# Electrical Circuits

1. Mar 16, 2013

### GreenPrint

Given the power that is needed to be output by the capacitor - P
the length of time that the capictor must give off that power - t
the maximum operation voltage that the capacitor can operate at- V
the dialectric constant of several materials - ε
the breakdown field of those several materials - $U_{d}$
and the mass density of those several materials - ρ
Design the capacitor that meets these requirements

Dialectic Material 1
ε= $4ε_{0}$
$U_{d} = 1 \frac{V}{nm}$
$ρ = 2.5 \frac{g}{cm^{3}}$

Dialectic Material 2
ε= $80ε_{0}$
$U_{d} = .03 \frac{V}{nm}$
$ρ = 1 \frac{g}{cm^{3}}$

Dialectic Material 3
ε= $200ε_{0}$
$U_{d} = .1 \frac{V}{nm}$
$ρ = 1 \frac{g}{cm^{3}}$

Metal 1
$ρ=2.7 \frac{g}{cm^{3}}$

Metal 2
$ρ=2.1 \frac{g}{cm^{3}}$

The machine which can make the capacitor has the following thickness restrictions
$10 nm < t_{dialectic}< 10 μm$
$10 μm < t_{dialectic}< 10 cm$

Equations
$C = \frac{εA}{d}$
$C = \frac{dQ}{dV}$
$U = \frac{dW}{dt}$
the maximum energy stored in a capacitor is below
$w = \frac{1}{2}εCd^{2}U_{d}^{2}$
$V = U_{d}d$
$U=CV(t)\frac{dV}{dt}$
$U=\frac{1}{2}CV^{2}$
$I=\frac{dQ}{dt}$
$hp ≈ 746 W$
$ε_{0} ≈ 8.85*10^{-12} \frac{F}{m}$
$m = 10^{9} nm$
$m = 10^{6} μm$
$m = 10^{3} cm$
$kg = 1000 g$

Attempt at a solution
I have no idea how to do this at all or where to begin. I just randomly chose material 3 for the dielectric material and started solving and go the following relations.

C = 3.082644628 F
Q = 678.1818182 C
$w = 10 ε_{0}Cd^{2}$

I don't know where to go from here or if I'm even on the right track at all. Thanks for any help that anyone can provide. It would be greatly appreciated.

2. Mar 17, 2013

### Simon Bridge

Can you relate electrical properties to material properties?
Do you know the equations for the discharge of a capacitor?

3. Mar 17, 2013

### GreenPrint

I don't know either of these equations. I could come up with the voltage as a function of time in the capacitor but because I don't know what else the capacitor is connected to I'm not sure how to go about this.

4. Mar 17, 2013

### haruspex

Fwiw, a dialectic is a rational dialogue as a means for resolving disagreements (as made famous by Plato). The word you are looking for is dielectric.

5. Mar 17, 2013

### Simon Bridge

Well, to get started you need to go through your notes and find the bits that relate things like the area of the plates and the distance between them and the amount of dielectric and so on to the capacitance.

Since the rate of discharge of an ideal capacitor depends also on the resistance connected to it so you'll have to use the information provided to work that out as well as what sort of supply voltage is intended. It's not just about equations: you need to draw diagrams too.

6. Mar 23, 2013

### GreenPrint

I got all those relations in post 1. I still don't see how it's solvable because you don't know the impedance of the rest of the circuit.

7. Mar 23, 2013

### Simon Bridge

The max power a capacitor is able to supply does not depend on the load.

8. Mar 23, 2013

### GreenPrint

But you don't know the charge though right?

P = (1/2)QV

9. Mar 24, 2013

### Simon Bridge

You know the maximum voltage the cap can operate at.

10. Mar 24, 2013

### Staff: Mentor

That formula gives energy (Joules), not power (Joules/sec).

If the capacitor is expected to be able to provided a certain power for a certain period of time, then there must be some fancy load that makes the energy draw constant despite changes in capacitor charge (thus potential difference). That means Ic*Vc = constant. More importantly, it also tells you total energy that you need to store, E = P*t, which is where capacitor energy equations come in.

There's another version of the capacitor energy equation which involves the capacitance value: E = (1/2)CV2. For the given E = P*t you can draw a C versus V curve, where V would then represent the maximum required working voltage of the device of capacitance C for this application. The max voltage tells you the minimum thickness of the given dielectric, too.

So, it's beginning to look like an exercise in linear programming. It's not clear to me from the problem statement whether this is an optimization problem, and if so, what the cost function would be. Does cost of materials, weight, or overall size matter?

