Equivalent capacitance problem

1. Dec 1, 2015

gracy

1. The problem statement, all variables and given/known data
Find the equivalent capacitance of the combination between A and B in the figure.

2. Relevant equations

For Equivalent Capacitance in series

$\frac{1}{C}$=$\frac{1}{C_1}$+$\frac{1}{C_2}$

For Equivalent Capacitance in parallel

$C$=$C_1$+$C_2$.....
3. The attempt at a solution
Three capacitors of capacitance 3μF ,5μF and 4μF are in parallel.Hence their equivalent capacitance is 12μF.
We can see the capacitor of capacitance 2 μF became isolated.The reduction process left it with only one lead connected to the circuit.Without both leads connected a component cannot carry current. Charge cannot move onto or off of a capacitor through one wire alone. The capacitor thus has no influence, no utility, as far as the circuit is concerned.

Here capacitors of capacitance 12μF and 6micro F are in series.Using the formula ofequivalent capacitance of capacitor in series I got the answer for equivalent capacitance between A and B as 4μF but the answer given in textbook is $\frac{10}{3}$μF.I want to know what went wrong.
Thanks.

2. Dec 1, 2015

azizlwl

It is a bridge circuit with 5 microF as the bridge.

3. Dec 1, 2015

Staff: Mentor

In this circuit, there are no capacitors in parallel.

4. Dec 1, 2015

gracy

But these three capacitors share two nodes (red and orrange)so they must be in parallel.

5. Dec 1, 2015

6. Dec 1, 2015

ehild

No, C and B do not belong to the same node. Use a different color for the lines connected to B.
There are neither parallel, nor series capacitors in this circuit. It is a bridge.

7. Dec 1, 2015

8. Dec 1, 2015

gracy

Then orange color wire would be touching whatever colored wire I 'll use for line connected to B.It is wrong because any wire segments that touch must be the same node (color).

9. Dec 1, 2015

ehild

The orange wire does not go though the 6 uF capacitor., and is not connected to B.

10. Dec 1, 2015

Staff: Mentor

A node is made up of wire, only wire, nothing but wire, highly conductive copper or silver wire. These wires can together make a join that has any shape, but the node is still made of nothing but wires connected together.

11. Dec 1, 2015

Staff: Mentor

Gracy, You have to sort out the identification of the nodes first, as pointed out by others. Then you will reach the conclusion as voiced by @NascentOxygen, that there are no opportunities for combining parallel capacitors. This is based on the topology of the circuit.

A slight re-arrangement of the diagram on the page will make this more obvious, but you'll need to make a try at it. Another person (@azizlwl) mentioned that the circuit is a "bridge". This is a well-known circuit configuration. Look up images of "bridge circuit" so you can get an idea of how such a circuit looks when the author isn't trying to confuse you for the sake of making an interesting puzzle.

As is often the case with these sorts of problems the author as tried to confuse you by drawing the circuit in an "inconvenient" way for recognition of helpful symmetries, but I'll let you know in advance that the author has also included a trick via the component values that will eventually make things simpler. But before we get there you need be clear about the nodes and the lack of parallel and serial simplification opportunities for this topology.

So see if you can re-draw the circuit in a way that shows the symmetry of a bridge circuit.

12. Dec 1, 2015

gracy

13. Dec 1, 2015

Staff: Mentor

14. Dec 1, 2015

gracy

But where I did this mistake?

15. Dec 1, 2015

gracy

Oh,yes I got it.Thanks.

16. Dec 1, 2015

gracy

If there is no parallel or series capacitor how can I solve it with the previous method?

17. Dec 1, 2015

gracy

Is it correct now?

18. Dec 1, 2015

Staff: Mentor

You cannot. That's the point of this particular problem. There are circuit configurations that cannot be reduced by the basic series and parallel reduction methods because there are no series or parallel opportunities to exploit.

When you run into such a situation you must resort to more powerful methods of circuit analysis, or, as is the case with this problem, look for the trick that the author incorporated by a clever choice of component values. Hint: it relies on symmetry, which is why the author disguised the circuit layout as he did so the symmetry would not be obvious. You should be able to re-draw the circuit to uncover the symmetry.

19. Dec 1, 2015

Staff: Mentor

20. Dec 1, 2015