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I am trying to find the voltage at V2 using the wire label in the Qucs-s simulator, powered by Ngspice. I have used Ohm’s law (V=IR) to work out the the voltage across R1 which was 1.66 volts;, but the simulator is claiming that there is around 8.3 volts at V2.

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I also tried using the voltage divider equation, but where did the simulator get the 8.3 V?

Russell Borogove
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Ekene Nwokedi
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    \$V_2 = 10V * \frac{300\Omega + 200\Omega}{100\Omega + 300\Omega + 200\Omega} = 8.333V \$ – G36 Apr 30 '23 at 16:54
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    Basically voltage across resistor R1 that you calculated is correct but, V2 is not the voltage across the resistor. V2 = 10V - drop across R1. – sai Apr 30 '23 at 16:59
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    what voltage were you expecting? – jsotola Apr 30 '23 at 17:01
  • The simulator will report the voltage at V2 relative to Ground. – Peter Bennett Apr 30 '23 at 17:12
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    Now I am old... while I know the value of simulation for both education and engineering, I still don't see enough value for me in order to learn how to simulate a circuit. And now someone SIMULATES A DAMN VOLTAGE DIVIDER !!! – fraxinus May 01 '23 at 08:48
  • You simply threw way the bit you wanted to keep, and kept the rest... –  May 01 '23 at 11:56

4 Answers4

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Yes, there's 1.67V across R1. And R1 sits next to +10V. So on the other side of it, at V2, is 10 - 1.67 = 8.33V.

hobbs
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Notice that the total input resistance is given by:

$$\text{R}_\text{in}=\text{R}_1+\text{R}_2+\text{R}_3=100+200+300=600\space\Omega\tag1$$

So, the input current is given by:

$$\text{I}_\text{in}=\frac{\text{V}_\text{in}}{\text{R}_\text{in}}=\frac{10}{600}=\frac{1}{60}\approx0.016\space\text{A}\tag2$$

So, the voltage \$\text{V}_2\$ is given by:

$$\text{V}_2=\text{V}_\text{in}-\text{V}_{\text{R}_1}=\text{V}_\text{in}-\text{R}_1\cdot\text{I}_\text{in}=10-100\cdot\frac{1}{60}=\frac{25}{3}\approx8.3\space\text{V}\tag3$$

Jan Eerland
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I redrew your circuit to make it easier to understand. Hope this helps visualize it: circuit

Rodo
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    if you are going to redraw the circuit to make it clearer, then draw the resistors vertically aligned ... V1 at top, ground at bottom ... that makes it easier to visualize voltage levels at the nodes – jsotola Apr 30 '23 at 17:57
  • https://i.imgur.com/gZemr3e.png – Hatman May 01 '23 at 20:36
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I've made two minor corrections to your statement here:

I am trying to find the potential at V2 using the wire label in the Qucs-s simulator, powered by Ngspice. I have used Ohm’s law (V=IR) to work out the the voltage across R1 which was 1.66 Ω V, but the simulator is claiming that there is around 8.3 volts at V2.

Yes, that's all correct.

The word "voltage" means the difference between the potential at one place and the potential at another. Specifically, "the voltage across R1" means the potential on one side of R1 minus the potential on the other side of R1.

The potential on the left side of R1 (which you've labeled as "vfrore") is 10 V, and the potential on the right side of R1, as the simulator is telling you, is 8.333 V. This means that the voltage across R1 is 10 V - 8.333 V, or 1.667 V. That matches the calculation that you did using Ohm's law.

Cassie Swett
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