My book says the following:

As noted previously, every method of circuit analysis must satisfy KVL, KCL, and the device iāv relationships. In developing the node-voltage equations in Eqs. (3ā4), 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. (3ā3). 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.

## Best Answer

Picture this:

^{simulate this circuit ā 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.