Why Can't I Solve This Circuit Resistance Problem?

  • Thread starter Thread starter gstatejoe
  • Start date Start date
  • Tags Tags
    Resistance
AI Thread Summary
The discussion centers on a circuit resistance problem involving three batteries and two resistors, where the user struggles to find the correct current values and potential difference. The user initially calculated the currents through the batteries and the potential difference but received incorrect results. Another participant suggests using Kirchhoff's laws to set up a system of equations based on the circuit, emphasizing the importance of defining the potential difference correctly. They provide a method to derive the equations needed to solve for the unknowns in the circuit. The conversation highlights the challenges of circuit analysis and the utility of systematic approaches in solving such problems.
gstatejoe
Messages
2
Reaction score
0
Hello! I have tried for roughly 3 hours now with no success on this one problem. My professor assigns webassign's (homework submittal process) for just review...which isn't graded or anything and I can't seem to get this one
correct.

[I cannot imitate the emf symbol which looks like a capital script E, so I have subsituted that symbol with 'E'.]

Here is the problem:
In the figure below, the resistances are R1 = 1.3, R2 = 1.5 , and the ideal batteries have emfs E1 = 2.0 V, and E2 = E3 = 4.5 V.

http://img81.imageshack.us/img81/9648/help6nt.gif

The problem asks for the following:
1.) What is the current through battery 1, 2 and 3
2.) and What is the potential difference Va - Vb ?


My answers were:
Batteries 1, 2, 3, = .74 A, .373 A, .373 A
Potential Difference = 3.941 V

Some of the equations I manipulated to get values:
Vb - Va = E2 - IR2 = R1 + (2R1)(2I)
I = (e2 - e1 / 4R1 + R2)

I did the problem in the book which was similar except it had different numbers, and I got that one right. But this one comes out wrong.
Can you guys tell me what I am doing wrong, and if you are able to solve the problem, what values did you get, and more importantly, how did you get them?

THANKS!o:)
Joseph
GSTATE!
 
Last edited by a moderator:
Physics news on Phys.org
Can't see your image 'cause it's pending approval...is there any way you can just host it on imageshack and post up a link??
 
Last edited by a moderator:
The trick is to call the potential difference between a and b, V.
This leads, for each branch, to an equation:
I1 = (V - E1) / 2 R1
I2 = (V - E2) / R2
I3 = (V - E3) / 2 R1
(that's the total potential difference per branch over the total resistance over each branch).
We have one further equation (Kirchhoff): I1 + I2 + I3 = 0.
This system of 4 equations has 4 unknowns: V, I1, I2 and I3.
cheers,
Patrick.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top