# Equivalent capacitance in a circuit

• Jahnavi
In summary, the homework statement says that there is a mistake in the circuit and that you need to find the equivalent capacitance.
Jahnavi

## The Attempt at a Solution

I think the circuit given above is incorrect due to the presence of the two resistors . I have never seen/done a problem where equivalent capacitance is to be calculated across two points in a circuit , having resistors along with capacitors .

Is it that there is a prining mistake such that the top two resistors are actually capacitors ? If that is the case then net capacitance can be calculated .

I might be wrong .

#### Attachments

• Eq capacitance~2.jpg
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I don't think the circuit is wrong.
You need to find the equivalent capacitance seen by a voltage source connected between points A and B.
One way to do it would be to assume some voltage VAB to find the total energy stored in the capacitors and use E=1/2CeqV2.

cnh1995 said:
One way to do it would be to assume some voltage VAB to find the total energy stored in the capacitors and use E=1/2CeqV2.

Are you sure ?

Jahnavi said:
Are you sure ?
Yes, I don't see why that's not a right approach.

cnh1995 said:
Yes, I don't see why that's not a right approach.

The energy stored depends on the potential difference applied . How would we find Ceq ?

Jahnavi said:
The energy stored depends on the potential difference applied . How would we find Ceq ?
As I said in #2, you can assume some value for the applied voltage, say V. Then find the total energy stored, say E.
Equivalent capacitance will be Ceq=2E/V2.
This is the general method.
In this particular problem, you can see a Wheatstone bridge, so you can solve it just by inspection.

Jahnavi
cnh1995 said:
In this particular problem, you can see a Wheatstone bridge, so you can solve it just by inspection.

I wonder how Wheatstone bridge is present when there is a combination of resistors and capacitors .

I did nodal analysis and concluded that potential difference across the middle capacitor is 0 .

The middle capacitor can be removed .This puts the bottom two capacitors in series . Equivalent capacitance will be 2μF i.e option 4) .

I hope you are also getting option 4) .

This is same if we treat this circuit as Wheatstone bridge

How did you identify that this is a Wheatstone bridge ?

Jahnavi said:
How did you identify that this is a Wheatstone bridge ?
By inspection.
You can replace capacitance C with capacitive reactance (-j/ωC) so that the bridge can be identified more easily.

If this were not a balanced bridge, you would indeed need nodal analysis and KVL to solve for capacitor voltages and total stored energy.

cnh1995 said:
You can replace capacitance C with capacitive reactance (-j/ωC)

What is ω in this case ?

Jahnavi said:
What is ω in this case ?
Any arbitrary value. It is just to see more clearly if the circuit is a balanced bridge or not.
In this case, you can see the capacitances in the lower branch and resistances in the upper branch will have the exact same voltage division making it a balanced bridge.

Jahnavi
I have a question about this one: In general, between points in a circuit such as this, a voltage ## V_{AB}(\omega) ## can be applied, and an impedance can be computed, ## Z(\omega) ##. For a simple ## RC ## circuit that is a resistor and capacitor in parallel, you get quite a different answer than a series ## RC ## configuration. Are they assuming in this problem that this is effectively a parallel combination of ## R ## and ## C ##, and asking what ## C ## is in this case? That would seem to be the case. (e.g. in the DC case, a DC current results in the steady state in this problem). The question is also though, is it possible to consider this circuit as being effectively a resistor and capacitor in parallel? i.e. does it behave identically to that particular case between the points ## A ## and ## B ##, if the only probe is between points ## A ## and ## B ## ?## \\ ## My background in circuit analysis is somewhat limited, so perhaps @cnh1995 knows how the EE's typically analyze this. ## \\ ## Edit: Additional item: In any real circuit, even if it is assumed to be a parallel combination of ## R ## and ## C ##, there necessarily must be a finite resistance ## R_C ## in the capacitor branch of the circuit. How do they take this into account? ## \\ ## Edit: See post 13 below.

