Isaac Physics Help!
https://isaacphysics.org/questions/resistance_and_power
Been trying tons of things for half an hour now, I can't do question 1 could someone help
Been trying tons of things for half an hour now, I can't do question 1 could someone help
Original post by porrige51122
https://isaacphysics.org/questions/resistance_and_power
Been trying tons of things for half an hour now, I can't do question 1 could someone help
Been trying tons of things for half an hour now, I can't do question 1 could someone help
Can you say what have you tried? Or better still show us your working.
Well for P1 it is a simple circuit so P1 = V^2/(R+r)
For P2 I am a little stuck. I seem to have gotten V = 2IR + Ir and RT = 2R +r but I don't think those are correct and I am having difficulty using kirchhoff's 2nd law and relating it to find P2.
For P2 I am a little stuck. I seem to have gotten V = 2IR + Ir and RT = 2R +r but I don't think those are correct and I am having difficulty using kirchhoff's 2nd law and relating it to find P2.
Original post by porrige51122
Well for P1 it is a simple circuit so P1 = V^2/(R+r)
For P2 I am a little stuck. I seem to have gotten V = 2IR + Ir and RT = 2R +r but I don't think those are correct and I am having difficulty using kirchhoff's 2nd law and relating it to find P2.
For P2 I am a little stuck. I seem to have gotten V = 2IR + Ir and RT = 2R +r but I don't think those are correct and I am having difficulty using kirchhoff's 2nd law and relating it to find P2.
For P2,
V = 2IR + Ir is the potential drop for one loop that consist of one cell with an internal resistance r and external resistance R.
The total power dissipated is
P2= 4I2R + 2I2r
Note that total effective resistance for the second circuit is R + 0.5r.
From here, you should be able to find the required ratio.
Original post by Eimmanuel
For P2,
V = 2IR + Ir is the potential drop for one loop that consist of one cell with an internal resistance r and external resistance R.
The total power dissipated is
P2= 4I2R + 2I2r
Note that total effective resistance for the second circuit is R + 0.5r.
From here, you should be able to find the required ratio.
V = 2IR + Ir is the potential drop for one loop that consist of one cell with an internal resistance r and external resistance R.
The total power dissipated is
P2= 4I2R + 2I2r
Note that total effective resistance for the second circuit is R + 0.5r.
From here, you should be able to find the required ratio.
hi I'm doing the same question and I still don't get it
Original post by anonymouspersona
hi I'm doing the same question and I still don't get it
Please post what have you done if you still need help.
Original post by Eimmanuel
Please post what have you done if you still need help.
ive only done what was written here:
P1 = V^2/(R+r)
P2= 4I2R + 2I2r
Note that total effective resistance for the second circuit is R + 0.5r.
Original post by anonymouspersona
ive only done what was written here:
P1 = V^2/(R+r)
P2= 4I2R + 2I2r
Note that total effective resistance for the second circuit is R + 0.5r.
P1 = V^2/(R+r)
P2= 4I2R + 2I2r
Note that total effective resistance for the second circuit is R + 0.5r.
In the second circuit, you can find an expression for the current in terms of the emf V and the total effective resistance such that you can write P2 in terms of V, R and r.
Then use the question info for the ratio of power to find the required ratio of R/r.
Here's a solution:
https://www.youtube.com/watch?v=TC1seXPLCjo&lc=UgxVcAz2357va8k44kR4AaABAg
The main idea is to treat the two internal resistances in the second circuit as a set of parallel resistors in series with the main resistor. At that point the solution becomes pretty easy... it's just that the Isaac Physics hints lead you down a road that is pretty complex.
The diagrams the video does makes it easier to understand.
https://www.youtube.com/watch?v=TC1seXPLCjo&lc=UgxVcAz2357va8k44kR4AaABAg
The main idea is to treat the two internal resistances in the second circuit as a set of parallel resistors in series with the main resistor. At that point the solution becomes pretty easy... it's just that the Isaac Physics hints lead you down a road that is pretty complex.
The diagrams the video does makes it easier to understand.
(edited 3 years ago)
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