0
$\begingroup$

In this circuit, we determine that the emf needs to be reversed and that its value is 108.75 V.

My confusion is that I thought the voltage between any two points must be the same irrespective of the path taken.

For instance, in the parallel group of 15.0 and 20.0 resistors, let's call the point on the right a and left b. From the top branch, V_ab = 108.75 - 15(4.25) = +45V. From the bottom branch, V_ab = (-2.25)20 = -45V.

I see how this is true from Kirchoff's loop rule, but it violates my previous intuition.

(A copy of the problem can be found in (PPT file): physics.wku.edu/~womble/phys260/ch26exp1.ppt)

enter image description here

$\endgroup$
0
0
$\begingroup$

Edit: I had previously misread which node was a and which was b in the question.

V_ab = 108.75 - 15(4.25) = +45V

If a is on the right, and E = -108.75, then you should have

$$ V_{ab} = -108.75 + (15\Omega)(4.25\mathrm{V}) = -45\mathrm{V}$$

$\endgroup$
7
  • $\begingroup$ The current in this lower branch is flowing from b to a. See (warning: powerpoint file) physics.wku.edu/~womble/phys260/ch26exp1.ppt for confirmation. $\endgroup$
    – sailor
    Mar 18 '15 at 16:44
  • $\begingroup$ @sailor, not if E = -108.75 V. $\endgroup$
    – The Photon
    Mar 18 '15 at 16:59
  • $\begingroup$ I am confused because the solution to the problem shows that the current is indeed -2.25 A in the lower branch, so I don't follow what you are trying to say $\endgroup$
    – sailor
    Mar 18 '15 at 17:13
  • $\begingroup$ On slide 10 it shows I_1 = -2.25 A. But notice that I_1 is from b to a. You want the current from a to b, so that is +2.25 A. $\endgroup$
    – The Photon
    Mar 18 '15 at 17:28
  • $\begingroup$ Ooops, I had a on the left and b on the right (see how much easier things would be if you just put the node labels on your schematic?). Will edit. $\endgroup$
    – The Photon
    Mar 18 '15 at 17:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.