction,"

In general, pick one node in the circuit and declare that node to be "ground" ("common reference node" or just "common" is a better name, but "ground" is what you will hear mostly). The voltage on that node is, by definition, 0 V.

Label all of the nodes in the circuit that join more than two components. Also label nodes that connect just two components if the voltages at those nodes are of interest to the problem (which is the case here).

While you CAN pick a random node as the common and everything will work out fine, a good choice can make things a lot easier and less error prone. Typically, nodes that are connected directly to a voltage source terminal (preferably the negative terminal) make decent choices.

In this problem, you have four nodes and can establish the voltage on three of them immediately. So you really only have to worry about the remaining node.

Then, for each component, assign the voltage across and current through that component a symbolic value (i.e., a variable name). Again, you CAN pick all of these randomly, you just have to then be consistent with your choices. If the final value turns out to be negative, then the actual voltage or current simply has the opposite polarity. But a couple of common conventions make things a lot easier and less error prone. First, follow the passive sign convention, which says that the direction of the current is chosen so that a positive current enters the positive terminal of the device. Second, assign the direction of the currents through all devices in the same branch to be the same.

So do that much, annotate your drawing, and then take another run at it.

#### Human2

Joined Dec 22, 2017
51
Okay wait 4 nodes ? I don't understand.

What do you mean by node ? By junction I mean a place where 3 or more branches intersect (like A).
By branch I mean a place between 2 junctions (in this shematic there are 3: C1,C2, - E1, - E2,C3

What is a node ? If by node you mean junction, how come I can't see four of them, I see only 2

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#### WBahn

Joined Mar 31, 2012
25,294
Think of a node as a place in the circuit where you can place a voltmeter probe.

A branch is a segment of the circuit consisting of components in series, whether it be one or many.

What you are calling a junction is also known as an "essential node". Nodes connecting only two components are considered non-essential because including them in the KVL/KCL equations describing the circuit accomplishes nothing.

#### shteii01

Joined Feb 19, 2010
4,647
Yeah. What is your common/ground/0 volts node?

#### Human2

Joined Dec 22, 2017
51
I was never thought to ground an essential node in shematics with capacitors, but I know about that from that voltage solving method where you ground 1 node and write eq. for other nodes.

How I was thought to solve this problem with capacitors is to pick a node, and with respect to that node write equation for charges.

And doing so in my notes I only see capacitors that are first to the node are only being put into equation, it his example:

Q1 + Q3 = 0

Charge from C1 is being taken into account here in this eq. but C2 is not, why is that?

Anyways I'd need to take first capacitor and its charge is a sign of which polarity is turned to the node A, in this case for C1 + is turned to A so its not -Q1 its just Q1.

So on the left side I have charges and on the right side I have 0, but if I have an assigment that has (don't know how to translate it to english) "start charge" then I write those charges to the right

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#### Human2

Joined Dec 22, 2017
51
I don't follow what you mean by, "only first closest capacitor in each branch that goes to that junction, his charge will be added into sumation with a sign thats turned to that junction,"

In general, pick one node in the circuit and declare that node to be "ground" ("common reference node" or just "common" is a better name, but "ground" is what you will hear mostly). The voltage on that node is, by definition, 0 V.

Label all of the nodes in the circuit that join more than two components. Also label nodes that connect just two components if the voltages at those nodes are of interest to the problem (which is the case here).

While you CAN pick a random node as the common and everything will work out fine, a good choice can make things a lot easier and less error prone. Typically, nodes that are connected directly to a voltage source terminal (preferably the negative terminal) make decent choices.

In this problem, you have four nodes and can establish the voltage on three of them immediately. So you really only have to worry about the remaining node.

Then, for each component, assign the voltage across and current through that component a symbolic value (i.e., a variable name). Again, you CAN pick all of these randomly, you just have to then be consistent with your choices. If the final value turns out to be negative, then the actual voltage or current simply has the opposite polarity. But a couple of common conventions make things a lot easier and less error prone. First, follow the passive sign convention, which says that the direction of the current is chosen so that a positive current enters the positive terminal of the device. Second, assign the direction of the currents through all devices in the same branch to be the same.

So do that much, annotate your drawing, and then take another run at it.
Okay so here are four nodes A, B, C and D:

I will never have currents in this assigment with capacitors, our professor and my book with assigments only have: charge, "start charge", voltage and capaciitance.

