# Equivalent capacitance homework problem

• gracy
In summary, the author attempted to solve a problem involving capacitors in series and parallel, but made a mistake in the diagram. Without parallel and series capacitors, the problem cannot be solved with the previous method.
gracy

## Homework Statement

Find the equivalent capacitance of the combination between A and B in the figure.

## Homework 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##...

## 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.

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

In this circuit, there are no capacitors in parallel.

NascentOxygen said:
In this circuit, there are no capacitors in parallel.
gracy said:
Three capacitors of capacitance 3μF ,5μF and 4μF are in parallel.

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

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

ehild said:
Use a different color for the lines connected to B
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).

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

gracy
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.

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.

gracy
gneill said:
a node does not continue through components.
But where I did this mistake?

Oh,yes I got it.Thanks.

gneill said:
You can if you do it properly
ehild said:
There are neither parallel, nor series capacitors in this circuit
If there is no parallel or series capacitor how can I solve it with the previous method?

Is it correct now?

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

gracy
gneill said:
You can if you do it properly.
gneill said:
You cannot.
gneill said:
When you run into such a situation you must resort to more powerful methods of circuit analysis
azizlwl said:
It is a bridge circuit with 5 microF as the bridge.
Bridge circuit is it the powerful method you are referring to?

gracy said:
Bridge circuit is it the powerful method you are referring to?
A bridge circuit is a particular circuit layout. The methods I refer to include Y-Delta and Delta-Y transformations, Kirchhoff's voltage and current laws (KVL and KCL), mesh analysis, and so on. I don't believe that you've covered any of these yet, and they are not strictly required for this particular problem thanks to the choice of component values.

gracy
gneill said:
The methods I refer to include Y-Delta and Delta-Y transformations, Kirchhoff's voltage and current laws (KVL and KCL), mesh analysis, and so on.
I want to learn all of these.Would you help me?

gracy said:
I want to learn all of these.Would you help me?
You can bring your questions to Physics Forums, yes. I can't guarantee that it will always be me that addresses your questions; there are lots of helpers.

gracy
In this particular problem which method shall I go for?Bridge circuit method?

gracy said:
In this particular problem which method shall I go for?Bridge circuit method?
"Bridge circuit" is not a method; It's a particular arrangement of components.

However, there are certain symmetries associated with that layout that you can take advantage of if the component values are conveniently chosen. Did you find out how a nicely drawn bridge circuit appears? Can you re-draw your circuit in that style?

gracy
gracy said:
gneill said:
You can if you do it properly.
gneill said:
You cannot.
Those statements do NOT contradict each other.

The first refers to determining if components form a parallel combination and determining if components form a series combination.

Applying those criteria to the configuration in this thread, tell you that none of the capacitors here form a parallel combination and none form a series combination. That's precisely what the second statement refers to.

gneill said:
Can you re-draw your circuit in that style?
Before doing that
Could you please give me some guidance what points we should keep in mind to draw Bridge circuit?

gracy said:
Before doing that
Could you please give me some guidance what points we should keep in mind to draw Bridge circuit?
Find an post a picture of a typical bridge circuit. It doesn't matter if its components are capacitors or resistors, it's the layout of components that is of interest.

Good. Now, can you see that the circuit layout in this problem matches the drawing above (except for the component values)?

If so, substitute in the component values from the problem at hand into the above drawing. Then we can work on seeing the symmetry.

A fine choice for an image. Now, note the layout. Imagine that the middle resistor R3 is removed for now (This resistor is called the "bridge" component, since it forms a bridge between the middles of the two parallel branches). On the left there are two resistors R1 and R4 that form a potential divider across the 24 V supply. On the right side is a similar pair, R2 and R5, which form another potential divider for the 24 V. So without R3 there you could easily determine the potentials at those two middle nodes with a voltage divider formula used on each pair.

Now, the potential that a potential divider produces at its center node depends upon ratios. If the components of two different potential dividers have the same ratio, both potential dividers will produce the same voltage division. In the above circuit, if R1/R4 = R2/R5 then the middle nodes of the bridge would be balanced (Ignore the actual values that were included on the picture, we're just looking at theory right now). This is the sort of symmetry that you look for in bridge circuits that can be helpful: Equal component ratios on either side of the bridge let you know that the unloaded (no R3) potential dividers would produce the same potential.

Now, if that were the case (equal potentials) then laying R3 back into the circuit would do nothing! It would have the same potential at both of its connections, so no current could flow through it. The circuit would be undisturbed in any way by reconnecting R3. So R3 might as well not be in the circuit (when the bridge is balanced).

Once you have digested that, see if you can follow jbriggs444's advice and match the components from your circuit to their equivalent locations on this new bridge diagram. For bonus points, determine if your bridge is balanced!

In order to understand this properly I searched for potential divider.And got the meaning as In electronics, a voltage divider (also known as a potential divider) is a passive linear circuit that produces an output voltage (Vout) that is a fraction of its input voltage (Vin).Now I don't know what is passive linear circuit ?I know what is linear element and passive element in circuits.Is the circuit consisting of these element called passive linear circuit ?

gracy said:
Now I don't know what is passive linear circuit ?

This is a passive linear circuit. It contains resistors as passive elements and battery as active element. Linear circuit has its output "proportional" to the input. For instance, resistor current is proportional to the voltage across it,but the V-I curve for a diode is not linear.

gneill said:
voltage divider formula
Vout=Vin [ ##\frac{C_1}{C_1+C_2}##]
Is this formula correct?
Of course in case of capacitors.

gneill said:
voltage divider formula
##\frac{V_1}{V_2}##=##\frac{R_1}{R_2}##
Is this formula correct?
Of course in case of resistance?

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