Electrical circuit with resistors problem

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The discussion centers on calculating the electric current through an ammeter in a circuit with resistors. The user references a textbook answer of I(A) = 6ε/13R but finds it incorrect for their specific circuit. They are encouraged to apply the loop and junction rules to analyze the circuit, emphasizing that the ammeter does not affect the circuit's behavior. Participants stress the importance of attempting the problem independently for better understanding. The conversation highlights the need for a deeper grasp of current division and circuit analysis techniques.
zeev55
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1. what is the Electric current that goes through the ampermeter (see the picture)
http://imageshack.us/photo/my-images/545/66683625.png"


[PLAIN]http://img545.imageshack.us/img545/4379/66683625.png

2. the current that goes through the 2R resistor is \frac{4\epsilon}{13R}.<br /> (I dident find it myself , it was the answer of a similar problem in which the subject was the resistor not the ampermeter.)



the answer given by the textbook: I(A)=\frac{6\epsilon}{13R}
 
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You'll have to use the loop rule and junction rule. At each junction of wire, the current in is equal to the current out. Also, if you add up the changes in potential within a closed loop in the circuit, they should add up to zero. For example, if you start at point A, you move through resistors or batteries throughout the loop and return back to point A, the sum of those changes in potential should be zero,.
 
It looks like the answer attributed to the book is not correct for this circuit. So if you're despairing of reproducing that answer, you can relax a bit. There is still hope!

Zeev55, what have you tried?
 
nothing worth writing in here.. I don't mean to sound rude but it would really help me if someone write the full answer to the problem.
thank you
 
zeev55 said:
nothing worth writing in here.. I don't mean to sound rude but it would really help me if someone write the full answer to the problem.
thank you

Alack and alas, that is not what we do here. We can help you to solve the problem, but not solve the problem for you.

Why not have a go at either loop equations or some other approach to the problem, and we can then see how to guide you.

It might (or might not!) be of interest to know that the problem can be solved with a bit of study of how the currents divide and add. Keep in mind that the Ammeter doesn't influence the circuit in any way: it just tells you the current flowing through it.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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