So I bet that my voltage equations are correct:
E1 = Q1/C1 + Q2/C2
-E2 -E1 = - Q3/C3

But from your post I didn't quiet get it how to write correctly charge equations ?

#### shteii01

Joined Feb 19, 2010
4,647
Okay so here are four nodes A, B, C and D:

View attachment 144487

I will never have currents in this assigment with capacitors, our professor and my book with assigments only have: charge, "start charge", voltage and capaciitance.

So I bet that my voltage equations are correct:
E1 = Q1/C1 + Q2/C2
-E2 -E1 = - Q3/C3

But from your post I didn't quiet get it how to write correctly charge equations ?
You still need to designate REFERECE node. Note. I DID NOT say ground. I said reference.

#### Human2

Joined Dec 22, 2017
51
You mean choose any node as reference node and use it to write charge equations ?

for A node:

Q1 + Q3 = 0 ???

What do you mean by reference node

#### MrAl

Joined Jun 17, 2014
7,180
Alright so I am studying this now and started this assigment but I dont understand it completley:

View attachment 144453

Alright so as far as I got it, I should chose a junction (in this case I chose A) and for that junction, only first closest capacitor in each branch that goes to that junction, his charge will be added into sumation with a sign thats turned to that junction,
and then I should write Voltage equations like this:
1. Choose a direction
2. Going through a loop, the sign of a voltage is the sign out of which the current from direction leaves.

So:

Q1 + Q3 = 0
E1 = Q2/C2 + Q1/C1
-E2 - E1 = - Q3/C3

And for U3 I got it correctly, its 36V.

for U2 and U1 values should be
U2 = 8 V
U1 = 4 V

but I got something else... So can someone please explain to me what mistakes am I making while writing equations either for charges
or voltages :/ ?
Hi,

This can be done another way too, but this kind of circuit is also interesting because we often assume the steady state solution when there could be existing anomalies for the start up conditions. In this circuit, it is the infinite current that would flow when the source E1 is turned on. After steady state is reached it's no longer a concern though.

But the other solution is to simply analyze the circuit replacing the caps with Laplace impedances, then use the final value theorem to find the voltages. It is surprisingly easy to do that way.
If you like, you can also simulate a real circuit by placing a small resistance in series with each cap or just one cap when two are in series. The results of each voltage come out the same in the steady state solution when the circuit is driven by only DC sources like batteries.

If you'd like to see a simple example, once you've gotten all your results straightened out i'll provide one here. The calculation does not have to involve charge, but you can of course compute that too after you find the voltages.

#### Human2

Joined Dec 22, 2017
51
Thank you for your time but I need to use the method with charge equations and voltage equations as stated above, I might be interested in that way also but I have exam on monday so I need to dstudy a lot of things.

Anyway, do you know what am I doing wrong with charge equation here ? How should it look like ?

#### MrAl

Joined Jun 17, 2014
7,180
Thank you for your time but I need to use the method with charge equations and voltage equations as stated above, I might be interested in that way also but I have exam on monday so I need to dstudy a lot of things.

Anyway, do you know what am I doing wrong with charge equation here ? How should it look like ?
Hi,

I think you did ok really. What do you think you did wrong?

You are defining voltages by the default definition which means they are defined differentially. That means that all of your voltage results will be differential as well, and i think you did that.

However, when we do circuits like this it is usually considered a little more "complete" if we define a ground node from which all other voltages can be defined. That's an absolute method of defining voltages, and that is a very common way to do it. That means you quote a voltage at a node, unless otherwise noted. For example, maybe node 1 equals 2v. That is always referenced to ground. You can always go back to writing out the differential voltage also, such as "the voltage across C1 is 4 volts", which might be written vC1=4 just for one example.
So the ground node is not mandatory, but a lot of people will be looking for one. More importantly you have to do it the way the course instructor wants you to do it.

So as far as i can see everything looks ok, unless you want to declare one node as the ground (zero volts) node and then you can state node voltages as well as individual component voltages with the proper notation. In this circuit the ground node would most likely be the lower most node in the circuit as that makes picturing the node voltages easy, but it's always up for grabs as to what you want to use according to what is easier for you or what the course demands.

#### WBahn

Joined Mar 31, 2012
25,294
You mean choose any node as reference node and use it to write charge equations ?

for A node:

Q1 + Q3 = 0 ???

What do you mean by reference node
Node A has three branches, what sums to zero is the charge that goes to Q1, to Q3, and to E1.