11. Mar 24, 2013

### GreenPrint

So I think I'm beginning to understand how to solve this problem. I just don't see how does the thickness of the metal have an impact on the capacitor since capacitance has to do with surface area and dosen't' involve the thickness of the metal?

Actually this dosen't make since. I'm getting that the smallest distance possible is 220 m between the plates
using V = $U_{d}d$

Last edited: Mar 24, 2013
12. Mar 24, 2013

### Staff: Mentor

Seems to me you can build a range of solutions given any choice of dielectric. So how do you go about "picking a material"? You can always increase the capacitance by making the area larger and decreasing the required voltage. The available thickness range of the dielectric may offer a way to select the range of solutions for each material, but without other constraining factors there isn't a way to pick a particular solution, or choose between materials.

13. Mar 24, 2013

### GreenPrint

I don't think it really matters. The assignment dosen't make it clear either but says that I should just compare different combinations so I think I'm supposed to take weight into consideration. It says nothing about size or cost.

So if I plot

C = $\frac{2Pt}{V}$. Am I just supposed to pick a point. I mean I would imagine the smallest capacitor is best so I should pick a voltage close that would minimize this value? Since the max voltage is 200, how does this help determine the minimum thickness of the capacitor?

$\frac{εA}{d} = \frac{2Pt}{V}$

I don't know the surface area or the distance between the capacitors. Thanks for all your help!

14. Mar 24, 2013

### Simon Bridge

Those are things you have to determine.

15. Mar 24, 2013

### GreenPrint

Ok well

$E = Pt$
E is energy in joules
P is power in watts
t is time in seconds

$E = \frac{1}{2}CV$
E is energy in joules
V is voltage in volts

$Pt = \frac{1}{2}CV$

Solving for capacitance

$C = \frac{2Pt}{V}$

I don't know what values to pick for voltage and capacitance and am guessing that this is what I'm supposed to figure out. Just thinking about this. The smallest the capacitance then the smaller the capacitor. So I'm assuming that I want the voltage to be as large as possible. So I'm going to choose the highest operating voltage of 220 V

$C = \frac{2(100 hp)(10 hr)(745.699822 W)(60 min)(60 s)}{(hp)(hr)(min)(220 V)} ≈ 2.440*10^{7} F$

So this is a rather large capacitor but is the smallest I can get I suppose.

So I know that

$C = \frac{εA}{d}$
C is the capacitance
ε is the dialectic constant of the dialectic
A is the area of the plates of the parallel capacitor
d is the distance between the plates of the capacitor

If I were to choose a certain material I would still have two unknowns so I'm going to utilize this equation

$V = V_{d}d$
V is the voltage
$V_{d}$ is the break down field
d is the distance between the plates

Solving this equation for d I get

$d = \frac{V}{V_{d}}$

Plugging this into

$C = \frac{εA}{d}$

I get

$C = \frac{εAV_{d}}{V}$

Going back to this equation

$C = \frac{2Pt}{V}$

I get

$\frac{εAV_{d}}{V} = \frac{2Pt}{V}$

which simplifies to

$εAV_{d} = 2Pt$

Assuming I just choose a material I'll have one unknown, the area of the metal plates. So I'll rearrange this equation for the area of the parallel plate capacitor and get the following equation

$A = \frac{2Pt}{V_{d}ε}$

I want the surface area to be as large as small as possible to make the capacitor weight less so I want the quantity $V_{d}ε$ to be as large as possible. Looking at the particular materials this quantity will be as large as possible for the polymer A. After making this decision I am able to solve for the surface area of the parallel plate capacitors.

$A = \frac{2(100 hp)(10 hr)(m)(nm)(60 min)(60 s)(m)(745.699822 W)}{200(8.85*10^{-12} F)(.1 V)(hr)(min)(m)(hp)10^{9} nm} ≈ 3.033*10^{10} m^{2}$

Am I doing something wrong as this is going to be a huge capacitor?

I'm going to just use square plates. I could use circles or whatever, but no matter the shape it's going to weight the same etc.

