How to simplify this capacitor circuit?

[Frigus]

Homework Statement

Find the equivalent capacitance.

Relevant Equations

Ceq(parallel)=1/C₁+1/C₂ +......1/Cn
Ceq(series)=C₁+C₂ +.....C₃

I tried to find logic behind how do we simplified the circuit as given for 2 hours and I unable to find any clue.
I can do it just by following rules but I am unable to get intuition.
I tried to make sense of it but how can we even make two points as one.

 

[etotheipi]

Components are in parallel if their terminals are connected to the same nodes. Notice that all three capacitors have both of their terminals connected to either node $2$ or node $3$. They are in parallel!

Also check your capacitor series/parallel equations, you have them the wrong way around . You should have$$C_{parallel} = \sum_i C_i$$ $$\frac{1}{C_{series}} = \sum_i \frac{1}{C_i}$$

 

[berkeman]

Homework Statement:: Find the equivalent capacitance.
Relevant Equations:: Ceq(parallel)=1/C₁+1/C₂ +...1/Cn
Ceq(series)=C₁+C₂ +...C₃

I tried to find logic behind how do we simplified the circuit as given for 2 hours and I unable to find any clue.
I can do it just by following rules but I am unable to get intuition.
I tried to make sense of it but how can we even make two points as one.

For me, it's easier to see if you orient the capacitors vertically instead of horizontally. Make the wire that goes from 2-4 the bottom wire, and the wire that goes from 1-3 the top wire. Can you see how each capacitor then goes between those two wires vertically? You could even add a little ground symbol to the bottom wire to make it look nicer...

 

[gneill]

Let's clean up your drawing just a bit:

Here's a useful "rule of thumb" for finding components that are in parallel. If you can draw a closed path through the two components using only the components themselves and the wire paths of the circuit, and NOT passing through any other components, then those two components are in parallel.

Consider $C_1$ and $C_2$ in the diagram above. We might draw such a closed path as follows:

Note that the loop in red transverses only the two components in question and the connecting wire paths. Thus $C_1$ and $C_2$ are in parallel.

Can you draw a similar loop for $C_2$ and $C_3$ (Or even $C_1$ and $C_3$)?

 

[LCSphysicist]

It is interesting try to find nodes in the circuit with the same potential, and try to rearrange it.

 

[Frigus]

Thanks to all of you for helping me,
I am studying from a book where they are solving capacitor Circuit using "Method of same potential" and I have 2 doubts about that method and that are
• How do we know that potential difference across C₄ or C₅ is different from potential difference across C₁ or C₂ or C₃,I can figure out how C₁,C₂ and C₃ have same potential.
•after finding all the capacitors that are in parallel how do we simplify the circuit further.

 

[gneill]

Since any contiguous wire network must be at the same potential throughout, you can identify "islands" of same potentials through the simple trick of coloring the contiguous paths with different colors. You could also simply place labels along the paths, but you may find the "coloring method" to have more visual impact, making results more obvious.

So for your circuit you might take out your highlighting pens and trace over all contiguous wire paths that you can find:

Looks like just two separate potentials suffice for the entire circuit.

 

[Frigus]

Since any contiguous wire network must be at the same potential throughout, you can identify "islands" of same potentials through the simple trick of coloring the contiguous paths with different colors. You could also simply place labels along the paths, but you may find the "coloring method" to have more visual impact, making results more obvious.

So for your circuit you might take out your highlighting pens and trace over all contiguous wire paths that you can find:

View attachment 266107

Looks like just two separate potentials suffice for the entire circuit.

Thanks sir,
It is way easier to find capacitors with same potential with this approach but still there is one doubt that is,
How do we solve circuit after finding capacitor which are in parallel.
Like in the case I have uploaded we can see that capacitor 1,2 and 3 are in parallel with each other and capacitor 4 and 5 are also in parallel with each other and capacitor 5 is single after that what do we have to do?
The book from which I have taken this circuit has simplified it but doesn't explained how do we have simplified.
Please help me to find that.
Thanks

 

[gneill]

Once you've identified parallel components it's time to re-draw your circuit to make the layout simpler to interpret. For example, you've found that capacitors 1,2 and 3 are in parallel. Redraw them to show this clearly:

Do the same for the others, connecting the groupings appropriately according to your color scheme.

 

[Frigus]

Once you've identified parallel components it's time to re-draw your circuit to make the layout simpler to interpret. For example, you've found that capacitors 1,2 and 3 are in parallel. Redraw them to show this clearly:

View attachment 266168

Do the same for the others, connecting the groupings appropriately according to your color scheme.

I tried to solve the circuit please check it if it is correct or not.

 

[gneill]

Okay, you're making progress. Your simplification of the drawing is good, but let's just touch it up a tad:

Take a look at the $C_4$ $C_5$ subcircuit. What effect do you think this subcircuit will have on the overall behavior of the circuit? Hint: how many wire paths lead to this subcircuit? Can current flow into and out of it?

 

[Frigus]

Okay, you're making progress. Your simplification of the drawing is good, but le't just touch it up a tad:

View attachment 266177

Take a look at the C4 C5 subcircuit. What effect do you think this subcircuit will have on the overall behavior of the circuit? Hint: how many wire paths lead to this subcircuit? Can current flow into and out of it?