#### WBahn

Joined Mar 31, 2012
25,294
Okay so here are four nodes A, B, C and D:

View attachment 144487

I will never have currents in this assigment with capacitors, our professor and my book with assigments only have: charge, "start charge", voltage and capaciitance.

So I bet that my voltage equations are correct:
E1 = Q1/C1 + Q2/C2
-E2 -E1 = - Q3/C3

But from your post I didn't quiet get it how to write correctly charge equations ?
Keep in mind that the movement of charge from one place to another IS a current. The only way for those capacitors to get charged is for a current to flow. It doesn't flow for very long, but it does flow.

You can sidestep this by assigning directions to the movement of charge instead (which, as already noted, IS a current no matter what we call it).

E1 = Q1/C1 + Q3/C3

E1 is the voltage at node A relative to node B, or Vab.

Q1/C1 is the voltage at node A relative to node C, or Vac.

Q3/C3 is the voltage at node D relative to node B, or Vbd.

So what you are claiming is that

Vab = Vac + Vdb

I see no basis for such a claim.

Consider the following:

What is the relationship between Q0, Q1 and Q3?

What is the voltage across the three capacitors in terms of the charges given above and their capacitances.

What does KVL require as you go around the left-hand loop? Around the right-hand loop?

EDIT: C2 should have been C3. But in doing the correction I realized that I misread the TS's post entire.

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#### Human2

Joined Dec 22, 2017
51
"Q2/C2 is the voltage at node D relative to node B, or Vbd." Umm.. Isn't Q2/C2 a voltage between node C and B ? I mean I don't get
it isn't the voltage at node D relative to node B Q3/C3 ?

Also Q0 = Q1 + Q3

now, Q1 = charges on C1, is (Q1 = Q of C1 + Q of C2) orrr
is charge in a branch equal for all capacitors in series ?

So then
E1 = Q1/C1 + Q1/C2 ?

EDIT: Yes thats right, I didnt even need charge equations in this assigment,

I got Q1 from first voltage eq. and Q3 from second voltage equation. therefor
I got voltages on capacitors:
U1 = 4V, U2 = 8V, U3 = 36V.

Thanks everyone for your time I appreciate it.

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#### MrAl

Joined Jun 17, 2014
7,180
"Q2/C2 is the voltage at node D relative to node B, or Vbd." Umm.. Isn't Q2/C2 a voltage between node C and B ? I mean I don't get
it isn't the voltage at node D relative to node B Q3/C3 ?

Also Q0 = Q1 + Q3

now, Q1 = charges on C1, is (Q1 = Q of C1 + Q of C2) orrr
is charge in a branch equal for all capacitors in series ?

So then
E1 = Q1/C1 + Q1/C2 ?

EDIT: Yes thats right, I didnt even need charge equations in this assigment,

I got Q1 from first voltage eq. and Q3 from second voltage equation. therefor
I got voltages on capacitors:
U1 = 4V, U2 = 8V, U3 = 36V.

Thanks everyone for your time I appreciate it.

Hi,

The way i took your post was to mean that each voltage was the voltage ACROSS some single element unless of course it was a sum.
The source of confusion seems to be that there might be a question of how you are really defining the voltages.
When we mention components alone, sometimes it can be a little unusual when we are used to seeing node voltages.

For example your:
E1 = Q1/C1 + Q2/C2

is perfectly fine because:
vC1=Q1/C1

and:
vC2=Q2/C2

In the above i used the notation ":vC1" to describe the voltage across C2.
It does seem intuitive that Q2 is the charge in C2 and not anywhere else.

#### WBahn

Joined Mar 31, 2012
25,294
"Q2/C2 is the voltage at node D relative to node B, or Vbd." Umm.. Isn't Q2/C2 a voltage between node C and B ? I mean I don't get
it isn't the voltage at node D relative to node B Q3/C3 ?

Also Q0 = Q1 + Q3

now, Q1 = charges on C1, is (Q1 = Q of C1 + Q of C2) orrr
is charge in a branch equal for all capacitors in series ?

So then
E1 = Q1/C1 + Q1/C2 ?

EDIT: Yes thats right, I didnt even need charge equations in this assigment,

I got Q1 from first voltage eq. and Q3 from second voltage equation. therefor
I got voltages on capacitors:
U1 = 4V, U2 = 8V, U3 = 36V.

Thanks everyone for your time I appreciate it.
I completely misinterpreted your post. I had it in my head that you were summing the charges at that node and didn't pay close enough attention.