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- Thread starter Kevin J
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- #2

berkeman

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Is this for schoolwork? What do you think the answer would be?

- #3

Dale

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Hint, for a voltage source the 2 Ohm internal resistance is in series with the 10 V.

- #4

Kevin J

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Don't worry, this is not a school hw, shouldn't the current be 5A? Since the current only flows through the 0 Ohms path?Is this for schoolwork? What do you think the answer would be?

- #5

Kevin J

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Ok, let me think, if the parallel resistance is (5×0)/5+0, this means the parallel resistance is equal to 0 Ohms, if the 2 Ohm resistance is series doesn't that mean 0+2 Ohms? Correct me if I'm wrongHint, for a voltage source the 2 Ohm internal resistance is in series with the 10 V.

- #6

Dale

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Yes, exactly!shouldn't the current be 5A?

- #7

Kevin J

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Thanks! People here are so nice, in physicsstackexchange, they are sometimes very rude, every question I asked is considered as a duplicate and they even banned me from their forum :(Yes, exactly!

- #8

Kevin J

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- #9

Kevin J

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The person on the web says the 10A flows through the short circuit(0 resistance path), but how do you even get this conculsion? If we take a look at the problem, the total parallel resistance would be (10×10×0)/10+10+0=0 Ohms, using V=I×R, doesn't it mean current is infinite?

*

To make it clearer if you look at the diagram drawn by me(brown pen), the current flowing through the short circuit path would be 5A, because the source has internal resistance (10V/2Ohms), it is very clear that I don't use the 5 Ohms resistor to calculate the current, so why do the person on the web used those resistors?

*

To make it clearer if you look at the diagram drawn by me(brown pen), the current flowing through the short circuit path would be 5A, because the source has internal resistance (10V/2Ohms), it is very clear that I don't use the 5 Ohms resistor to calculate the current, so why do the person on the web used those resistors?

- #10

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BUT perhaps the person on the web says something else which you didn't notice, perhaps he is saying that the voltage source has internal resistance 5Ohm?

OR perhaps if originally the circuit is not shorted, then we have 10A current flowing. When we short circuit it, then there will be a small transient state during which the current will rise from 10A to infinite. At the start of the transient state the current flowing through the short circuit will be 10A and soon it will become infinite. Is this what the person at the web says?

- #11

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Which person? Can you post a link to the video?The person on the web

- #12

Kevin J

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Which person? Can you post a link to the video?

Here's the link, was there something I missed?

- #13

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The ideal case: I agree with the video that current will be infinite, because an ideal source can keep steady voltage even for infinite current through the source.

The practical case: I partially agree with the video. The current in this case will be finite(it will not be 10A as the video says) but very large, it will start from 10A and rise to the large finite value (which will be the maximum current that the non ideal voltage source supports). Then one of the following things can happen because the voltage source is not ideal and can support up to a maximum value of current

1) The voltage source might catch fire

2) The voltage source might drop its voltage and so the current will drop back to a lower value

3) If the voltage source is protected, then just the fuse will intervene and the current will drop to zero

4) In general some other unpredictable thing , like the short circuit path catching fire (because it will become extremely hot due to the large current) e.t.c

- #14

CWatters

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1) All real world voltage sources have some internal resistance and he should show that on his circuit.

2) All real world short circuits have some resistance and he should show that on his circuit.

It is the combination of these two resistances that determines/limits how much current will flow.

Edit: Delta mentions other reasons why a practical circuit might behave differently that are also correct.

- #15

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Same reason why I sought refuge here in PF. I still have an account in physics StackExchange, but I go there once in a long while to chat with old friends who are reluctant to come here.Thanks! People here are so nice, in physicsstackexchange, they are sometimes very rude, every question I asked is considered as a duplicate and they even banned me from their forum :(

- #16

Kevin J

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and how do you count the battery's maximum current, is it it's voltage divided by it's internal resistance?

The ideal case: I agree with the video that current will be infinite, because an ideal source can keep steady voltage even for infinite current through the source.

The practical case: I partially agree with the video. The current in this case will be finite(it will not be 10A as the video says) but very large, it will start from 10A and rise to the large finite value (which will be the maximum current that the non ideal voltage source supports). Then one of the following things can happen because the voltage source is not ideal and can support up to a maximum value of current

1) The voltage source might catch fire

2) The voltage source might drop its voltage and so the current will drop back to a lower value

3) If the voltage source is protected, then just the fuse will intervene and the current will drop to zero

4) In general some other unpredictable thing , like the short circuit path catching fire (because it will become extremely hot due to the large current) e.t.c

- #17

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Maybe you could take as an upper bound that value (EMF/internal resistance) , because for this value the voltage drop at the battery's terminals will become zero but I think most batteries can support maximum current that it is quite smaller than this value.and how do you count the battery's maximum current, is it it's voltage divided by it's internal resistance?

- #18

russ_watters

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Try that math again...The person on the web says the 10A flows through the short circuit(0 resistance path), but how do you even get this conculsion? If we take a look at the problem, the total parallel resistance would be (10×10×0)/10+10+0=0 Ohms...

- #19

Kevin J

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What do you mean try the math again?0 divided by 20 is 0Try that math again...

- #20

russ_watters

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0+10 is 10What do you mean try the math again?0 divided by 20 is 0

Unless you wrote it wrong, you're doing the order of operations wrong.

- #21

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The usual problem when we don't use latex, we forget to put parentheses and this can mess a lot things. He means (10x10x0)/(10+10+0)Try that math again...

Though again for three resistances in parallel the total resistance is ##\frac{R_1R_2R_3}{R_1R_2+R_1R_3+R_2R_3}## in this case it should be ##\frac{10\cdot10\cdot0}{10\cdot10+10\cdot0+10\cdot0}##

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