# Ed's Radio (AA6ED) – Kirchhoff's Law

## Kirchhoff's Law – Node‐Voltage Method

Node‐Voltage Analysis is almost the reverse of the Branch Current Method. Using the Node‐Voltage Method, you begin by specifying branch currents at a principle node in terms of voltage drops. As defined earlier, a principle node is a junction of three or more components. The key to this method is that the voltage drop between two principle nodes is common to two different loop equations. Thus, solving the node voltage drop makes it easy to solve the two adjacent loop equations. This usually makes the Node‐Voltage method a simpler way of analyzing a circuit using Kirchhoff's Law than the Branch Current Method.

## Example: Solve a Circuit by the Node‐Voltage Method

We will revisit the circuit from figure 8 but this time solve it using the Node‐Voltage Method. The circuit, shown below in figure 9, has two principle nodes, labled A and B, with a voltage drop between them. We will focus on solving this voltage drop across R3 which will then allow us to solve the voltage drops across R1 and R2.

We begin by specifying the branch currents at principle node A. We have I1 and I2 entering the node with I3 leaving. That gives our familiar `I1 + I2 = I3` current equation. Now, however, we will attempt to specify those currents in terms of voltage drops. Since I1 is the only current flowing through R1, we can use Ohm's Law to rewrite I1 as VR1 / R1. Similarly, I2 is the only current flowing through R2, so we know that I2 = VR2 / R2. Lastly, we have I3 = VR3 / R3. From these three equations, we can rewrite our current equation in terms of voltage drops:

`I1 + I2 = I3`
`(VR1 / R1) + (VR2 / R2) = (VR3 / R3)`

We want to solve for the node voltage, VR3, but we still have two other unknowns: VR1 and VR2. We can use Kirchhoff's Voltage Law to specify each of these in terms of VR3. We know that:

`VR1 = V1 - VR3` and `VR2 = V2 - VR3`

From these, we can rewrite our current equation yet again as:

`((V1 - VR3) / R1) + ((V2 - VR3) / R2) = (VR3 / R3)`

Substituting known values from figure 9, we get:

`((130 - VR3) / 20) + ((20 - VR3) / 5) = (VR3 / 10)`

We now have one equation with one variable and can solve for VR3. Simplifying, we get:

`VR3 = 30 volts`

We can now return to our two loop equations and insert the value of VR3 to solve for VR1 and VR2:

`VR1 = V1 - VR3`
`VR1 = 130 - 30 = 100 volts`

`VR2 = V2 - VR3`
`VR2 = 20 - 30 = -10 volts`

Lastly, we solve for the branch currents using Ohm's Law across each resister:

`I1 = VR1 / R1 = 100 / 20 = 5 amps`
`I2 = VR2 / R2 = -10 / 5 = -2 amps`
`I3 = VR3 / R3 = 30 / 10 = 3 amps`

As you can see, we arrive at the same values found by using the Branch Current Method. This includes the negative values for I2 and VR2 which indicate the I2 current actually flows in the opposite direction from that we first assumed.