# Finding node voltage of a “Y” resistor network

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.

simulate this circuit – Schematic created using CircuitLab

• Why not use superposition? Then you've just got a pair of voltage dividers. – Null Nov 26 '14 at 2:27
• 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? – disorder Nov 26 '14 at 3:12
• 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. – Null Nov 26 '14 at 3:18

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:

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

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

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:

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

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