# How to find voltage and current in the given circuit?

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?

• 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? May 11, 2017 at 17:37
• 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.
– jonk
May 11, 2017 at 21:10

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:

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.

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

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

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

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.

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.

• 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. May 17, 2017 at 11:01