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I need to find node voltage at A. This seems like a trivial question, but when I set up the resistive voltage divider equations they are not equal to each other. Am I doing something wrong?

My equations are \$V_A = V_1(R3/(R3+R1)) = V_2(R3/(R3+R2))\$

schematic


I tried solving this problem with an LTSpice simulation and with the superposition method, but my answers did not match (although very close to each other). Is this because the simulator uses an approximation? Can someone compare results?

Schematic

Simulation

Superposition Method

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  • \$\begingroup\$ What's the current through R1 with respect to Va? What's the current through R2 with respect to Va? Current through R3 wrt these two currents? \$\endgroup\$ – BEE Jan 31 at 0:35
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No, you can't use the voltage divider formula this way. R3 is not in series with either of the other two resistors.

You could use superposition or the mesh-current method.

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  • \$\begingroup\$ After simulating the circuit on LTSpice, Va was shown to be 1V. However, when I used the superposition method, I got Va = 0.967V. \$\endgroup\$ – Hector Jan 31 at 18:38
  • \$\begingroup\$ Show. your. work. \$\endgroup\$ – Elliot Alderson Jan 31 at 19:50
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One way to check your solution is to use a free SPICE simulator like LTSPICE to sketch your circuit.

enter image description here

Then use the netlist file generated and an online symbolic circuit analysis tool, CircuitNav, to solve the equations to check your answer.

enter image description here

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  • \$\begingroup\$ After simulating the circuit on LTSpice, Va was shown to be 1V. However, when I used the superposition method, I got Va = 0.967V. \$\endgroup\$ – Hector Jan 31 at 18:38
  • \$\begingroup\$ @Hector, I'd like to know if you know the thevenin and norton equivalent circuit methords yet. If not, how about KCL? Or, could you please show your current work, then we start from there. \$\endgroup\$ – X J Jan 31 at 18:55
  • \$\begingroup\$ Of course I do. \$\endgroup\$ – Hector Jan 31 at 22:55
  • \$\begingroup\$ @Hector, good job! If you use fractions instead of approximated decimals, you will get exactly 1V. \$\endgroup\$ – X J Feb 1 at 2:15
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If you have two voltage sources (\$V_1\$ and \$V_2\$), each with their series resistances, (\$R_1\$ and \$R_2\$), then tied together at a point called \$V_x\$, you find:

$$V_x=\frac{V_1\cdot R_2+V_2\cdot R_1}{R_1+R_2}$$

The above situation is simply a resistor divider between two voltage sources. Easy.

If you have three voltage sources (\$V_1\$, \$V_2\$, and \$V_3\$), each with their series resistances, (\$R_1\$, \$R_2\$, and \$R_3\$), then tied together at a shared point called \$V_x\$, you find:

$$V_x=\frac{V_1\cdot R_2\cdot R_3+V_2\cdot R_1\cdot R_3+V_3\cdot R_1\cdot R_2}{R_1\cdot R_2+R_2\cdot R_3+R_1\cdot R_3}$$

The above situation is simply three resistors starred together at \$V_x\$ with each driven by separate voltage sources.

This is exactly your situation. So you can use the above "Thevenin source combining" equation for your needs. (Note that your \$V_3=0\:\text{V}\$, though.)


The above process generalizes, readily:

$$\begin{align*} V_x&=\frac{\sum^N_{i=1}\left[V_i\cdot\prod^N_{j\ne i} R_j\right]}{\sum^N_{i=1}\left[ \prod^N_{j\ne i} R_j\right]}\\\\R_{x}&=\frac{\prod^N_{i=1} R_i}{\sum^N_{i=1}\left[ \prod^N_{j\ne i} R_j\right]} \end{align*}$$

Where \$V_x\$ is the resulting Thevenin voltage and \$R_x\$ is the resulting Thevenin resistance (which you don't need, but I'm providing anyway.)

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