Last edited:
@Jahnavi and @cnh1995 I solved this one, but there is really a trick to it. In order to solve it, I wrote out the conservation of current/charge at the lower middle point between the 3 capacitors. I assumed for simplicity ## V_A=6 ## , ## V_B=0 ##, making the voltage ## V_C=4 ## between the two resistors. Calling the voltage ## V_D ## at the lower middle, I solved the system of equations of ## Q_1 ## ## Q_2 ##, ## Q_3 ## and ## V_D ## as unknowns. I called the charge on the middle capacitor ## Q_1 ## and the left one ## Q_2 ## and the right one ## Q_3 ##. To simplify further I left the ## \mu=10^{-6} ## off of the value for ## C ##, and called the unknowns ## x,y,z, w ##. Anyway, upon solving, ## x=Q_1=0 ##, and the rest of it could be found by substituting into the original equations. The result was then ## V_D=4 ## and ## Q_2=Q_3=12 ##. ## \\ ## By the values given to the two lower capacitors, comparing to the two upper resistors, it takes the middle capacitor out of the loop. (I didn't spot this arrangement that could be seen by inspection, so taking the long route, I found the charge on the center capacitor is zero). ## \\ ## The final result is there are two capacitors in series on the lower branch which, with the formula for two capacitors in series, results in the answer. ## \\ ## @cnh1995 I see in your post 11 that you recognized it was a balanced bridge. :)

Jahnavi and cnh1995
My background in circuit analysis is somewhat limited, so perhaps @cnh1995 knows how the EE's typically analyze this. \\
I don't remember having come across such problems, so I can't answer your question satisfactorily.
However, I vaguely remember a similar problem in my book that was solved using Laplace transform, where you replace capacitance C by its Laplace equivalent 1/sC. I will get back once I find that problem in my book.

cnh1995 said:
I don't remember having come across such problems, so I can't answer your question satisfactorily.
However, I vaguely remember a similar problem in my book that was solved using Laplace transform, where you replace capacitance C by its Laplace equivalent 1/sC. I will get back once I find that problem in my book.
@cnh1995 The questions I had about this in post 12 I basically answered in post 13. I didn't realize what you had meant by your statement that the bridge was balanced (posts 6 and 9) until I solved the system of equations and I was surprised to find ## Q_1=0 ##. Once I saw that, I understood your previous posts. If the bridge were not balanced, coming up with a result for the equivalent capacitance might have been much more lengthy: e.g. working through the differential equations to get the time constant ## \tau=RC ## of the system, when the voltage between points ## A ## and ## B ## is removed. ## \\ ## I give you "+1" for spotting that the bridge was balanced without solving the system of equations. :)

cnh1995
cnh1995

## 1. What is equivalent capacitance in a circuit?

Equivalent capacitance in a circuit is the combined capacitance of two or more capacitors in a circuit. It represents the total amount of charge that can be stored in the circuit and is measured in farads (F).

## 2. How is equivalent capacitance calculated in a series circuit?

In a series circuit, the equivalent capacitance is calculated by adding the reciprocal of each individual capacitor's capacitance and then taking the reciprocal of the sum. This can be represented by the formula: 1/Ceq = 1/C1 + 1/C2 + ... + 1/Cn, where Ceq is the equivalent capacitance and C1-Cn are the individual capacitances.

## 3. What about in a parallel circuit?

In a parallel circuit, the equivalent capacitance is calculated by adding the individual capacitances. This can be represented by the formula: Ceq = C1 + C2 + ... + Cn, where Ceq is the equivalent capacitance and C1-Cn are the individual capacitances.

## 4. How does equivalent capacitance affect the overall capacitance of a circuit?

The equivalent capacitance in a circuit determines the total amount of charge that can be stored. Generally, the larger the equivalent capacitance, the larger the overall capacitance of the circuit. This means that the circuit can hold more charge and have a longer discharge time.

## 5. Can the equivalent capacitance of a circuit ever be less than the individual capacitances?

No, the equivalent capacitance of a circuit will always be greater than or equal to the individual capacitances. This is because adding more capacitors in a circuit will always increase the total amount of charge that can be stored, resulting in a larger equivalent capacitance.

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