How do I solve using nodal analysis when there is a battery before the resistor?

Question

I need to use node analysis on the following circuit.

I am confused as to how to solve node 1 as there is a battery of 3V between the node and 4 ohm resistor. Should I solve for it?

Answer 1

First, if you are going to use nodal analysis you must select and clearly indicate which node is the reference node.

Again assuming that you are required to use nodal analysis: If you have a voltage source in your circuit that is not connected to the reference node (either directly or through other voltage sources) then you must create a supernode that encloses both of the nodes of the voltage source itself.

You can write one equation for the currents in and out of the supernode, and another equation (by inspection) for the voltage between the two nodes of the supernode.

In your specific circuit, you need to name a new node at the right end of the 3V source. Create a supernode comprising that node and V1. Start writing equations, and off you go.

Answer 2

You will need to use the superposition theorem. That is replace the voltage sources with a short subsequently and then add the resulting values.

This explains it in detail:

https://www.allaboutcircuits.com/textbook/direct-current/chpt-10/superposition-theorem/

Also: https://en.wikipedia.org/wiki/Superposition_theorem

Answer 3

Well, we have the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \text{I}_2=\text{I}_1+\text{I}_3\\ \\ \text{I}_1=\text{I}_2+\text{I}_6\\ \\ \text{I}_5=\text{I}_4+\text{I}_3\\ \\ \text{I}_4=\text{I}_5+\text{I}_6 \end{cases}\tag1 $$

When we use and apply KVL, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{10-\text{V}_1}{3}\\ \\ \text{I}_2=\frac{\text{V}_1}{8}\\ \\ \text{I}_2=\frac{\text{V}_3-\text{V}_2}{4}\\ \\ \text{I}_4=\frac{\text{V}_3}{12}\\ \\ \text{I}_5=\frac{6-\text{V}_3}{14}\\ \\ \text{V}_2-\text{V}_1=3 \end{cases}\tag2 $$