Circuit Theory

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!

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.

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....

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!)

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.

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

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

I could very well be wrong, but the appropriate reply would seem to be: "Your face is wrong."