# Using KVL in nodal analysis

My book says the following:

As noted previously, every method of circuit analysis must satisfy KVL, KCL, and the device i–v relationships. In developing the node-voltage equations in Eqs. (3–4), it may appear that we have not used KVL. However, KVL is satisfied because the equations $v_1 = v_1$, $v_2=v_A-v_B$, and $v_3=v_B$ were used to write the right side of the element equations in Eqs. (3–3). The KVL constraints do not appear explicitly in the formulation of node equations, but they are implicitly included when the fundamental property of node analysis is used to write the element voltages in terms of the node voltages.

I don't see why this is part of the above excerpt is true:

The KVL constraints do not appear explicitly in the formulation of node equations, but they are implicitly included when the fundamental property of node analysis is used to write the element voltages in terms of the node voltages.

How is KVL used to write the element voltages in terms of the node voltages? Note that I'm not asking how to write the element voltages in terms of the node voltages. I'm asking why the author claims KVL plays a role in this.

Picture this: simulate this circuit – Schematic created using CircuitLab

It looks pretty obvious that $V_a = V_1 + V_2$, but this is technically a form of KVL. If you take a loop clockwise, you've got:

$V_a - V_1 - V_2 = 0$

which is the same thing that your intuition tells you.

Essentially, whenever you make one of those substitutions, you're making a loop and applying KVL around it. The only weird thing is that not all of the pieces of your loop need to be components in the circuit - it's perfectly valid to make one of them a voltage jump, like this example.

EDIT:

You asked for another circuit. Here's my best shot: simulate this circuit

If we go up the voltage supply and down the 'jump' in loop 1, we get $V_a - V_1 = 0$ or $V_1 = V_a$.

If we 'jump' to V1, move across R1, and 'jump' from V2 to ground, we get $V_1 - V_{R1} - V_2 = 0$ or $V_{R1} = V_1 - V_2$.

If we 'jump' to V2 and move back down R2, we get $V_2 - V_{R2} = 0$ or $V_{R2} = V_2$.

Hopefully those three examples make sense.

• I added a more realistic example. Does it help? – Greg d'Eon Sep 21 '14 at 1:25

How is KVL used to write the element voltages in terms of the node voltages?

A KCL equation at a node is formally a sum of all the current variables entering set equal to the sum of all the current variables exiting the node. For example

$$i_{R1} + i_{R2} = i_{S1} + i_{R3}$$

But, in order to solve for the node voltages, the currents must be expressed in terms of the node voltages.

This is where KVL enters the picture. If, for example, $i_{R1}$ is the current through a resistor that is connected between node A and node B and we designate the current through the resistor to be from node A to node B, then by KVL, the voltage across the resistor is

$$v_{R1} = v_A - v_B$$

and so the current through the resistor is

$$i_{R1} = \frac{v_{R1}}{R_1} = \frac{v_A - v_B}{R_1}$$

Thus, KVL is explicitly used to write the voltage across the resistor

$$V_A = V_{R_1} + V_B$$

in order to write an equation for the current variable in the KCL equation.