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I am trying to find the voltage at NODE1. My initial approach was to find the current starting from the \$15\$V source then going through \$R_1\$ and \$R_3\$ to ground. Then calculating the current from \$14\$V source through \$R_2\$ and \$R_3\$ to ground. Then adding the two currents and using Ohm's law with total current and \$R_3\$ to find voltage from NODE1 to ground. This seems plausible to me only because of KCL (total current entering a node must equal current exiting node). However my answer is incorrect as I have proved both in SPICE and on breadboard.

What am I missing? I would appreciate any suggestions that don't totally give me the answer.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Why not use superposition? Then you've just got a pair of voltage dividers. \$\endgroup\$ – Null Nov 26 '14 at 2:27
  • \$\begingroup\$ That did it. Can I assume then, for example, the 14V source "sees" the 15V source as a short? And thus the 14V source "sees" a total resistance of 105k? \$\endgroup\$ – disorder Nov 26 '14 at 3:12
  • \$\begingroup\$ Exactly. And the resistances will give you the voltage from the voltage divider equation. Then do the same for the 15V source, and add up the two contributions. \$\endgroup\$ – Null Nov 26 '14 at 3:18
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The easiest way to do this is to use superposition. Set one source to \$0\$ (i.e. shorted) and find the node voltage due to the other source, then set the other source to \$0\$ and find the node voltage due to the first source. Add the two contributions for the node voltage due to both sources.

For the case with the \$14\$V source set to \$0\$ (finding the contribution of the \$15\$V source) the circuit looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

\$R_2\$ and \$R_3\$ are in parallel since \$V_2\$ is shorted, and \$V_x\$ can be found using a simple voltage divider:

$$V_x = \frac{R_2||R_3}{R_2||R_3 + R_1}15\text{V}$$

It's a similar process for finding \$V_y\$, the contribution of the \$14\$V source to the node voltage (with the \$15\$V source set to \$0\$).

Then by superposition $$V_{\text{NODE}1} = V_x + V_y$$

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Here is a very useful trick you can use for n resistors connected to n voltage sources:

schematic

simulate this circuit – Schematic created using CircuitLab

$$V_\mathrm{M1} = \left (\frac{V_1}{R_1} + \frac{V_2}{R_2} + \cdots + \frac{V_n}{R_n} \right ) \times (R_1 || R_2 || \cdots || R_n)$$

Where $$R_1 || R_2 || \cdots || R_n = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n} }$$

So, it can also be written:

$$V_\mathrm{M1} = \frac{\frac{V_1}{R_1} + \frac{V_2}{R_2} + \cdots + \frac{V_n}{R_n} } {\frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n} }$$

In your case:

  • R1 = 10K, V1 = +15V
  • R2 = 10K, V2 = 0V
  • R3 = 100K, V3 = +14V

$$R_1 || R_2 || R_3 = \frac{100}{21}\ \mathrm{kΩ} $$

and I'll leave the rest for you...

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