Can the Voltage in an RC Circuit be Found by Examining a Picture?

In summary, you need to find the source voltage for a circuit by either knowing the value of a current or voltage somewhere in the circuit, or by solving an equation. Once you find the source voltage, you can use V=IR to find the resistance and voltage across each load.
  • #1
lepp1892
21
0
I have a circuit I need to find a voltage on. Is there a way to figure it out on the provided picture in here?

Thanks
 

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  • #2
Maths can usually give the answer.

• The voltage between which two points?

• Are you wanting to know the amplitude of a sinewave, omitting the transient component?

• Is this a homework exercise?
 
  • #3
1) I want to find out the source voltage.

2) I don't know, maybe...

3) No, I am trying to figure this out for my job. I am an Mechanical Engineer that is trying to figure out EE stuff and I am a bit lost. ha
 
  • #4
lepp1892 said:
1) I want to find out the source voltage.
Typically, you apply a known source voltage. Apparently not in your case, so you'd better explain what this circuit applies to.

You must know the value of a current or voltage somewhere in this circuit, in order to be able to determine the source voltage. Having marked it as 50Hz, I suspect you do know the source voltage.

Anyway, explain how this came about.
 
  • #5
Without a source voltage...you will not be finding any output voltages.

Once you do find the source voltage...you simply apply V=IR

The resistance (reactance) of your capacitors will be (1/JWC)...W in this case will be 2∏*50.

You will need to use vector math because of the "J" in the equation. Combine resistances of each parallel load...then do a voltage divider to find the voltage across each load.

Or...you can find the combined total impedance for all the loads...find the current using V=IR...then use a current divider thru each parallel load. First idea will be the easier of the two.

For each prospective parallel branch...the voltage will be the same across the resistor and capacitor. The voltages across the two prosective branches will be different however.
 
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  • #6
hmm...

How about knowing the max voltage that will be across both parallel branches seperatly? Would that give me enough knowledge to then find the source voltage?

This isn't exactly a circuit equation. I am actually trying to find breakdown voltages across multiple layers of materials and I came across an article that had this as a way to solve it but it didnt go into much detail about solving it.
 
  • #7
lepp1892 said:
hmm...

How about knowing the max voltage that will be across both parallel branches seperatly? Would that give me enough knowledge to then find the source voltage?

This isn't exactly a circuit equation. I am actually trying to find breakdown voltages across multiple layers of materials and I came across an article that had this as a way to solve it but it didnt go into much detail about solving it.

Yes, the voltages across both parallel branches will be related to the voltage coming from the source. You can work backwards to solve for the source voltage. Note that the maximum voltage across any individual parallel branch does not necessarily correspond to the maximum voltage from the source at the same time. You have reactive components, so you will have phase differences.

But after reading your other information, I think a breakdown voltage will essentially change the circuit you have drawn. Can you provide the reference you were using?
 
  • #8
Ok well the first branch has a voltage of 30 kV and the second branch has a voltage of 4 kV.
 
  • #9
lepp1892 said:
Ok well the first branch has a voltage of 30 kV and the second branch has a voltage of 4 kV.

Then your voltage source is 34 kV
 
  • #10
Its really just adding the 2 voltages together?
 
  • #11
lepp1892 said:
Its really just adding the 2 voltages together?

Yes, but you could be omitting information if you are not so sure what is going on and where you got the 30kV and 4kV numbers.

You just use kirchhoffs voltage law (second item in the link):

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmlaw.html

Its just like pressure drops in a hydraulic or thermodynamic system.
 
  • #12
What I meant by omitting information is that because your voltage source is AC, its voltage is 34kV only at the instant in time that the voltage across both // groups is 30kV and 4kV. This does not necessarily correspond to the maximum voltage from the AC source. The maximum voltage from the AC source is when the sum of the two // groups voltages are a maximum.
 
  • #13
I got the 30 and 4 from the dielectric strength of the materials. The dielectric strength of the dielectric layer is 30 kV/mm and it is going through 1 mm and the air gap is 4 mm and the dielectric strength of air is about 1 kV
 
  • #14
lepp1892 said:
I got the 30 and 4 from the dielectric strength of the materials. The dielectric strength of the dielectric layer is 30 kV/mm and it is going through 1 mm and the air gap is 4 mm and the dielectric strength of air is about 1 kV

Dielectric strength just says what voltage is needed to breakdown the dielectric. It can't be used for circuit analysis unless you have more information.

I think we'd be able to help you a lot more if you provide the reference you are using for this method.
 
  • #15
Not buying the 30 and 4 numbers.

Give us a voltage source and the problem is easily solved.

Let's say the voltage source is 480 volts. Then you will get something like 240 volts across the first load and 200 volts across the second load...for example.

Let's now say the voltage is 13,200. You will then get something like 7200 volts across first branch and 6000 volts across second branch...for example.

Huge difference.

Caps are defined by C*(dv/dt)=i(t)

In other words...the changed in voltage over time multiplied by the capacitance equals the current. The voltage source and frequency determine the behavior of the capacitors...not the other way around.
 
  • #16
hmm... well I am trying to find what voltage it should take to break through multiple materials so I am unsure what the final voltage should be. from my initial testing i am getting 33-38 kV. I am just trying to find a way to calculate it to verify my results.
 
  • #17
And I can't seem to find my source of the information I got. I thought I had saved the article but I guess I didn't
 
  • #18
lepp1892 said:
hmm... well I am trying to find what voltage it should take to break through multiple materials so I am unsure what the final voltage should be. from my initial testing i am getting 33-38 kV. I am just trying to find a way to calculate it to verify my results.

