## 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 R_{3} which will then allow us to solve the voltage drops across R_{1} and R_{2}.

We begin by specifying the branch currents at principle node *A*. We have I_{1} and I_{2} entering the node with I_{3} leaving. That gives our familiar

current equation. Now, however, we will attempt to specify those currents in terms of voltage drops. Since I`I _{1}` +

`I`=

_{2}`I`

_{3}_{1}is the only current flowing through R

_{1}, we can use Ohm's Law to rewrite I

_{1}as V

_{R1}/ R

_{1}. Similarly, I

_{2}is the only current flowing through R

_{2}, so we know that I

_{2}= V

_{R2}/ R

_{2}. Lastly, we have I

_{3}= V

_{R3}/ R

_{3}. From these three equations, we can rewrite our current equation in terms of voltage drops:

`I _{1}` +

`I`=

_{2}`I`

_{3}`(``V`_{R1} / `R`_{1}) + (`V`_{R2} / `R`_{2}) = (`V`_{R3} / `R`_{3})

We want to solve for the node voltage, V_{R3}, but we still have two other unknowns: V_{R1} and V_{R2}. We can use Kirchhoff's Voltage Law to specify each of these in terms of V_{R3}. We know that:

and `V _{R1}` =

`V`-

_{1}`V`

_{R3}`V`_{R2} = `V`_{2} - `V`_{R3}

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

`((`

`V _{1}` -

`V`) /

_{R3}`R`) + ((

_{1}`V`-

_{2}`V`) /

_{R3}`R`) = (

_{2}`V`/

_{R3}`R`)

_{3}Substituting known values from figure 9, we get:

`((130 - `

`V _{R3}`) / 20) + ((20 -

`V`) / 5) = (

_{R3}`V`/ 10)

_{R3}We now have one equation with one variable and can solve for V_{R3}. Simplifying, we get:

`V _{R3}` = 30 volts

We can now return to our two loop equations and insert the value of `V _{R3}` to solve for

`V`and

_{R1}`V`:

_{R2}`V _{R1}` =

`V`-

_{1}`V`

_{R3}`V`_{R1} = 130 - 30 = 100 volts

`V _{R2}` =

`V`-

_{2}`V`

_{R3}`V`_{R2} = 20 - 30 = -10 volts

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

`I _{1}` =

`V`/

_{R1}`R`= 100 / 20 = 5 amps

_{1}`I`_{2} = `V`_{R2} / `R`_{2} = -10 / 5 = -2 amps

`I`_{3} = `V`_{R3} / `R`_{3} = 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 I_{2} and V_{R2} which indicate the I_{2} current actually flows in the opposite direction from that we first assumed.