bunthulhu
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Hi, again

This time it's circuits that have got me stumped, surprise surprise!

Starting off with capacitors...


1. A 4 microFarad capacitor and a 6 microFarad capacitor are connected in parallel across a 660V supply line. The charged capacitors are disconnected from the line and from each other, and then reconnected to each other with terminals of unlike sign together. Find the final charge on each and the voltage across each.

I'm hoping there's an easy way of doing this and I'm just being silly :p:



2. In the figure below, C1 = C5 = 8.4 microFarads and C2 = C3 = C4 = 4.2 microFarads. The applied potential is Vab = 220V.
a) What is the equivalent capacitance of the network between points a and b?
b) Calculate the charge on each capacitor and the potential difference across each capacitor.



Okay, so I think the problem is that I'm not very good at splitting up circuits so I can't work out which bits are series or parallel or whatever


Okay, normal circuity bits!


3. You are asked to determine the resistance per meter of a long piece of wire. You find a battery, voltmeter, and an ammeter. You put the leads from the voltmeter across the terminals of the battery and the meter reads 12.6V. You cut off a 20m length of wire and connect it to the battery, with an ammeter in series with it to measure the current in the wire. The ammeter reads 7A. You then cut off a 40m length of wire and connect it to the battery, again with the ammeter in series to measure the current. The ammeter reads 4.2A. What is the resistance of one meter of wire?

Um, I tried to create some sort of simultaneous equations but I think I'm missing a key concept because I'm getting an impossible answer along the vein of 1 = 2


4. A material of resistivity p is formed into a solid, truncated cone of height h and radii r1 and r2 at either end. (Below)
Calculate the resistance of the cone between the two flat end faces.



I tried integrating the expression for one disk between the limits r1 and r2 but I just ended up with a 1/r1 - 1/r2 on the end of my answer which is wrong, it's meant to be over ph/(pi r1 r2)


5. The potential difference across the terminals of a battery is 8.4V when there is a current of 1.5A in the battery from the negative to the positive terminal. When the current is 3.5A in the reverse direction, the potential difference becomes 9.4V.
a) What is the internal resistance of the battery?
b) What is the emf of the battery?

I can get b) from a) but I don't know how to start part a)!


6. A 12.6V car battery with negligible internal resistance is connected to a series combination of a 3.2 ohm resistor that obeys Ohm's law and a thermistor that does not obey Ohm's law, but instead has a current-voltage relation V = 3.8I + 1.3I^2. What is the current through the 3.2 ohm resistor?

!


7. Ahh, Kirchoff's Laws, the bane of my life :p: I understand how to 'traverse a loop' and add all the bits and bobs but I find it difficult to know what to do about the currents sometimes, i.e. expressing one in terms of the others and things like this? For example this problem was giving me some grief - "Find the currents":





And lastly...this is mostly related to the problem I have with splitting complicated circuits into series and parallel components. If anyone could give me some tips I would REALLY appreciate it

This is the sort of thing I've been getting stuck on:

If an ohmmeter is connected between points a and b, what will it read?



All numbers in Ohms



Thaaaaaaaaaaaaaaaaanks
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laeti
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Just had a quick look through and for question 4, if you combine the fractions you'll get a r_1r_2 as the denominator, which might solve your problem. Sorry I can't help with the rest, gotta go out!
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bunthulhu
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(Original post by laeti)
Just had a quick look through and for question 4, if you combine the fractions you'll get a r_1r_2 as the denominator, which might solve your problem. Sorry I can't help with the rest, gotta go out!
Hey, thanks for the reply I did think about that but that would leave me with r2 - r1 on the top, which isn't meant to be there
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laeti
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OK I think question 4 might be to do with the fact that as the radius is changing the length is also changing so the resistance is a function of two variables, and I think you might have to do something like a double integral? But I've not done those before so I'm not too sure!

For question 6, try using the fact that the voltage supplied by the battery = the sum of the voltages across each of the components, so that 12.6 = V_{resistor} + (3.8I + 1.3I^2), and you'll end up with a quadratic in I.
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bunthulhu
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(Original post by laeti)
OK I think question 4 might be to do with the fact that as the radius is changing the length is also changing so the resistance is a function of two variables, and I think you might have to do something like a double integral? But I've not done those before so I'm not too sure!