I think after some time of connecting it current will stop flowing into it because after some time electrostatic equilibrium will be established.

 

[gneill]

I think after some time of connecting it current will stop flowing into it because after some time electrostatic equilibrium will be established.

Without a complete circuit, no current can flow. We're not looking at electrostatic charges in basic circuit theory. Also, if you look at your original diagram there's a shunt (short circuit across the $C_4$ $C_5$ pair:

 

[Frigus]

Also, if you look at your original diagram there's a shunt (short circuit across the C4 C5 pair

I can't understand that if their is a short circuit across C₄₅ then why not across C₁₂ as current could have move through green wire which is above the capacitor C₁₂.

 

[gneill]

I can't understand that if their is a short circuit across C₄₅ then why not across C₁₂ as current could have move through green wire which is above the capacitor C₁₂.

Note that capacitors $C_{12}$ are not in series as are $C_{45}$. The node between $C_1$ and $C_2$ is shared with another connection.

 

[Frigus]

Note that capacitors $C_{12}$ are not in series as are $C_{45}$. The node between $C_1$ and $C_2$ is shared with another connection.

I tried to apply ohm's law on C₄₅ and after completing the loop potential difference came out to be 0 so V term in ohm's law became 0 and due to which I understood that current is 0 But their is some energy stored in capacitor which is the purpose of connecting it to circuit then why we neglect it.
Honestly speaking while I was simplifying capacitors I just connected them just to match the colours across them but I didn't understood why I have done that and also I can't understand how do we knew that which capacitors are in series in circuit like this,if it is simple circuit then we can clearly see that equal current will pass through it but in this it seems chaos.
Thanks .

 

[Merlin3189]

I like Gneill's ideas.
It's like working out the cct for a PCB. The components are all over the place, but all that matters is which tracks they connect to. So I just take a track (here a line)and colour it and all the tracks it connects to. However wiggly on the PCB, all that is in one colour is one line in the cct. Half the components are usually connected to one or other or both of two lines - the black and red power lines. Then you've usually just got a few short tracks for the other colors.
In this problem, as soon as Gneill coloured the two tracks in the first cct, it's obvious. Then in the larger cct, he uses the left to right convention, so it is equally obvious that C4 and C5 go nowhere. A bit of experience with real cct diagrams will soon build your intuition. (Though, you'd be better off with old hand drawn ones. Computers generate some weird diagrams!)

Hemant. When simplifying a diagram like this, don't worry what the components are (caps, resistors, inductors.)
They all have two connections and current that goes in one end comes out the other. Thinking about storage will not help here. It's just finding the path through. C4 & C5 would do nothing, whatever sort of component they were, because there's no current that goes through them.

 

[gneill]

I tried to apply ohm's law on C₄₅ and after completing the loop potential difference came out to be 0 so V term in ohm's law became 0 and due to which I understood that current is 0 But their is some energy stored in capacitor which is the purpose of connecting it to circuit then why we neglect it.

There will be no energy stored in $C_4$ or $C_5$ since no current will have moved in it. Those capacitors are bypassed by the short circuit which will carry any current past them.

Honestly speaking while I was simplifying capacitors I just connected them just to match the colours across them but I didn't understood why I have done that and also I can't understand how do we knew that which capacitors are in series in circuit like this,if it is simple circuit then we can clearly see that equal current will pass through it but in this it seems chaos.
Thanks .

Same colors mean same potential. All points with the same potential can be considered to be the same node, so by connecting the capacitors according to the color scheme you've maintained the circuit topology (how things are connected) despite rearranging the layout visually.

The rules for parallel and series connections for components are fairly simple. If you go by the color scheme, components that share the same two colors on their leads are in parallel. If one color connects just two components then they are in series (In our example the yellow color is shared only by $C_4$ and $C_5$).

 

[Frigus]

There will be no energy stored in C4 or C5 since no current will have moved in it. Those capacitors are bypassed by the short circuit which will carry any current past them.

C4 & C5 would do nothing, whatever sort of component they were, because there's no current that goes through them.

If I try to apply ohm's law then there is some current as we can see that V term is not 0 as P wire has some potential and Q wire is at 0 potential and their is always some resistance so we get some some current.

I can't figure out what is wrong in this.

 

[etotheipi]

This is how I would think of redrawing the circuit:

I have coloured the path of the current through the arrangement in that lower redrawing in orange. You can also deduce that the potential difference across $C_4$ and $C_5$ is zero, since they are connected in parallel with a loop of zero resistance.

Hopefully that is now sufficient to do whatever analysis you were after!

 

[Frigus]

This is how I would think of redrawing the circuit:

View attachment 266479

I have coloured the path of the current through the arrangement in that lower redrawing in orange. You can also deduce that the potential difference across $C_4$ and $C_5$ is zero, since they are connected in parallel with a loop of zero resistance.

Hopefully that is now sufficient to do whatever analysis you were after!

Thanks,
New abilities unlocked.

 

[Frigus]

For me, it's easier to see if you orient the capacitors vertically instead of horizontally. Make the wire that goes from 2-4 the bottom wire, and the wire that goes from 1-3 the top wire. Can you see how each capacitor then goes between those two wires vertically? You could even add a little ground symbol to the bottom wire to make it look nicer...

Today I clearly understood what you was saying.