Recent content by eddysd

OK so this is what I got: -λ4r2\frac{dT}{dr} + \dot{q}4∏r2dr = ρc4∏r2\frac{dT}{dτ}dr -4∏r2(λ\frac{dT}{dr} + \frac{d}{dr}(λ\frac{dT}{dr})dr) Is this correct? Since the flow is steady the time derivative \frac{dT}{dτ}=0 But then when I rearrange everything I get...

Homework Statement where λ= thermal conductivity \dot{q}= dissipation rate per volume Homework Equations qx=-kA\frac{dT}{dx} The Attempt at a Solution I don't know where to start from to be honest, so any help would be greatly appreciated

Yes sorry, typo, that is the correct equation! Do you think you could tell me how to get from that to the equation for V?

Having a bit of trouble with this equation, I need to find V explicitly and this would obviously be done by factorising and integrating, but I can't seem to factorise it correctly. I have what I think is the correct answer but can't do the steps to get there. Any help would be greatly...

No, this is what I was trying to do, but can't think of a way without knowing the current

So now I have worked out the Thevenin equivalent resistance to be 270Ω, but I still don't know how to work out the equivalent voltage. This isn't actually for coursework or homework, but is revision for an exam, so I would be grateful if anyone could give me a step by step method to work it out.

Thanks, that is a much better way of looking at it than my lecturer's!

OK, so would I add R1 and R2 then take the parallel with R3? Even if I have worked out Rth, since I don't know the current how would I work out Vth?

I think the idea is that the load is on the right of the Vout and not actually shown in that image judging by the lecturers previous notes, I apologise for not mentioning this in the original post. The lecturer told me that you follow the current round, and when there is a choice of direction...

I used 1/R = 1/R2 + 1/R3 and then added that to R1.

Homework Statement Work out the Thevenin Equivalent of this circuit: https://www.physicsforums.com/attachment.php?attachmentid=46120&stc=1&d=1334164751 The Attempt at a Solution I worked out the Rth to be 300Ω, not sure if this is correct, and then I have no idea how to work out Vth Any...

Thanks for the help

Really simple problem but: If you have a circuit with a power supply, a potentiometer and an output voltage taken from midway down the potentiometer and just after it; would this act like a potential divider with the potentiometer acting as two resistors?

Thanks HallsofIvy, couldn't have hoped for a better answer!

I know this is really simple, but it's been a while since I studied maths, but when you have something to the power 3/2, say x for example, would it be sqrt(x^3) or (sqrt(x))^3?