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About LinkBacksReally 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!
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!
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
Last edited by BenW; 19-04-2006 at 07:35 PM.
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?
Last edited by herulach; 19-04-2006 at 08:37 PM. Reason: spelling
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.
"Well, there was your Uncle Tiberius who died wrapped in cabbage leaves but we assumed that was a freak accident."
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!
"Well, there was your Uncle Tiberius who died wrapped in cabbage leaves but we assumed that was a freak accident."
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?
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.
Last edited by BenW; 19-04-2006 at 09:10 PM.
So on another question i'm given this circuit:
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.
"Well, there was your Uncle Tiberius who died wrapped in cabbage leaves but we assumed that was a freak accident."
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.
"Well, there was your Uncle Tiberius who died wrapped in cabbage leaves but we assumed that was a freak accident."
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