For question 6, try using the fact that the voltage supplied by the battery = the sum of the voltages across each of the components, so that 12.6 = V_{resistor} + (3.8I + 1.3I^2), and you'll end up with a quadratic in I.
Oh dear, I think you might be right, but I can't work out how to find the length in terms of anything that doesn't involve lots of square roots!

Good idea, I had tried that and it didn't seem to work but now it looks like it does work but only if you take -b + sqrt(b^2 - 4ac). Maybe you have to take the positive because otherwise you'd get a negative current?

Thanks for the help
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Drummy
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I'll do question 1 for you

a) calculate the charge on each capacitor when connected to the 660V supply using Q = CV (2.64mC and 3.96mC)

b) When they are now connected positive to negative SOME of the charge will cancel and you will be left with 3.96 - 2.64 = 1.32mC shared between them.

c) calculate the combined capacitance of the parallel combination ( 10 microF) then use V = Q/C to calculate the voltage across this combination.
So V = 1.32 mC / 10 microF = 132 V.

This is therefore the voltage across each capacitor as they are in parallel. Use Q = CV to find the charge on each one (which should add up to 1.32 mC)

Question 2 isn't as hard as it looks

Imagine it as 3 capacitors in series. C1, a bit in the middle then C5.

The bit in the middle can be looked at as 2 in parallel. C2 and the combination of C3 and C4.

So...
Work out C3 and C4 combined.
Work out the middle bit combination.
Work out the whole lot.

To get the charges. use Q = CV for the combination.
This MUST be equal to the charge on C1, on C5 and on the middle bit as the charge is just passed down the line.
You can then get the voltage on C1 and C5. The total voltages minus these is the voltage across the middle bit etc....
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bunthulhu
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(Original post by Drummy)
I'll do question 1 for you

a) calculate the charge on each capacitor when connected to the 660V supply using Q = CV (2.64mC and 3.96mC)

b) When they are now connected positive to negative SOME of the charge will cancel and you will be left with 3.96 - 2.64 = 1.32mC shared between them.

c) calculate the combined capacitance of the parallel combination ( 10 microF) then use V = Q/C to calculate the voltage across this combination.
So V = 1.32 mC / 10 microF = 132 V.

This is therefore the voltage across each capacitor as they are in parallel. Use Q = CV to find the charge on each one (which should add up to 1.32 mC)

Question 2 isn't as hard as it looks

Imagine it as 3 capacitors in series. C1, a bit in the middle then C5.

The bit in the middle can be looked at as 2 in parallel. C2 and the combination of C3 and C4.

So...
Work out C3 and C4 combined.
Work out the middle bit combination.
Work out the whole lot.

To get the charges. use Q = CV for the combination.
This MUST be equal to the charge on C1, on C5 and on the middle bit as the charge is just passed down the line.
You can then get the voltage on C1 and C5. The total voltages minus these is the voltage across the middle bit etc....

Ooooooooh I think I love you, thanks so much, I understand capacitors a lot better now

If I could just ask one thing... in the first question why do the charges on the capacitors partly cancel one another? And what would happen if you connected them back to back (i.e. negative to negative)?

Thanks again!
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Drummy
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np m8

If you connect a positively charged plate to a negatively charged plate then a bit of cancelling is going to take place and whichever is bigger will "win". Imagine electrons flowing from a negative plate to a plate where there is a defficiency of electrons. After this cancelling then the remaining charge will be shared.

If you connected them positive to positive, negative to negative then the charge would slosh around a bit until the voltage across the capacitors was equal. As V = Q/C the ratio of the charges would be the ratio of the capacitances.

V_1 = V_2 so \frac{Q_1}{C_1} = \frac{Q_2}{C_2} therefore \frac{Q_1}{Q_2} = \frac{C_1}{C_2}

The total amount of charge would be the same.
Here's a good analogy. Imagine two buckets with water in at different heights. A big bucket carries a lot of water (charge). The height of the water represents the voltage. Imagine joining the two buckets with a pipe near their bottoms. What will happen?
(this analogy is either genius or a load of C R AP as I haven't really thought about it a lot!)
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bunthulhu
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(Original post by Drummy)
np m8

If you connect a positively charged plate to a negatively charged plate then a bit of cancelling is going to take place and whichever is bigger will "win". Imagine electrons flowing from a negative plate to a plate where there is a defficiency of electrons. After this cancelling then the remaining charge will be shared.

