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I’m very confused when I have to find voltage at different resistors terminals. How do I do it? Series-parallel circuit. For example v01 at R2 terminals and V02 at R4 terminals

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    \$\begingroup\$ Can you calculate the total current draw from the 12 V supply? Can you calculate how much current is flowing in each branch? \$\endgroup\$
    – winny
    Commented Aug 22, 2023 at 18:39
  • \$\begingroup\$ I know that the total current is 6A, I just used Ohms Law, But I thought I need to use voltage divider. Anyway, I dont know How to calculate current in each branch. do I need to use Kirchoff? \$\endgroup\$ Commented Aug 22, 2023 at 18:43
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    \$\begingroup\$ The total current will be 6mA = 0.006A not 6A. Yes, you need to use Kirchhoff's. \$\endgroup\$
    – G36
    Commented Aug 22, 2023 at 18:48
  • \$\begingroup\$ Of course, you can use voltage divider \$V_{01} = V_I \frac{R_2||(R_3 + R_4)}{R_1 + R_2||(R_3 + R_4} = 6V\$ and \$V_{02} = V_{01}\frac{R_4}{R_3 + R_4 } = 3V\$ \$\endgroup\$
    – G36
    Commented Aug 22, 2023 at 18:53
  • \$\begingroup\$ Everything is a voltage divider if you allow difficult enough expressions, but in this case you could solve it as just voltage dividers. If I were your teacher, I wouldn’t mind. Please try. \$\endgroup\$
    – winny
    Commented Aug 22, 2023 at 18:54

2 Answers 2

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You can approach this from the right to the left using elemental voltage divider methods. Or set up two equations using KCL and solve that using Cramer's rule, if solving by hand or just using a solver, otherwise. And that's not the limit of approaches you can use, by any stretch.

Let's go from right backwards to the left.

On the right side, you have a voltage divider formed by \$R_3\$ and \$R_4\$. Here, \$V_{O_2}=V_{O_1}\cdot\frac1{1+\frac{R_3}{R_4}}\$ and its load impedance is \$R_{O_2}=R_3+R_4\$.

Next, you have a voltage divider formed by \$R_1\$ and \$R_2\$ but with an added load (in parallel with \$R_2\$) of \$R_{O_2}\$. So here, \$V_{O_1}=V_{I}\cdot\frac1{1+\frac{R_1}{R_2\,\vert\vert\, R_{O_2}}}\$.

So the complete symbolic-form answer is:

$$\begin{align*} V_{O_1}&=V_{I}\cdot\frac1{1+\frac{R_1}{R_2\,\vert\vert\, \left(R_3+R_4\right)}} \\\\ V_{O_2}&=V_{O_1}\cdot\frac1{1+\frac{R_3}{R_4}}=V_{I}\cdot\frac1{1+\frac{R_1}{R_2\,\vert\vert\, \left(R_3+R_4\right)}}\cdot\frac1{1+\frac{R_3}{R_4}} \end{align*}$$

The main thing to note is that the right-side divider is itself unloaded, so it solves as a simple voltage divider problem, but that this right-side divider does load down the left-side divider, and that fact has to be accounted for when working out the left side.

In your approach, you find that the total load impedance is \$2\:\text{k}\Omega\$. If so, then you should recognize that the source current will be \$\frac{12\:\text{V}}{2\:\text{k}\Omega}=6\:\text{mA}\$. This means that \$R_1\$ must drop \$6\:\text{V}\$ across it, leaving \$V_{O_1}=6\:\text{V}\$. That current will equally split between \$R_2\$, which is \$2\:\text{k}\Omega\$ and the series pair, \$R_3+R_4=2\:\text{k}\Omega\$. So there will be \$3\:\text{mA}\$ through \$R_3+R_4\$ and each will drop exactly \$3\:\text{V}\$. So, \$V_{O_2}=3\:\text{V}\$. So, your approach works fine.

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Here's how I "see" such circuits:

schematic

simulate this circuit – Schematic created using CircuitLab

Then

schematic

simulate this circuit

Ok, so VO1 is a mid-point of a 1k:1k voltage divider from 12V - so VO1=6V.

Now that we know VO1, we can use that fact:

schematic

simulate this circuit

Once we know VO1, we stick a voltage source in there - it reflects the potential present at that point in the circuit. Adding this source to the original circuit changes nothing, since no current flows through that source. VO1 is now held fixed at 6V by the source, and we can remove any components in parallel with that source - they don't change any voltages.

After removing R2, there then is a current flowing through VO1, but so what - it's an ideal voltage source. The voltage is fixed no matter the current. Then we're left with VO2 being is 6V divided down by another 1k:1k divider - and is 3V.

The values in this circuit were chosen specifically to facilitate the transformations I've performed above. It's not a coincidence that the numbers "work out" so nicely. It's by design in this case. That's what authors of teaching materials - be it textbooks or teachers - deal with :)

This circuit is a "cut down" version of the R-2R ladder used in some DAC converters.


If you'd want to simulate the circuit, only one variant should be left, and the others removed from the schematic, since:

  • SPICE disallows repeated reference designators, and
  • the nodes labeled GND (triangle symbol), VO1 and VO2 are respectively connected together between the circuit variants as shown.

That's why simulation won't work as-is. But as soon as you delete two circuits that you don't care about from a simulation, and leave one left, everything will work as expected. CircuitLab does "dynamic" DC analysis - it shows DC operating point without having to run simulation. Just mouse over points in the circuit and voltages and/or currents will be displayed.

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  • \$\begingroup\$ Thank you everyone for the answears!! It did clarify things for me \$\endgroup\$ Commented Aug 24, 2023 at 17:58

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