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I have the following circuit:

enter image description here

The solution is this:

$$v_3=-1~\mathrm{V}\\ v_5=\frac{-2}{3}~\mathrm{V}$$

I need the calculations to find \$v_3\$.

Here follows my attempt:

enter image description here

Nodes Method

$$\left[\begin{matrix} \frac{1}{3}+\frac{1}{2+1} &0 \\\\ 0&\frac{1}{2}+1 \\\\ \end{matrix}\right] \left[\begin{matrix} V_a \\\\ V_b \\\\ \end{matrix}\right] = \left[\begin{matrix} 1\:\text{A} \\\\ -1\:\text{A} \\\\ \end{matrix}\right]$$

This solves to:

$$ \left[\begin{matrix} V_a=\frac{3}{2}\:\text{V} \\\\ V_b=\frac{-2}{3}\:\text{V} \\\\ \end{matrix}\right]$$

So $$V_{5}=V_b=\frac{-2}{3}V$$

$$V_{3}=?$$

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    \$\begingroup\$ Where is the ground? \$\endgroup\$
    – Lundin
    Jan 27, 2023 at 14:31
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    \$\begingroup\$ "Solve the circuit" is an awfully generic title. Can you please edit it to something more significant? Thank you. \$\endgroup\$ Jan 27, 2023 at 15:57
  • \$\begingroup\$ Where is \$v_A\$ and \$v_B\$? \$\endgroup\$ Jan 27, 2023 at 16:55
  • \$\begingroup\$ The ground is where I circled node O. $$V_{a}$$ and $$V_{b}$$ are the tensions respectively from O to A and from O to B. \$\endgroup\$ Jan 28, 2023 at 16:28
  • \$\begingroup\$ I see you changed the title from "Solve the circuit" to "Find the voltages in the circuit". I think that is just as vague. What about instead: "What are the voltages in this circuit with 5 resistors and a current source?" \$\endgroup\$ Jan 28, 2023 at 19:02

6 Answers 6

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This problem hinges on two rules: one, that voltage through parallel paths is equivalent and two, that current through parallel paths sums to the overall current. You should start by combining the far left resistors (they are in series, so add together). Then, look at the two left parallel paths and make a system of equations using V=iR of both paths. Because V is the same (Rule 1), you can set i1R1=i2R2. Both paths are 3ohms, so you're left with i1=i2. Because i1+i2=itotal, set 2i=1A -> i = 0.5A. Then find the voltage across the resistor using V=iR -> V=0.5A*2ohm (Note that because the voltage is dropping with respect to the direction of the arrow, V3 and V5 will be negative). Now use the same strategy for the two righthand parallel paths.

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    \$\begingroup\$ Voltage is "across" and current is "through". \$\endgroup\$
    – Andy aka
    Jan 27, 2023 at 15:43
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Hints to help the OP

As in your last question you could simplify things making it easier on the eye (and thus the brain): -

enter image description here

Then for V3, reduce the circuit on the left to find the voltage across the 3 ohm resistor: -

enter image description here

It's an easy step to find V3 now.

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This is A LOT EASIER if each component has a reference designator.

Reduce the circuit by combining the two resistors on the right and the three resistors on the left to single resistance values.

Use permutations of Ohm's Law to solve for all voltages, currents, and resistance values.

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  • \$\begingroup\$ That is actually way overcomplicating it. Because there's a current source, you only need to combine the two far left resistors to get the answers. \$\endgroup\$
    – InBedded16
    Jan 27, 2023 at 14:53
  • \$\begingroup\$ If I combine the resistors on the left I think that I lose the V3 information and the same thing with the resistor on the right where I lose the information on V5 \$\endgroup\$ Jan 27, 2023 at 14:54
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    \$\begingroup\$ @user3576105 that is not necessarily true. A big part of circuits is combining components to find characteristics such as voltage or current, and then un-combining them to apply those characteristics. It is just unnecessary to do so here. \$\endgroup\$
    – InBedded16
    Jan 27, 2023 at 15:01
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RHS:

The source current splits 2:1 in favour of the 1 ohm resistor as 2/3 A.

LHS:

The source current splits equally between the two branches as 1/2 A.

Now the voltages are easily calculated by Ohm’s Law.

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Start by de-obfuscating the schematic

A common ploy by textbook makers is to write schematics that are intentionally drawn cluttered and confusing, in order to force you to think about what is actually happening.

This ability to fix schematics is an important skill. So let's re-draw the schematic.

schematic

simulate this circuit – Schematic created using CircuitLab

Now you can see a bunch of things plain as day.

  • You can see the 1 and 2 ohm resistors in series are simply a 3-ohm resistor. Series resistances simply add. Now you have 2 functional 3-ohm resistors side by side. Parallel resistances split the current in proportion to their conductance (1/resistance) and they're equal, so half the current goes down each leg.

  • Since the resistors are the same, figuring out the total resistance of two paralleled 3-ohm resistors is easy. Now you know current and that half of current goes up each resistor chain.

  • Now that we have voltage drop from bottom to top rail, the leftmost pair of resistors is simply a resistor ladder.

  • Conductance uses the symbol G and the unit siemens. G=1/R. R=1/G. V=I/G. VG=I.

  • The two resistors on the right are paralleled. Parallel conductances simply add. So convert to conductance (0.5 siemens and 1 siemens respectively). The sum is 1.5 siemens. Convert back to resistance giving 0.666 ohms. Betcha they didn't teach you that!

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No one used matrices, yet.

To start, redraw your schematic to show current flow from left to right:

schematic

simulate this circuit – Schematic created using CircuitLab

Construct the matrix from the above:

$$\left[\begin{matrix} \frac1{R_1}+\frac1{R_3}+\frac1{R_4}+\frac1{R_5}&\frac{-1}{R_1}&\frac{-1}{R_3} \\\\ \frac{-1}{R_1}&\frac{1}{R_1}+\frac{1}{R_2}&\frac{-1}{R_2} \\\\ \frac{-1}{R_3}&\frac{-1}{R_2}&\frac{1}{R_2}+\frac{1}{R_3} \end{matrix}\right] \left[\begin{matrix} V_1\vphantom{\frac{-1}{R_3}} \\\\ V_2\vphantom{\frac{-1}{R_3}} \\\\ V_3\vphantom{\frac{-1}{R_3}} \end{matrix}\right] = \left[\begin{matrix} 0\:\text{A}\vphantom{\frac{-1}{R_3}} \\\\ 0\:\text{A}\vphantom{\frac{-1}{R_3}} \\\\ 1\:\text{A}\vphantom{\frac{-1}{R_3}} \end{matrix}\right]$$

This solves as:

$$ \left[\begin{matrix} V_1=\frac23\:\text{V}\vphantom{\frac{-1}{R_3}} \\\\ V_2=\frac76\:\text{V}\vphantom{\frac{-1}{R_3}} \\\\ V_3=\frac{13}6\:\text{V}\vphantom{\frac{-1}{R_3}} \end{matrix}\right]$$

Now work out your voltage differences of interest. I don't know what part of the world you come from, but different places put different meaning on the arrows you drew. Regardless, the answers will either be both positive or both negative depending upon your choice.

No simplifications were used: all of the unknown nodes were included. This allows the direct extraction of any question.

(Of course, \$V_0=0\:\text{V}\$. The rank-nullity theorem requires that assignment and also requires ignoring the ground node when preparing the matrix. The ground node decides which rows and columns of the Laplacian to remove to create the reduced version and explains some of the details of the left-side matrix.)

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