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Current in Open Circuits

by dancergirlie
Tags: circuits, current, voltage
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dancergirlie
#1
Jan24-11, 10:15 PM
P: 200
1. The problem statement, all variables and given/known data

Looking at the open circuit demonstrated in figure 2 (see attachment), what is the current through the ammeter? What is the voltage at point p?

2. Relevant equations



3. The attempt at a solution

I am assuming if it an open circuit, I would think that the current would be zero (I might be wrong though). Meaning that since the current is zero, so is the voltage.

I have a feeling I might be wrong though...

Any help would be great!
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gneill
#2
Jan24-11, 10:36 PM
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P: 11,625
What's the voltage across the terminals of a 12V battery that's not connected to a circuit? What current is it producing?
dancergirlie
#3
Jan24-11, 10:51 PM
P: 200
Wouldn't the voltage be 12V? However there is no resistance, so using ohm's law: I=V/R, with 0 resistance, that would make the current infinite, which doesn't make any sense...

gneill
#4
Jan24-11, 11:00 PM
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P: 11,625
Current in Open Circuits

Quote Quote by dancergirlie View Post
Wouldn't the voltage be 12V? However there is no resistance, so using ohm's law: I=V/R, with 0 resistance, that would make the current infinite, which doesn't make any sense...
With no connection the resistance would be infinite; no conductivity.

So in your circuit, what's the total resistance between the battery terminals?
dancergirlie
#5
Jan24-11, 11:02 PM
P: 200
Well since it is not connected then it would be infinity

Using my Ohm's law argument that would give us our voltage of 12 volts across the battery

and 12/inf is approx 0, so the current would read out zero.

Am I right? If so then thank you so much!!!
gneill
#6
Jan24-11, 11:11 PM
Mentor
P: 11,625
Right; The open circuit dangling off of the positive battery terminal would all be at the battery's potential, and no current would be flowing. By the way, I chose 12V as an example. Check the actual battery voltage specified in your problem.
dancergirlie
#7
Jan24-11, 11:13 PM
P: 200
Yeah I noticed that, mine would be 5V :)

Thank you so much, that makes total sense!


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