# 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.
gneill said:
R1/R4 = R2/R5
Why R1/R4 = R2/R5 and not R1/R2=R4/R5 ?

gracy said:
Why R1/R4 = R2/R5 and not R1/R2=R4/R5 ?
Isn't it one and the same expression?

gneill said:
you could easily determine the potentials at those two middle nodes with a voltage divider formula used on each pair.
You mean at A and B?

cnh1995 said:
Isn't it one and the same expression?
I think these two are same in the sense that equality will still hold,but let's say
R1/R4 = R2/R5=x
Then R1/R2=R4/R5=y
Am I correct?

gracy said:
I think these two are same in the sense that equality will still hold,but let's say
R1/R4 = R2/R5=x
Then R1/R2=R4/R5=y
Am I correct?
Yes.

So what in this case?Can I apply either of the two equations?

gracy said:
So what in this case?Can I apply either of the two equations?
Yes but the equations are true only for a "balanced" bridge. For this circuit, you'll need KCL and KVL.

gneill said:
the potential that a potential divider produces at its center node depends upon ratios.
gracy said:
So what in this case?Can I apply either of the two equations?
No,I don't think we can apply either of them.We have to use the one @gneill mentioned.But I am still confused ,how I will the decide the ratios in other questions of this type.I think we should look for resistors/capacitors in series considering there is no resistance /and capacitors in between.As in here we assumed R3 is not there.

gracy said:
No,I don't think we can apply either of them.
That's because the "ratios" aren't equal. If they were equal, Va would be equal to Vb i.e. Vab would be 0.

cnh1995 said:
Vab would be 0.
And bridge would be balanced?

gracy said:
And bridge would be balanced?
Yes. That's why the middle resistor is equivalent to an open switch in the balanced condition. It doesn't matter if it is there or not.

gracy said:
And bridge would be balanced?
cnh1995 said:
Yes.
And that's what we want.But with this formula R1/R2=R4/R5
It is not possible
that's why we can't use it,right?

gracy said:
And that's what we want.But with this formula R1/R2=R4/R5
It is not possible
that's why we can't use it,right?
Actually, with that formula, we 'check' if the bridge is balanced or not. If it is, the calculations become simpler. If it isn't, we have to use the methods gneill mentioned earlier.

cnh1995 said:
with that formula
This one
R1/R4 = R2/R5 ?

gracy said:
This one
R1/R4 = R2/R5 ?
Right.

gneill said:
match the components from your circuit to their equivalent locations on this new bridge diagram.
How to do that?How I will know this capacitor should come in between i.e which capacitor would form C3 ?Which of them would be C1,C2,C4?And moreover there is no voltage in my actual problem .

gracy said:
How to do that?How I will know this capacitor should come in between i.e which capacitor would form C3 ?Which of them would be C1,C2,C4?And moreover there is no voltage in my actual problem .
You'll need some practice for "spotting" the bridge in the circuit. Same goes for star-delta conversions in more complex circuits.
gracy said:
moreover there is no voltage in my actual problem .
Always assume a voltage source between the points between which the equivalent resistance or capacitance is asked. Equivalent resistance or capacitance between two points is actually the equivalent resistance or capacitance "seen" by a voltage source connected between those points.

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gracy said:
How to do that?How I will know this capacitor should come in between i.e which capacitor would form C3 ?Which of them would be C1,C2,C4?And moreover there is no voltage in my actual problem .
In the given circuit,after assuming the voltage source,if you trace the path of current through the capacitor branches and observe the 'splitting' and 'reunion' of the currents carefully, you'll realize its a bridge.

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cnh1995 said:
Always assume a voltage source between the points between which the equivalent resistance or capacitance is asked. Equivalent resistance or capacitance between two points is actually the equivalent resistance or capacitance "seen" by a voltage source connected between those points.
cnh1995 said:
After assuming the voltage source,if you trace the path of current through the capacitor branches and observe the 'splitting' and 'reunion' of the currents carefully, you'll realize its a bridge.
Applying these I drew a diagram (Sorry it is not neat)

As my bridge is not horizontal it is vertical hence formula to check whether it is balanced or not would be
##\frac{C1}{C2}##=##\frac{C3}{C4}## and not C1/C4 = C2/C5

Putting values

##\frac{2}{4}##=##\frac{1}{2}##

##\frac{3}{6}##=##\frac{1}{2}##

Hence the bridge is balanced.
Right?

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gracy said:
Applying these I drew a diagram (Sorry it is not neat)
View attachment 92753
As my bridge is not horizontal it is vertical hence formula to check whether it is balanced or not would be
##\frac{C1}{C2}##=##\frac{C3}{C4}## and not C1/C4 = C2/C5
You can label your components in any order you wish. If you are going to refer to them in formulas you should show their labels on your circuit diagram.
Putting values

##\frac{2}{4}##=##\frac{1}{2}##

##\frac{3}{6}##=##\frac{1}{2}##

Hence the bridge is balanced.
Right?
Right!

Which means you can remove the "bridge" component without affecting the circuit. Then look to see if any parallel or series reduction opportunities have appeared as a result.

Will there be any change in formulas in the following two cases
1)bridge is vertical
2)bridge is horizontal

(1) (2)

gracy said:
Will there be any change in formulas
No. You can rotate either of them to make them look alike.

But I am still confused ,how I will the decide the ratios in other questions of this type?

gracy said:
But I am still confused ,how I will the decide the ratios in other questions of this type?
You can remember it this way-ratio of resistors on one side of the bridge=ratio of resistors on the other side of the bridge. But it is important to spot the bridge first.

gracy
But then I get confused which one will be denominator and which one will be numerator.

gracy said:
But then I get confused which one will be denominator and which one will be numerator.
It doesn't matter which should be numerator or denominator. Just pick any ratio on one side, use the same on the other side.

Let say if I would have taken
C1/C2=C4/C3
After putting the values
I could have thought the bridge is not balanced.

gracy said:
But then I get confused which one will be denominator and which one will be numerator.
You want to see if the two voltage dividers in the bridge circuit will produce the same voltage division. So a simple procedure is to make ratios of the components that make up each voltage divider. Compare the ratios. You can't go wrong if you select corresponding pairs (use symmetry and common sense).

gneill said:
You can't go wrong if you select corresponding pairs (
You mean if I have selected C1 as numerator for one potential divider I shall select the one which is adjacent to it but at the other side of bridge as numerator of second potential divider,Right?

gracy said:
Let say if I would have taken
C1/C2=C4/C3
After putting the values
I could have thought the bridge is not balanced.
You can call the two resistors on one side as 'upper' and 'lower'. So, upper/loewer on one side=upper/lower on the other side.

gracy
cnh1995 said:
You can call the two resistors on one side as 'upper' and 'lower'. So, upper/loewer on one side=upper/lower on the other side.
Its just a memory trick..To understand the bridge balance conceptually, you should analyse it using voltage divider.

upper/lower on one side=upper/lower on the other side.And if bridge is of this sort

upper left/upper right on one side=lower left/lower right on the other side

gracy said:

upper/lower on one side=upper/lower on the other side.And if bridge is of this sort

upper left/upper right on one side=lower left/lower right on the other side
Right. Once you're fluent, you won't need this trick.

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