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enter image description here

I need to find voltage V(ab) and current i(0) in the circuit. The task should be solved by mesh-current analysis.

So,What did I try?

Let's notice that there are 3 meshes. It means , using KVL , I can write 3 equations and find current. My equations are following:

$$20x+20(x-y)+30(x-z)=80$$ $$20(y-x)+20y+30(y-z)=80$$ $$30(z-x)+30z+30(z-y)=0$$

$$where\quad x,y,z \quad are \quad currents $$

After this , I can find z(that is current of right mesh) from systems of equations. And I suppose that z is the i(0). And to find V(ab) , I have to multiply 30 to z. Is my solution right? If not, how to solve this problem?

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  • \$\begingroup\$ Why not do as you say in your questions, come up with answers and then plug them back into the circuit to see if they are right? \$\endgroup\$
    – Tyler
    Commented May 11, 2017 at 17:37
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    \$\begingroup\$ Yes, I think you have it laid out. Took me only a few seconds to work out which currents were x, y, and z in your equations but I had made the right assumptions and so matched your results right away. Using symmetry I can ignore the shared \$20\:\Omega\$ resistor between the two supplies, yielding Thevenin of \$48\:\textrm{V}\$ and \$12\:\Omega\$ on each side. This means \$\frac{48\: \textrm{V}+48 \: \textrm{V}}{12 \: \Omega+30\:\Omega+12\: \Omega}=1 \frac{7}{9}\: \textrm{A}\$ as the result. \$\endgroup\$
    – jonk
    Commented May 11, 2017 at 21:10

2 Answers 2

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I have tried to solve this circuit in the general case, without knowing the values for the various resistances. Just for the fun of it of course. I have applied the Extra-Element Theorem (EET, see https://en.wikipedia.org/wiki/Extra_element_theorem) with one limit though, \$V_1=V_2\$. I have used the following labels: enter image description here

The first thing is to select the extra-element, the one that bothers you or would make the analysis simpler if it were either open-circuited or replaced by a short. Here, I adopted \$R_2\$ as the extra element that I will remove (open-circuit it) from the network. I will then calculate the voltage \$V_{ab}\$ without it. This becomes my reference voltage, \$V_{ref}\$ and the final voltage applying the EET will be defined as

\$V_{ab}=V_{ref}\frac{1+\frac{R_n}{R_2}}{1+\frac{R_d}{R_2}}\$

If calculate \$V_{ref}\$ using superposition, you have

\$V_{ref}=V_1\frac{R_6}{R_6+R_5+R_4||(R_1+R_3)}(1+\frac{R_1+R_3}{R_1+R_3+R_4})\$

The second thing is to reduce the excitation voltage to 0 V, meaning you replace both sources \$V_1\$ and \$V_2\$ by a short circuit. Then, you look at the resistance offered by \$R_2\$'s terminals, again, locally applying the EET with \$R_6\$ as the extra element in this sub-circuit. enter image description here

You should find

\$R_d=(R_5+R_1||(R_3+R_4))\frac{1+\frac{R_3||((R_5||R_1)+R_4)}{R_6}}{1+\frac{R_5+R_4||(R_1+R_3)}{R_6}}\$

The last part is to find the resistance offered by \$R_2\$'s terminals when the response \$V_{ab}\$ is a null, implying that \$V_a=V_b\$. The last sketch is here enter image description here

You install a test current source \$I_T\$ which delivers across its terminals a voltage \$V_T\$. \$\frac{V_T}{I_T}\$ is the resistance you want. If you solve that circuit correctly, then you have

\$R_n=\frac{R_3(2R_1+R_5)}{2(R_1+R_3)+R_4}\$

The voltage across terminals \$a\$ and \$b\$ is finally defined as:

\$V_{ab}=V_1\frac{R_6}{R_6+R_5+R_4||(R_1+R_3)}(1+\frac{R_1+R_3}{R_1+R_3+R_4})\frac{1+\frac{\frac{R_3(2R_1+R_5)}{2(R_1+R_3)+R_4}}{R_2}}{1+\frac{(R_5+R_1||(R_3+R_4))\frac{1+\frac{R_3||((R_5||R_1)+R_4)}{R_6}}{1+\frac{R_5+R_4||(R_1+R_3)}{R_6}}}{R_2}}\$

This is a quite ugly result and it assumes that both sources are equal to form 1 single injection when nulling the response. The calculation sheet is here enter image description here

while the Mathcad using the numerical values of the original sketch gives \$V_{ab}=53.333\;V\$ and \$I=1.777\;A\$

enter image description here

which is the result elegantly found by jonk yesterday. I am not sure in this case the EET is the best approach, but the general expression was derived almost by inspection, except for the \$R_n\$ part which required some efforts. The EET is part of the Fast Analytical Circuits Techniques (FACTs - my book) that allow you to derive transfer functions quickly and obtain results in a low-entropy format. You can have a look at my 2016 APEC seminar to know more about the subject.

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Yes, you are on the right track.

This network is also easy to solve by inspection, after noticing the symmetry. I found I0 in a few seconds with a calculator. You could compare answers from the two methods.

You can also take any set of answers and see if they are self-consistant. With Vab known, all other voltages are immediately know. You can use those and the resistances to see if your current calculations make sense.

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  • \$\begingroup\$ Downvoter, What exactly do you think is wrong, misleading, or badly written. Remember, directly giving answers to homework problems is frowned upon here, for good reason. \$\endgroup\$ Commented May 17, 2017 at 11:01

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