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Really really stuck on this piece of homework....just messes up my head :(
Anyway, these are some of the questions....anyone able to help? I've got some basic answers but just want to check. There are more later if anyone wants to help!
http://img424.imageshack.us/img424/6...sicshwk6za.png
V= IR, 1/R=1/R1+1/R2+...
any more help than that is cheating imo
Thanks. just realised i made a silly error! Used V = I/R instead of V = IR!!
Maybe thats why i got 0.003Amps as one answer!
EDIT: my minds playing up....i did have it right. So are these the correct answers?
a. 2k + 1k = 3kΩ
b. I = V/R I = 9/3000 I = 0.003A
c. The 1k resister resists 0.001A so the 4k resistors have 0.002/2 going through them which is 0.001A
d. Power Dissipated = VI = 3x0.001 = 0.003
e. If the battery has an internal resistance of 0.6Ω then the current output is reduced. If there is less available current then less will go through the resistors. The resister will dissipate less power
except for e, they look ok, it doesnt decrease the current, at least not directly, it just drops the voltage a bit, so you effectively have a lower voltage battery. you sure they should be k ohm?
what level is this? i tohught it would GCSE cw but no way!
The diagrams have rectangles on the circuit with '1k' and '4k' above them etc. so i asume they're k ohm. Thanks for the help with e
Jack, its AS level
Looks my AS work on Electrons and Photons... as for the internal resistance, I always remember the formula:
emf = V + v
Where V is the potential difference of the external circuit, and v is the potential difference 'lost' overcoming the internal resistance. emf stays the same, v increases (from zero), so V must decrease. This means each individual component has a lower voltage. Since V = IR, and R is also staying the same, I must drop by the same proportion as V. So, internal resistance does make the current drop, but not that directly.
:confused: That went in one eye and out the other!
herulach, surely if the resistance increases then the voltage would INCREASE to give the same current?
no, the easiest way to think about it is just sticking another resistor in series with the battery.
so the current drops, not the voltage increases?
basically the internal resistance drops the voltage before the thing even leaves the battery, so the voltage you get out of it is the emf - Ir (where I is the current drawn in the rest of the system, and r the internal resistance) IIRC theres a good explanation on scienceworld.wolfram.com in the physics section
The potential difference cannot just increase - it is limited by the emf of the cell. Are you familiar with the equation emf = V + v? It is quite simple - emf is essentially the voltage that the cell generates for the circuit, and that voltage is all sucked up by the potential difference of the various components (represented by V) and the internal resistance of the circuit (represented by v).
The cell will produce a constant emf i.e. per coulomb of charge, it will be giving out the same number of joules. We know that v has just increased, so, for emf to remain the same, V has to decrease by the same quantity. If V has decreased, then the potential difference of each individual component has decreased.
The resistance of each component is also going to stay the same. We know that V = IR, so I = V / R. R is the same, but V has decreased. This means that I also decreases. Hopefully that made more sense!
So putting it in to real world (and something we're all familiar with) a higher load on a computers PSU drops the voltage on a certain rail.
Only really makes sense to me when i try and put it with a real world situation
EDIT: Thanks mike, that makes perfect sense!
So does this look good for e?
Quote:
e. If the battery has an internal resistance of 0.6Ω then the current output would remain constant but the voltage would drop. The resistors will resist the same amount of current as before. The resister will dissipate less power as the p.d is smaller.
So on another question i'm given this circuit:
http://img413.imageshack.us/img413/6...icshwk15qe.png
I've done all but one question. I got to the last question and it says how much electrical energy is converted to heat by the red resistor in 7.5 mins?
I've based all my answers on nothing going through the right half of the circuit as restistance is greater than on the left
The current will drop since, in each component, the potential difference has dropped while the resistance remains the same. (I = V / R) Obviously, you need to use the reasoning explained above as well.
With a drop of both current and potential difference, then power will also drop in the components.
You can split the circuit into two 'loops'. The first loop has total resistance of 1 + 10 + 10 = 21kOhms, while the second loop has total resistance of 10 + 4 + 4 = 18kOhms. Assuming no internal resistance, for the second loop:
V = IR
9 = 18,000I ...
You have the current through the red resistor, and you have the resistance. Therefore, you can work out power.
I really need to see my teacher about this. Thanks for all the help :)
^^ not convinced that's right, Mike, you don't have 9V across that parallel component, and I'm not even certain that considering 9V across the whole 29k loop will give you the correct answer as parallel circuits are funny things...my method (right or wrong) follows.
Whoa, no, just because resistance is more on the right doesn't mean there won't be no flow, it means there will be proportionally less!Quote:
Originally Posted by BenW
Work out the total resistance of the circuit by considering the sum of the serial parts (10k + 1k) and the parallel bits (ends up to be ~6400Ohm), so total resistance of the circuit is 11kohm + ~6400Ohm = ~17400Ohm.
Use V=IR to get the current flowing through the serial bits (The bottom 10k and the top 1k resistor). turns out to be about .0005A
You can use the ratio of the resistances (So 10/28 of the current goes through the red bit) between the parallel parts to determine how much current flows through the right half of the parallel part (where our red resistor lies), then use P=VI=(I^2)R, and E = Pt (where t is in seconds).
Hope your teacher can give you a more intuitive way of dealing with this :)
Hmm, thinking about it, you're right about the bit on the right not taking 9V! But the way I would do is to work out the resistance of the parallel bit (the two 10ks and two 4ks). You can then work out the potential difference that will flow through the entire parallel bits. From the potential difference that flows through the entire parallel bits, you can split this up between the three bits (4k, 4k and 10k). Then just use Power = V^2/R for the 4k component.
I haven't pulled a calculator out, but that looks right.Quote:
Originally Posted by BenW
Resistor. Don't take the chance w/ typos in case you get an old school exam marker;). Anyway- no. In any single loop circuit the current flowing is the same in every component unless they're in parallel. The current drawn from the battery is .003A; the current going through the 1K resistor is .003A; the current going through each of the 4k resistors, given that they're identical, is .0015A- half of .003A. Current is charge x time- I.E. the flow of electrons. The number of electrons coming out of the battery is the same as the number going back in. The power 'generated' is due to the potential difference, not because electrons are somehow disappearing. So, Anyway:Quote:
c. The 1k resister resists 0.001A so the 4k resistors have 0.002/2 going through them which is 0.001A
d: The two 4k resistors in parallel appear to the circuit as one 2k resistor. Given that W=IsquaredR (forgive my failure to do a proper squared symbol):
.003 x .003 x 1000 = .009W.
I think. I am drunk and this is taxing my brain. I'm pretty certain that my reasoning is correct at least. My mum is a physics teacher and I got an A in every module of my physics A-level. How soon age turns one's brain to mush:(:(.
As for e:- they want you to work all the values out again- to quantify what's going to happen, not to just tell them which way the values will go in a handwavy fashion. It's not hard though, just include the .6ohm in the current calculations. It's not going to make much difference in the grand scheme of things TBH, the battery is hardly being stressed by the circuit. If the total resistance of the circuit was something like 10 ohms, like in a radio control car, the internal resistance of the battery suddenly becomes a big factor. In this case it's negligible- but they still want you to do the sums again.
Edit: typo corrected :o:o
raves correct, i didnt spot that. youd never think i had a degree (nearlY) in this crapsome times