I have the following circuit.

enter image description here

It is given that:

  • \$R_1=10\Omega\$
  • \$R_2=5\Omega\$
  • \$R_3=3\Omega\$
  • \$R_4=8\Omega\$
  • \$R_5=12\Omega\$
  • \$B_1=10V\$
  • \$B_2=5V\$

So using KCL on the two nodes gives;


\$I_3+I_5-I_4=0 \$

Applying the KVL is where I'm having trouble. The left loop and right loop are pretty straight foward.

\$ 10-10I_1-5I_2=0\$


However, for the middle loop I got;

\$ -5I_2-3I_3-8I_4=0\$

But it should be \$5I_2-3I_3-8I_4=0\$ (or \$-5I_2+3I_3+8I_4=0\$ depending on chosen convention). This gave the correct values when I checked using software. So whats wrong with my first equation? Looking at the middle loop \$I_2,I_3,I_4\$ are all travelling from + to - so shouldn't all the voltages across the resistors being negative?

  • \$\begingroup\$ Reduce the number of currents to three. \$\endgroup\$
    – Chu
    Apr 10 at 16:29

2 Answers 2



You are getting the wrong answer because you have not accounted for the fact that I_2 is perceived in the negative direction from the current loop in the central cell. I_3 and I_4 have the same sign for a current which flows in the clockwise direction, but the sign for I_2 flows in the counter-clockwise direction for the middle cell.

Fixing this sign yields the correct equation.

Alternative analysis

I find it very helpful to analyze KVL as a full loop, as below. I have defined three loop currents: $$I_\alpha, I_\beta, I_\gamma$$

I wrote the loops as a function of those currents, and then I provided the transformation for the currents $$ I_1, I_2, I_3, I_4, I_5$$:

enter image description here

In my opinion, this is a much more reliable way to write the eqns, as it does not require you to remember to analyze each current leg for the appropriate sign.


Using Current Branch Method, you have to select the reference direction of each branch current first, you could select like that: enter image description here

Then you will get those equations: \begin{equation} \ -10+10I_1+5I_2=0; \ -5+12I_5+8I_4=0; \ -5I_2+3I_3+8I_4=0 \end{equation}


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