Equivalent Capacitance of Complex Circuit

AI Thread Summary
The discussion focuses on determining the equivalent capacitance of a complex circuit, emphasizing the importance of symmetry in the arrangement. Participants explore the application of Kirchhoff's Law, noting that for symmetric circuits, points at the same potential can be connected as a single node. The solution involves recognizing that the voltage across certain capacitors can be zero, simplifying the analysis. The correct equivalent capacitance is found to be 4/3 C, but there is uncertainty about the method used to arrive at this answer. For asymmetrical circuits, a more comprehensive approach involving Kirchhoff's voltage and current laws is recommended.
megaspazz
Messages
13
Reaction score
0

Homework Statement



Determine the equivalent capacitance of the combination
shown in Figure P26.75. Suggestion: Consider the symmetry
involved.
prob-26-75.png


Homework Equations



Parallel is: C1+C2+C3+...
Series is: (C1^-1 + C2^-1 + C3^-1 + ...)^-1
q = CV

The Attempt at a Solution



I have no idea how to do it for a complex circuit. I think it might have something to do with Kirchhoff's Law, but the book never mentioned it and neither did the professor, so I'm thinking it can do it without. And I am not sure what "the symmetry involved" is supposed to mean.
 
Physics news on Phys.org
Do you see any change if you turn the figure upside down?
Is there any reason that both "C" capacitors are different or the both 2C capacitors are?

If they are equivalent, so is the charge and voltage across them.What is the voltage across the 3C capacitor then?

ehild
 
ehild, thank you so much! Since the voltage across it is zero, it's like the 3C capacitor isn't there, right?

Solving the simple circuit then gives 4/3 C, which is the correct answer. Is my way of solving it correct? Or did I simply get the right answer the wrong way?

Many thanks!

Also, this isn't a problem I have to do, but what if the circuit were asymmetrical? Then what would you do?
 
Your solution is correct.

It is a very useful trick for symmetric circuits, that we can connect those points which are at the same potential with a single wire, so the points become a single node. Symmetrically equivalent points are at the same potential.

In general problems, you need to apply Kirchhoff's voltage Law and also the equivalent of Current Law, but with charges.

Every capacitor has its own charge, q at one plate and -q on the other plate.
The net charge at a node is zero. When the capacitors are connected in series, they all have the same charge.
And you know the relation between charge and voltage: q=CV.

Set up all equations and solve.

ehild
 
can you do Kirchoff's Law for this problem? How would you take into account the initial part before the loops and the last wire after the loops?
 
Connect a voltage source to the terminals with arbitrary emf E. Write up the equations in term of E, and solve for the whole charge Q on the connected plates.
Te circuit can be représented by an equivalent capacitor connected to the same voltage source and you can apply C(equivalent)=Q/Eehild
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top