Finding the current in a circuit

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Homework Help Overview

The discussion revolves around a circuit analysis problem involving current, resistance, and electromotive force (Emf). Participants are tasked with finding the current in a resistor, the value of the resistor, and the Emf, as well as analyzing the effects of cutting the circuit at a specific point.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply Kirchhoff's laws to determine the current and resistance values, while also exploring the implications of cutting the circuit. Some participants question the assumptions made about current distribution after the circuit is cut, and others suggest reconsidering the current values used in calculations.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on the implications of cutting the circuit. There is a recognition of differing interpretations regarding current values, particularly after the circuit is altered, but no consensus has been reached on the correct approach to finding the current through the battery.

Contextual Notes

There is mention of a discrepancy between the calculated current and a value provided in a textbook, which raises questions about the assumptions made in the calculations. The discussion reflects a learning process where participants are clarifying their understanding of circuit behavior under different conditions.

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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 :)
 

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