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I am reading the book "Fundamentals of Electric Circuits 6th Ed." by Sadiku and Alexander and one of the practice problems (Practice Problem 3.7) I did not get right. The 4A current source and 3-ohm resistor branch form a supermesh. Hence, I got the following equations:

-24 + 5(i1-i3) + 10(i2-i3) + 20i2 = 0

5(i3-i1) + 5i3 + 10(i3-i2) = 0

i2 + 4 = i1

Solving these simultaneous equations, I have i1 = 4.8A

i2 = 0.8A

i3 = 1.6A

which are different from the mesh current values that the authors got. How did they arrive at that answer and where did I go wrong?

Practice Problem 3.7

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    \$\begingroup\$ Seems that the I1 - I2 really need to be 4A. \$\endgroup\$
    – DKNguyen
    Commented Sep 16, 2021 at 18:31

3 Answers 3

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I get the following four equations:

$$\begin{align*} 0\:\text{V}+24\:\text{V}-5\:\Omega\cdot\left(I_1-I_3\right)-V_{4\text{A}}-3\:\Omega\cdot 4\:\text{A}&=0\:\text{A} \\\\ 0\:\text{V}+3\:\Omega\cdot 4\:\text{A}+V_{4\text{A}}-10\:\Omega\cdot\left(I_2-I_3\right)-20\:\Omega\cdot I_2&=0\:\text{A} \\\\ 0\:\text{V}-5\:\Omega\cdot\left(I_3-I_1\right)-5\:\Omega\cdot I_3-10\:\Omega\cdot\left(I_3-I_2\right)&=0\:\text{A} \\\\ I_1-I_2&=4\:\text{A} \end{align*}$$

where \$V_{4\text{A}}\$ is the voltage across the current source.

I used the + sign on the top node and - on the bottom node of the current source. As shown below:

enter image description here

No need for any super- anything.

These solve out (using sympy):

var('i1 i2 i3 v4a')
eq1 = Eq( 0 + 24 - 5*(i1-i3) - v4a - 3*4, 0 )
eq2 = Eq( 0 + 3*4 + v4a - 10*(i2-i3) - 20*i2, 0 )
eq3 = Eq( 0 - 5*(i3-i1) - 5*i3 - 10*(i3-i2), 0 )
eq4 = Eq( i1 - i2, 4 )
solve( [ eq1, eq2, eq3, eq4 ], [ i1, i2, i3, v4a ] )

{i1: 24/5, i2: 4/5, i3: 8/5, v4a: -4}

Check your work against mine to see if you agree. Then check all this against the stated solution.


Toss the textbook into the trash can. And with it, the 8th edition of "Electronic Principles" by Malvino & Bates. And probably dozens of other terrible books on the topic of electronics. I've no idea how they get so bad, but they do. Even after 8 editions! The world is flush with garbage. I'm guessing it has to do, in part, with grad students. But I'm not sure.

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All equations are correct.

The author is definitely wrong as i1-i2 does not equal to 4 if we consider the solution given by the author.

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  • \$\begingroup\$ how did you arrive at that equation? I'm confused. \$\endgroup\$
    – Flash
    Commented Sep 16, 2021 at 18:37
  • \$\begingroup\$ -24+5(i1-i3)+3(i1-i2) = 0. \$\endgroup\$ Commented Sep 16, 2021 at 18:39
  • \$\begingroup\$ I got i1=1.6, i2=-2.4, i3=-0.8 which are still different from the answer of the authors. Also, from what I understood, shouldn't the supermesh be excluded from the kvl equations? \$\endgroup\$
    – Flash
    Commented Sep 16, 2021 at 18:45
  • \$\begingroup\$ Why not use a emulator (it's good practice to learn how to use one) and verify the results? \$\endgroup\$
    – Tyler
    Commented Sep 16, 2021 at 18:46
  • \$\begingroup\$ @Tyler Speaking of which: tinyurl.com/ygrpmncl \$\endgroup\$
    – DKNguyen
    Commented Sep 16, 2021 at 18:47
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This is how I would do it

enter image description here

Mesh one

$$-24V + (I_1 - I_3)5\Omega + (I_1 + I_2 - I_3)10\Omega + (I_1 + I_2)20\Omega = 0$$

Because we have a constant current source in mesh two we know that \$I_2 = -4A\$.

So we move on to mesh 3

$$(I_3 - I_1)5\Omega + I_3 \:5\Omega + (I_3 - I_2 - I_1)10\Omega = 0$$

And the solution is:

enter image description here

\$I_1 = 4.8A\$ and \$I_3 = 1.6A\$

a link

It looks like the author's solution is wrong.

And this method will work if one "mesh current" will enclose the same current source. So, we meet this requirement because in my example only one mesh current (I2) flows through 4A current source.

And for example, if you want to know what "actually current" is flowing through the resistor, for example, 20R. We look at the diagram is see that:

\$I_{20\Omega} = I_2 + I_1 = -4A + 4.8A = 0.8A\$

A similar question Circuit analysis using Mesh current method

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  • \$\begingroup\$ I2 cannot be equal to -4 A right? \$\endgroup\$ Commented Sep 16, 2021 at 19:07
  • \$\begingroup\$ @Jonathan_the_seagull Why not? The minus sign tells us that the current is flowing in the opposite direction to the one we have assumed at the beginning. \$\endgroup\$
    – G36
    Commented Sep 16, 2021 at 19:12
  • \$\begingroup\$ we arrived at the same i1 and i3 values. however, i got 0.8 for my i2. instead of i2=4A, shouldn't it be i1-i2=4? \$\endgroup\$
    – Flash
    Commented Sep 16, 2021 at 19:13
  • \$\begingroup\$ 4A is the sum of i1 and (-i2). This was the point of the whole discussion. \$\endgroup\$ Commented Sep 16, 2021 at 19:13
  • \$\begingroup\$ @PatrickGabuna "shouldn't it be i1-i2=4?" Not in my version. Notice that in my version the I1 current is flowing in a different path than in the original schematic. \$\endgroup\$
    – G36
    Commented Sep 16, 2021 at 19:37

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