What is the purpose of the voltage divider and two different dielectrics? Are you trying to compare one to another?

If you just want to test what voltage it takes to break through the materials, can't you just place the materials in series with a variable voltage source and increase it til they break down and write down that voltage?
 
  • #19
lepp1892 said:
hmm... well I am trying to find what voltage it should take to break through multiple materials so I am unsure what the final voltage should be. from my initial testing i am getting 33-38 kV. I am just trying to find a way to calculate it to verify my results.

Hmmmm...I don't think you are looking at things correctly.

Incidentally...why do you want to find the voltage across the loads? Why does that interest you? If you tell us why you are interested in voltage across load...that might help us solve your question.
 
  • #20
DragonPetter said:
What is the purpose of the voltage divider and two different dielectrics? Are you trying to compare one to another?

If you just want to test what voltage it takes to break through the materials, can't you just place the materials in series with a variable voltage source and increase it til they break down and write down that voltage?

Well I can do that and I have been doing that. I just want to see if I can verify my results with Math since I am being told that the numbers I am getting seem to be a bit high.

psparky said:
Hmmmm...I don't think you are looking at things correctly.

Incidentally...why do you want to find the voltage across the loads? Why does that interest you? If you tell us why you are interested in voltage across load...that might help us solve your question.

We are trying to test our parts dielectric strength and we need to have an air gap so we can use less tooling so we don't have to have a different test fixture for every part.
 
  • #21
lepp1892 said:
hmm... well I am trying to find what voltage it should take to break through multiple materials so I am unsure what the final voltage should be. from my initial testing i am getting 33-38 kV. I am just trying to find a way to calculate it to verify my results.
Your breakdown tests appear to be spot on. If the dielectric were rated with a breakdown of 30kV then you'd expect this to be conservative, so finding it can reliably withstand at least 33kV would confirm this.

Examining your RC model at 50Hz, we can overlook the presence of both resistances because they have little effect in comparison with capacitances of the values you give. Your air gap capacitance is so much smaller than the dielectric capacitance, that the air gap will carry 98% of the applied voltage until the air breaks down and conducts. This will occur at around 4kV.

For applied voltages greater than 4kV the air gap is effectively conducting, leaving all of the applied voltage impressed across your dielectric. With your dielectric's rated breakdown strength of 30kV then for applied voltages exceeding 30kV you will find the dielectric is subject to breaking down.

https://www.physicsforums.com/images/icons/icon2.gif Essentially, the air gap is not interfering with your voltage tests.

When the applied voltage is removed, the air gap capacitor repairs itself; alas, the plastic dielectric does not. :smile:
 
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  • #22
Ok, I think I get what you are saying.

So long as the air gap is not large enough to have a higher breakdown voltage then the Dielectric, the air gap doesn't matter?
 
  • #23
We are trying to test our parts dielectric strength and we need to have an air gap so we can use less tooling so we don't have to have a different test fixture for every part.

Do you have any idea what this means?

This means that I have no idea what I'm talking about:)

Interesting way to look at things...
 
  • #24
NascentOxygen said:
Your breakdown tests appear to be spot on. If the dielectric were rated with a breakdown of 30kV then you'd expect this to be conservative, so finding it can reliably withstand at least 33kV would confirm this.

Examining your RC model at 50Hz, we can overlook the presence of both resistances because they have little effect in comparison with capacitances of the values you give. Your air gap capacitance is so much smaller than the dielectric capacitance, that the air gap will carry 98% of the applied voltage until the air breaks down and conducts. This will occur at around 4kV.

For applied voltages greater than 4kV the air gap is effectively conducting, leaving all of the applied voltage impressed across your dielectric. With your dielectric's rated breakdown strength of 30kV then for applied voltages exceeding 30kV you will find the dielectric is subject to breaking down.

https://www.physicsforums.com/images/icons/icon2.gif Essentially, the air gap is not interfering with your voltage tests.

When the applied voltage is removed, the air gap capacitor repairs itself; alas, the plastic dielectric does not. :smile:

Won't the air gap have some resistance at this point when it is ionized, rather than acting strictly like a short?
 
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  • #25
Well I removed the air gap and my results didnt get any lower so I'm thinking that the air gap doesn't really do anything...
 
  • #26
DragonPetter said:
Won't the air gap have some resistance at this point when it is ionized, rather than acting strictly like a short?
Some, but negligible here.
 

1. What is an RC circuit?

An RC circuit is a type of electrical circuit that contains both a resistor (R) and a capacitor (C). It is commonly used in electronic devices to control the flow of current and store electrical energy.

2. How do you calculate the voltage in an RC circuit?

The voltage in an RC circuit can be calculated using the equation V = V₀ * (1 - e^(-t/RC)), where V₀ is the initial voltage, t is the time, R is the resistance, and C is the capacitance.

3. What is the time constant in an RC circuit?

The time constant in an RC circuit is the product of the resistance (R) and capacitance (C) in the circuit. It represents the amount of time it takes for the capacitor to charge to 63% of its maximum voltage.

4. How does the voltage change over time in an RC circuit?

In an RC circuit, the voltage initially increases rapidly as the capacitor charges, then slows down and approaches a maximum voltage as the capacitor becomes fully charged. The rate at which the voltage changes is determined by the time constant.

5. How does the resistance and capacitance affect the voltage in an RC circuit?

The resistance (R) affects the rate at which the capacitor charges and discharges, while the capacitance (C) determines how much charge the capacitor can hold. Higher resistance will result in a slower charging rate and lower capacitance will result in a lower maximum voltage in the circuit.

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