$A = l^{2}$
A is the area
l is the length

Solving for the length I get

$l = \sqrt{A} = \sqrt{3.033*10^{10} m^{2}} ≈ 1.742*10^5 m$

Going back to this equation

$V = V_{d}d$

I can now solve for the distance between the capacitors

$d = \frac{V}{V_{d}} = \frac{(220 V)(nm)m}{(.1 V)10^{9}nm} = 2.2*10^{-6} m$

Going back to the restrictions on the machine the dialectic must be between 10 nm and 10 μm. Checking that these conditions are met I convert these values to meters

$\frac{10 (nm)m}{10^{9} nm} = 1*10^{-8} m$
$\frac{10 (μm)m}{10^{6} μm} = 1*10^{-5} m$

So it looks like I meet these restrictions or am I doing something wrong as this is a huge capacitor.

Also what does the thickness of the metal plates have to do with anything? Why does the assignment mention that it has a restriction on it's thickness. My guess is it just make it as small as possible and to use the metal which weighs the least. I'm kind of confused.

Last edited: Mar 24, 2013
16. Mar 24, 2013

### Simon Bridge

... see post #4.
What electrical property of a metal depends on it's thickness?
Do you have an equation that relates some electrical property to a thickness or a length?

You are now on the right track -
You have been provided with a great deal of information to digest all at once ... just take it one step at a time.

17. Mar 24, 2013

### GreenPrint

Oh lol I always spell that wrong.

I honestly don't know if I know the answer to this question. My first thought was charge. However all the charge accumulates on the faces of the plates so it has nothing to do with this. I thought perhaps the electric field, but that's only inside the dielectric =). It's been a semester since I have taken electromagnetism. I'm not sure what property you are refereeing to.

Oh the resistance of the plates? This is dependent on the thickness of the metal, but what does this resistance matter? Hm. I'm not really sure about this. I just assumed the plates have zero resistance.

Other than this property I don't think I'm aware of any others.

18. Mar 24, 2013

### Simon Bridge

Is that a good assumption considering that you have been supplied with the resistivities of different metals?
OR are you expecting to use those in a different context?

The DC resistance of the capacitor will affect it's performance. i.e. the maximum current flows when you short-circuit the capacitor (after charging it to it's maximum rated voltage).

Any real capacitor is modeled by an ideal capacitance in series with a small resistance.

19. Mar 24, 2013

### GreenPrint

Hm very interesting. Lets see here.

$P = \frac{V^{2}}{R}$
P is the power - this is given in the problem to be 100 hp
V is the voltage - I choose 220 V
R is the resistance

Solving for resistance

$R = \frac{V^{2}}{P} = \frac{(220 V)^{2}(hp)}{(110 hp)(745.699822 W)} ≈ .649 Ω$

$R = \frac{ρl}{A}$
R is the resistance
ρ is the resistivity of the metal
A is the area - this was already calculated to be $3.033*10^{10} m$
l is the length (in this case the thickness)

So solving for thickness I got

$l = \frac{AR}{ρ}$

Now I only have two metals which are the aluminum and graphite, I also got to take into consideration their weight densities when choosing a metal. I can't just pick a metal because I would have to calculate the weight after first calculating the thickness.

$ρ_{R, aluminum} = 2.82*10^{-8} Ωm, ρ_{aluminum} = 2.7 \frac{g}{cm^{3}}$
$ρ_{R, graphite} = 1*10^{-5} Ωm, ρ_{graphite} = 2.1 \frac{g}{cm^{3}}$

I'm going to convert the restrictions on the metal now
$10 μm = 1*10^{-5} m$
$\frac{(10 cm)m}{100 cm} = .1 m$

I also be using the following formula to calculate mass

m = Vρ
m is the mass
V is the volume
ρ is the mass density

or in the case of a rectangular 3d shape

m = Alρ
A is the face area
l is the thickness
ρ is the mass density

I will then use the following formula to calculate the weight
$F_{g} = mg$
$F_{g}$ is the force of gravity on the object
m is the mass of the object
g is gravity (I will use 9.81 $\frac{m}{s^{2}}$)

Combing these equations I get

$F_{g} = Alρg$

For aluminum
$l = \frac{AR}{ρ} = \frac{(3.033*10^{10} m^{2})(.649 Ω)}{(2.82*10^{-8} Ωm)} ≈ 6.980*10^{17} m$

Hm dosen't seem like aluminum will work on this one...

For graphite
$l = \frac{AR}{ρ} = \frac{(3.033*10^{10} m^{2})(.649 Ω)}{(1*10^{-5} Ωm)} ≈ 1.968 m$

It looks like I'm screwed or did I do something wrong because this dosen't meet the thickness requirements that the machine can make?

20. Mar 25, 2013

### Simon Bridge

So check your figures - you may have to go back and try a different design.