If you connected them positive to positive, negative to negative then the charge would slosh around a bit until the voltage across the capacitors was equal. As V = Q/C the ratio of the charges would be the ratio of the capacitances.

V_1 = V_2 so \frac{Q_1}{C_1} = \frac{Q_2}{C_2} therefore \frac{Q_1}{Q_2} = \frac{C_1}{C_2}

The total amount of charge would be the same.
Here's a good analogy. Imagine two buckets with water in at different heights. A big bucket carries a lot of water (charge). The height of the water represents the voltage. Imagine joining the two buckets with a pipe near their bottoms. What will happen?
(this analogy is either genius or a load of C R AP as I haven't really thought about it a lot!)
Oooh...equal water levels

That makes loads of sense, thanks a gabillion
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Kyle_S-C
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For question 4, I think you need to consider disc shaped elements \Delta x thick and work out the little \Delta R components from each disc element and integrate them together.

For that you basically have:

\Delta R = \frac{\rho \Delta x}{A}

Next express A in terms of x, the distance from the the bigger end:

A = \pi \left(r_2 - \frac{x\left(r_2-r_1\right)}{h}\right)^2

From here my calculus gets a bit hazy and my physics supervisor would most certainly shoot me for being uncertain and/or wrong (i.e. don't take this as true):

\Delta R = \frac{\rho \Delta x}{\pi \left(r_2 - \frac{x\left(r_2-r_1\right)}{h}\right)^2}

As \Delta x \rightarrow dx, \Delta R \rightarrow dR.

Hence, you can integrate R from 0 to R, where R is the total resistance and x from 0 to h.

That means you get:

R = \frac{\rho}{\pi} \int_{0}^{h} \frac{1}{\left(r_2 - \frac{x\left(r_2-r_1\right)}{h}\right)^2}\, dx

That gives the pretty result:

R = \frac{\rho h}{\pi r_1 r_2}

If by some miracle of fate it turns out that that (third time around) was the answer you were looking for, I can explain stuff. If not, maybe there's an error somewhere that you can spot?

Edit: Oooh, just spotted that the answer corresponds.
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Drummy
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being incredibly bored I've had a look at question 5.

I'm a bit confused about the direction of the currents (from negative to positive is a bit ambigiuos in this context) nevertheless a bit of Kirchoff's 2nd gives us

8.4 = E + 1.5r and
9.4 = E - 3.5r

subtracting we get -1 = -3.5r -1.5r so 1 = 5r so r = 0.2 ohms

bunging this in either equation above gives E = 9.1V

as always I reserve the right to be wrong
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bunthulhu
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(Original post by Kyle_S-C)
For question 4, I think you need to consider disc shaped elements \Delta x thick and work out the little \Delta R components from each disc element and integrate them together.

For that you basically have:

\Delta R = \frac{\rho \Delta x}{A}

Next express A in terms of x, the distance from the the bigger end:

A = \pi \left(r_2 - \frac{x\left(r_2-r_1\right)}{h}\right)^2

From here my calculus gets a bit hazy and my physics supervisor would most certainly shoot me for being uncertain and/or wrong (i.e. don't take this as true):

\Delta R = \frac{\rho \Delta x}{\pi \left(r_2 - \frac{x\left(r_2-r_1\right)}{h}\right)^2}

As \Delta x \rightarrow dx, \Delta R \rightarrow dR.

Hence, you can integrate R from 0 to R, where R is the total resistance and x from 0 to h.

That means you get:

R = \frac{\rho}{\pi} \int_{0}^{h} \frac{1}{\left(r_2 - \frac{x\left(r_2-r_1\right)}{h}\right)^2}\, dx

That gives the pretty result:

R = \frac{\rho h}{\pi r_1 r_2}

If by some miracle of fate it turns out that that (third time around) was the answer you were looking for, I can explain stuff. If not, maybe there's an error somewhere that you can spot?

