Equivalent capacitance homework problem

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The discussion revolves around finding the equivalent capacitance between points A and B in a complex circuit involving capacitors. Participants clarify that the circuit is a bridge configuration, which complicates the application of simple series and parallel capacitance formulas. It is emphasized that proper identification of nodes is crucial, as misidentifying them can lead to incorrect conclusions about capacitor arrangements. The conversation also highlights the need for more advanced circuit analysis methods, such as Kirchhoff's laws, when straightforward reduction is not possible. Ultimately, understanding the bridge circuit's symmetry is key to solving the problem effectively.
  • #31
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!
 
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  • #32
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 ?
 
  • #33
gracy said:
Now I don't know what is passive linear circuit ?
php?image=http%3A%2F%2Fwww.electronicsteacher.com%2Fdirect-current%2Fdc-network-analysis%2F00486.png

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.
 
  • #34
gneill said:
voltage divider formula
Vout=Vin [ ##\frac{C_1}{C_1+C_2}##]
Is this formula correct?
Of course in case of capacitors.
 
  • #35
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?
 
  • #36
gneill said:
R1/R4 = R2/R5
Why R1/R4 = R2/R5 and not R1/R2=R4/R5 ?
 
  • #37
gracy said:
Why R1/R4 = R2/R5 and not R1/R2=R4/R5 ?
Isn't it one and the same expression?
 
  • #38
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?

ab.png
 
  • #39
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?
 
  • #40
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.
 
  • #41
So what in this case?Can I apply either of the two equations?
 
  • #42
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.
 
  • #43
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.
 
  • #44
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.
 
  • #45
cnh1995 said:
Vab would be 0.
And bridge would be balanced?
 
  • #46
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.
 
  • #47
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?
 
  • #48
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.
 
  • #49
cnh1995 said:
with that formula
This one
R1/R4 = R2/R5 ?
 
  • #50
gracy said:
This one
R1/R4 = R2/R5 ?
Right.
 
  • #51
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 .
 
  • #52
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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  • #53
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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  • #54
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)
BRIDGE.png

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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  • #55
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.
 
  • #56
Will there be any change in formulas in the following two cases
1)bridge is vertical
2)bridge is horizontal
gsed_0001_0017_0_img4261.png
bridge+resistor.gif



(1) (2)
 
  • #57
gracy said:
Will there be any change in formulas
No. You can rotate either of them to make them look alike.
 
  • #58
But I am still confused ,how I will the decide the ratios in other questions of this type?
 
  • #59
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.
 
  • Like
Likes gracy
  • #60
But then I get confused which one will be denominator and which one will be numerator.
 

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