I've been trying to solve this using the method the prof. taught us, and I happen to know the answer but I can't reach it no matter how many times I've tried. The circuit in question is below:


I am asked to use Maxwell's circulating current theorem to find out the current at the $20  \Omega$ resistor. My method was to write out all the loop equations:

$20 = 15I_1 + 10I_2$

$10 = 15I_1 + 25I_2 + 15I_3$

$10 = 15I_2 + 35I_3$

then solve by method of elimination. My answers are as follows:

$I_1 = 2.44A$

$I_2 = -1.66A$

$I_3 = -0.43A$

The answer for $I_3$ is $-0.57$.

Am I on the right path? If not, can someone point out where I am going wrong and why? Thanks


As I understand the circulating current theorem, you're really missing the simplicity of the method. You don't assign each resistor its own current. Call the current through the small, rectangular loop on the far left $I_\ell$ ($\ell$ for left). Let that current be positive when it runs clockwise. Let the current in the central loop be $I_c$, and for the right loop be $I_r$. The loop equation for left-hand loop is then

$$20 = 5I_\ell + 10(I_\ell - I_c)$$

The current through the 10-ohm resistor is a combination of the two loops' currents, as this is where they make contact. As I said, the currents are not labeled by resistors, but rather by loops. This makes each loop more complicated, but it saves you one extra equation (otherwise, you would've needed four currents for four resistors).

See if you can find the equations for the other two loops. The central one will likely be trickiest. Then try to express the current in the resistor you're interested in as a combination of these three currents (you already have to do this anyway to set up the system of equations).

If this isn't the basic idea behind how you learned the circulating current approach, then I apologize.

  • $\begingroup$ As far as I can tell the three loops are: $$ 1) 20=5I1 + 10(I1-I2) $$ $$ 2) 10 = 10(I1-I2) + 15(I2+I3) $$ $$ 3) 20I3 + 15(I2 + I3) $$ What I am really stuck on is finding out the value of the current at the 20ohm resistor. $\endgroup$ – Christy McGrory Dec 14 '12 at 7:36
  • $\begingroup$ The third line you just wrote is not an equation. Setting up this equation tells you the current through the 20 ohm resistor (because you must find its expression in terms of the 3 loop currents to set up this equation). In addition, I think you have an inconsistent sign convention for your currents (this tends to be tricky). $\endgroup$ – Muphrid Dec 14 '12 at 15:40

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.