Edit: Oooh, just spotted that the answer corresponds.

Wow, fantastic, thanks!! I'd never have gotten there on my own hehe.

Pretty please, could you just explain 2 things to me? I'm not quite sure how you got your expression for area, with the x and h and things in there, and also how did you know what your integral integrates to?

Thanks again!
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bunthulhu
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(Original post by Drummy)
being incredibly bored I've had a look at question 5.

I'm a bit confused about the direction of the currents (from negative to positive is a bit ambigiuos in this context) nevertheless a bit of Kirchoff's 2nd gives us

8.4 = E + 1.5r and
9.4 = E - 3.5r

subtracting we get -1 = -3.5r -1.5r so 1 = 5r so r = 0.2 ohms

bunging this in either equation above gives E = 9.1V

as always I reserve the right to be wrong
Hey that's right! I think you made a little mistake on the last part - if you bung it in it actually gives E = 8.7V which is the right answer, yaaaay thank you!!!
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Drummy
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errrrr.... yeah, thats what I said. 80)
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Kyle_S-C
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For the area, you basically apply:

A = \pi r^2 where r is the radius at a distance x from the larger end. Then you need to find an expression for this general r in terms of the variables you have.

To find this, consider the complete cone which the shape is taken from and say that it has a height d. You can use similar triangles to say that at a distance x from the base of this cone, r is given by r = \frac{r_1 \left(d - x\right)}{d} = r_1 - \frac{r_1 x}{d}. However, d is a variable which isn't included in the problem, so you need to eliminate it.

For r = r_2 you're just considering the case where x = h, which means that r_2 = \frac{r_1 \left(d - h\right)}{d} Then you rearrange to express d in terms of r_1, r_2 and h. This gives you \frac{1}{d} = \frac{r_1 - r _2}{h r_1} so you can go back to the expression for general r and get that r = r_1 - \frac{x \left(r_1 - r_2\right)}{h}. This then subs in to the area expression to give you the area expression (though in my original working, I got r_1 and r_2 the wrong way around I think).

As for the integral, if you express it as \left(r_2 - \frac{x \left(r_2 - r_1\right)}{h}\right)^{-2} it becomes clearer that you're just raising the power by 1 and dividing by the differential of what's in the brackets (which is a constant anyway). Of course, that suggests I did it that way, which I didn't. I cheated and used Calc101 to be honest. Oh, and bear in mind that the two rs are the wrong way round.
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Robob
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(Original post by Kyle_S-C)
For the area, you basically apply:

A = \pi r^2 where r is the radius at a distance x from the larger end. Then you need to find an expression for this general r in terms of the variables you have.

To find this, consider the complete cone which the shape is taken from and say that it has a height d. You can use similar triangles to say that at a distance x from the base of this cone, r is given by r = \frac{r_1 \left(d - x\right)}{d} = r_1 - \frac{r_1 x}{d}. However, d is a variable which isn't included in the problem, so you need to eliminate it.

For r = r_2 you're just considering the case where x = h, which means that r_2 = \frac{r_1 \left(d - h\right)}{d} Then you rearrange to express d in terms of r_1, r_2 and h. This gives you \frac{1}{d} = \frac{r_1 - r _2}{h r_1} so you can go back to the expression for general r and get that r = r_1 - \frac{x \left(r_1 - r_2\right)}{h}. This then subs in to the area expression to give you the area expression (though in my original working, I got r_1 and r_2 the wrong way around I think).

As for the integral, if you express it as \left(r_2 - \frac{x \left(r_2 - r_1\right)}{h}\right)^{-2} it becomes clearer that you're just raising the power by 1 and dividing by the differential of what's in the brackets (which is a constant anyway). Of course, that suggests I did it that way, which I didn't. I cheated and used Calc101 to be honest. Oh, and bear in mind that the two rs are the wrong way round.
You're wrong
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Kyle_S-C
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You're wrong
I could very well be wrong, but the appropriate reply would seem to be: "Your face is wrong."
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Robob
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(Original post by Kyle_S-C)
I could very well be wrong, but the appropriate reply would seem to be: "Your face is wrong."
Sh'up, 79 yeah
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