# Isaac Physics Help!

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#1
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
0
3 years ago
#2
(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
Can you say what have you tried? Or better still show us your working.
0
#3
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.
0
3 years ago
#4
(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,
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.
1
11 months ago
#5
(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.
hi I'm doing the same question and I still don't get it
0
11 months ago
#6
(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.
0
11 months ago
#7
(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.
0
11 months ago
#8
(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.
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.
0
11 months ago
#9
Here's a solution:

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.
Last edited by beniki; 11 months ago
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