Finding the current in a circuit

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The discussion focuses on solving a circuit problem involving current, resistance, and electromotive force (Emf). The user successfully calculated the current in resistor R as 2A and determined the Emf to be 42V, with resistance R found to be 5Ω. When the circuit is cut at point X, the current through the 28V battery was initially miscalculated; however, it was clarified that the current should be recalculated based on the new circuit configuration. Ultimately, the correct current through the bottom branch was determined to be 3.5A after adjusting the calculations. The conversation highlights the importance of accurately reassessing circuit conditions when changes are made.
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Homework Statement



[PLAIN]http://img156.imageshack.us/img156/2420/physicsm.jpg

sorry for the image, i made it very quickly

in the following circuit find

a) the current in R
b) R
c) Emf
d) if we cut the circuit in X, then what would be the current that goes through the 28V battery

Homework Equations



V = I*R

The Attempt at a Solution



a)
i used the first kirchhof law and i found that

I3 = I2 - I1 = 6 - 4 = 2A

b)i used the second law of kirchhof and found that

Emf - 18 - 24 = 0 => Emf = 42 V

c)again i used the second law and found that

28 - 42 + 24 - I3*R = 0 => R = 5 Ω

d) i don't know how to do it, please help
 
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If you cut the circuit at X then there will be no current in that branch of the circuit, and you are just left with the top and bottom branch. A much simpler problem because you have already calculated R.
(I haven't checked your answers to the other parts of the question.)
 
Stonebridge said:
If you cut the circuit at X then there will be no current in that branch of the circuit, and you are just left with the top and bottom branch. A much simpler problem because you have already calculated R.
(I haven't checked your answers to the other parts of the question.)

hi, thanks for the answer

but if we have only the bottom then it will be

28 - 3*6 - I*5 = 0 => 28 - 18 - I*5 = 0 => 10 = 5I => I = 2

but the book has 3.5 A :frown:
 
kliker said:
hi, thanks for the answer

but if we have only the bottom then it will be

28 - 3*6 - I*5 = 0 => 28 - 18 - I*5 = 0 => 10 = 5I => I = 2

but the book has 3.5 A :frown:

The current through the bottom resistance is no longer equal to 6A after you
cut the circuit at X.
 
kliker said:
hi, thanks for the answer

but if we have only the bottom then it will be

28 - 3*6 - I*5 = 0 => 28 - 18 - I*5 = 0 => 10 = 5I => I = 2

but the book has 3.5 A :frown:

Why are you using a current of 6A in the 3ohm resistor, and an unknown current of I? There is only one current now and this is the new current you need to find.
 
ok I understand it now, thanks everyone

so the answer will be

Vab = I*(5+3)

but E = Vab

so I = 28/8 = 3.5

thanks again :)
 
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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