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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_1-I_2-I_3=0\$

\$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\$

\$5-12I_5-8I_4=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?

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

2 Answers 2

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Problem

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.

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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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