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# Isaac Physics - Internal resistance in series watch

1. https://isaacphysics.org/questions/i...ance_in_series

Somebody, please enlighten me...

I know that E=IR+Ir
my current working out looks something like this:

Current for 1 cell = V1/(20+r)
Current for 2 cells = 1.5V1/(20+2r)

E=(V1/(20+r))*20 + (V1/(20+r))*r

2E=(1.5V1/(20+2r))*20 + (1.5V1/(20+2r))*2r
I'm not sure if you multiply by r or 2r

(V1/(20+r))*20 + (V1/(20+r))*r = (1.5V1/(20+2r))*20 + (1.5V1/(20+2r))*2r
This is where I get stuck...
2. (Original post by Mowahid)
https://isaacphysics.org/questions/i...ance_in_series

Somebody, please enlighten me, my head hurts too much after trying to solve this for the past hour.

I know that E=IR+Ir
my current working out looks something like this:

Current for 1 cell = V1/(20+r)
Current for 2 cells = 1.5V1/(20+2r)

E=(V1/(20+r))*20 + (V1/(20+r))*r

2E=(1.5V1/(20+2r))*20 + (1.5V1/(20+2r))*2r
I'm not sure if you multiply by r or 2r

(V1/(20+r))*20 + (V1/(20+r))*r = (1.5V1/(20+2r))*20 + (1.5V1/(20+2r))*2r
This is where I get stuck...

For one cell case, the p.d. across the external resistor R is V1, so you can use it to find the current through the cell.

You can do the same to find the current through the two cells for the two cells case.

For the two cells case,

2E = I2 (R + 2r)
3. (Original post by Mowahid)
https://isaacphysics.org/questions/i...ance_in_series

Somebody, please enlighten me, my head hurts too much after trying to solve this for the past hour.

I know that E=IR+Ir
my current working out looks something like this:

Current for 1 cell = V1/(20+r)
Current for 2 cells = 1.5V1/(20+2r)

E=(V1/(20+r))*20 + (V1/(20+r))*r

2E=(1.5V1/(20+2r))*20 + (1.5V1/(20+2r))*2r
I'm not sure if you multiply by r or 2r

(V1/(20+r))*20 + (V1/(20+r))*r = (1.5V1/(20+2r))*20 + (1.5V1/(20+2r))*2r
This is where I get stuck...
Did you quote my reply? But I cannot see your reply.
4. Just wanted to say thanks for the help, didn't really know how

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