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I recently asked this question:

Find the currents and voltages in the circuit shown in Fig. 2.28. enter image description here Answer: \$v_1 = 6 \ \text{V}\$, \$v_2 = 4 \ \text{V}\$, \$v_3 = 10 \ \text{V}\$, \$i_1 = 3 \ \text{A}\$, \$i_2 = 500 \ \text{mA}\$, \$i_3 = 2.5 \ \text{A}\$

It seems to me that we have two loops here. For the first loop, we have \$-10 \ \text{V} + 2i_1 + 8i_2 = 0\$. For the second loop, we have \$4i_3 - 6 \ \text{V} - 8i_2 = 0\$, where I have \$8i_2\$ by Ohm's law instead of \$-8i_2\$, because the current \$i_2\$ is actually going from + to - for the \$8 \ \Omega\$ resistor (and despite this being counter to the clockwise direction of the loop, my understanding is that this results in a positive current for the purpose of the Ohm's law calculation). Finally, if we designate the top middle node, then we have that \$i_1 - i_2 - i_3 = 0\$. Is my reasoning here correct?

I was informed that my reasoning is correct. However, when I perform the calculations, it seems clear that I'm missing something (an equation somewhere):

$$-10 \ \text{V} + v_1 + v_2 = 0 = -10 \ \text{V} + 2i_1 + 8i_2 \ \Rightarrow i_2 = \dfrac{10 \ \text{V} - 2i_1}{8} = \dfrac{5 \ \text{V} - i_1}{4}$$ $$-6 \ \text{V} - v_2 + v_3 = 0 = -6 \ \text{V} - (8i_2) + 4i_3 = -6 \ \text{V} - 8i_2 + 4i_3$$ $$i_1 - i_2 - i_3 = 0$$

$$-6 \ \text{V} - 8\left( \dfrac{5 \ \text{V} - i_1}{4} \right) + 4i_3 = 0 \ \Rightarrow -6 \ \text{V} - 10 \ \text{V} + 2i_1 + 4i_3 = 0 \\ \Rightarrow 2i_1 = 16 \ \text{V} - 4i_3 \ \Rightarrow i_1 = 8 \text{V} - 2i_3$$

$$(8 \ \text{V} - 2i_3) - i_2 - i_3 = 0 \ \Rightarrow 8 \ \text{V} - 3i_3 - i_2 = 0 \ \Rightarrow i_2 = 3i_3 - 8 \ \text{V}$$

What am I missing here?

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  • \$\begingroup\$ I don't want to go through all that, right now (busy.) But all you have is a simple unknown node. The outflowing currents are x/2 + x/4 + x/8 and the inflowing currents are 10/2 + (-6)/4 + 0/8. Set them equal (KCL) and the only solution is x = 4 V, right? Every other possible approach must result in the same answer. \$\endgroup\$
    – jonk
    Apr 15, 2022 at 18:04
  • \$\begingroup\$ A little tip: You have three linear independent equations for three variables (\$i_1, i_2, i_3\$), therefore the system has a unique solution, i.e. you're not missing any equations. \$\endgroup\$
    – some_user
    Apr 15, 2022 at 18:24
  • \$\begingroup\$ You set things up right at the top. Just checked. See here. \$\endgroup\$
    – jonk
    Apr 15, 2022 at 18:34

3 Answers 3

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Please see the answer below.Be careful about sign conventions.

enter image description here

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To answer your question, I believe the last equation you are missing is the current summing equation.

$$i_1 = i_2 + i_3$$

In addition to the two equations you already have:

$$6 = 4 \cdot i_3 - 8 \cdot i_2$$ $$10 = 2 \cdot i_1 + 8 \cdot i_2$$

You now have all the information you need to solve the system of equations.

You can also solve the problem using the current summing approach from the get go, as there is only one unknown voltage in the system. Let's call the voltage at the top node where the currents meet Vx, and the voltage at the bottom node can be 0V or ground. Our resulting current summing equation is:

$$i_1 = i_2 + i_3$$

Expanding this using ohms law we have:

$$i_1 = \frac{10 - v_x}{2}$$ $$i_2 = \frac{v_x + 6}{2}$$ $$i_3 = \frac{v_x}{8}$$

Plugging this in we have: $$\frac{10 - v_x}{2} = \frac{v_x + 6}{2} + \frac{v_x}{8}$$

Simplifying:

$$40 - 4 \cdot v_x = 2 \cdot v_x + 12 + v_x$$ $$28 = 7 \cdot v_x$$ $$v_x = 4V$$

From here it is very easy to work backwards and obtain all of the other currents using the original equations above...

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  • \$\begingroup\$ I've written the current summing equation as \$i_1 - i_2 - i_3 = 0\$. \$\endgroup\$ Apr 16, 2022 at 16:24
  • \$\begingroup\$ @ThePointer that is an equivalent statement yes. \$\endgroup\$ Apr 17, 2022 at 17:57
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To resolve the problem in your approach, you need to use \$i_1 - i_2 - i_3 = 0\$ and solve for \$i_2\$, plug that in your last equation (\$i_2 = 3i_3 - 8\$) and use \$i_1 = 8 - 2i_3\$ from the line above. Then you can solve for \$i_3\$.

More general: You have three equations: $$\begin{align} -10+2i_1+8i_2 &= 0 \quad(i) \\ 4i_3 - 6 -8i_2 &= 0 \quad (ii)\\ i_1 - i_2 -i_3 &= 0 \quad (iii) \end{align} $$ Since there are three lin. ind. equations for three variables, there exists a unique solution.
You could solve this system in many ways, I'll outline one possible way here:

  1. Solve (i) for \$i_2\$
  2. Plug the solution from before in (ii) and solve for \$i_3\$
  3. Plug the solutions from before in (iii) and you get a solution for \$i_1\$
  4. Get the solutions for \$i_2\$ and \$i_3\$ from the Identities derived in 1. and 2.
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  • \$\begingroup\$ I already used \$i_1 - i_2 - i_3 = 0\$ in $$(8 \ \text{V} - 2i_3) - i_2 - i_3 = 0 \ \Rightarrow 8 \ \text{V} - 3i_3 - i_2 = 0 \ \Rightarrow i_2 = 3i_3 - 8 \ \text{V},$$ so using it in the result would be double-counting. \$\endgroup\$ Apr 16, 2022 at 16:34
  • \$\begingroup\$ Ah yes, I didn't see that. But I don't understand what you mean by double-counting, the equation doesn't lose its validity when applied. \$\endgroup\$
    – some_user
    Apr 17, 2022 at 17:20
  • \$\begingroup\$ You're right – I had a misunderstanding of the substitution process. \$\endgroup\$ Apr 17, 2022 at 17:23
  • \$\begingroup\$ Of course there was some useless effort in my approach, as you have already used the equation. You can use the three equations you have derived \$i_2 = 3i_3 - 8, \; i_1 = 8 - 2i_3, \; i_2 = 20 - 4i_1\$ to solve for the currents. \$\endgroup\$
    – some_user
    Apr 17, 2022 at 17:25

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