Circuit Analysis Using KCL Method: Determining Voltages at Nodes Vx and Vy, Slides of Circuit Theory

A step-by-step guide on how to determine the voltages at nodes vx and vy in a circuit using the kcl (node voltage) method. The circuit diagram, equations, and the process of solving for the unknown voltages.

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2021/2022

Uploaded on 07/05/2022

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Circuit Analysis Using KCL (node voltage) Method
Below is a circuit to analyze. We shall now determine the voltage at each node.
The first step in the analysis is to label all the nodes except for the common node. Often, the
common node is the one connected to the negative terminal of the voltage source. More often
than not, it appears as a common lead across the bottom of a circuit diagram.
Secondly, label the currents entering or leaving each node. The direction of the current arrows
is totally arbitrary. If you choose the wrong direction, the current magnitude will simply be
negative.
The next step is to write the KCL equation for each node except the common node, or for
nodes that have a voltage source between the node and common. For node Vz, we see by
inspection that its value is 10V.
For node Vy, the KCL equation is: -I1 + I2 - I3 + 2 = 0
At node Vx, the KCL equation is: I3 + I4 - 2 = 0
The last step in setting up the equations to solve is to utilize Ohm’s law to replace the currents
with V/R. (remember I = V/R) For example:
I1 = (Vy - Vz) / R1 => I1 = (Vy - 10) / 5
Why did we use (Vy - 10)? Since we chose I1 to be flowing from right to left, we imply by the
passive sign convention that Vy is > Vx. (Current flows from higher potential to lower poten-
tial.) If our current arrow is in the right direction, this will yield a positive current as we
expect.
Continuing with the other currents:
I2 = (0 - Vy) / R2 => I2 = -Vy/10
Vz Vy Vx
I1
I2
I3
I4
common node
node Z node Y node X
pf2

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Page 1

Circuit Analysis Using KCL (node voltage) Method

Below is a circuit to analyze. We shall now determine the voltage at each node.

The first step in the analysis is to label all the nodes except for the common node. Often, the common node is the one connected to the negative terminal of the voltage source. More often than not, it appears as a common lead across the bottom of a circuit diagram. Secondly, label the currents entering or leaving each node. The direction of the current arrows is totally arbitrary. If you choose the wrong direction, the current magnitude will simply be negative. The next step is to write the KCL equation for each node except the common node, or for nodes that have a voltage source between the node and common. For node Vz, we see by inspection that its value is 10V.

For node Vy, the KCL equation is: -I1 + I2 - I3 + 2 = 0

At node Vx, the KCL equation is: I3 + I4 - 2 = 0

The last step in setting up the equations to solve is to utilize Ohm’s law to replace the currents with V/R. (remember I = V/R) For example:

I1 = (Vy - Vz) / R1 => I1 = (Vy - 10) / 5

Why did we use (Vy - 10)? Since we chose I1 to be flowing from right to left, we imply by the passive sign convention that Vy is > Vx. (Current flows from higher potential to lower poten- tial.) If our current arrow is in the right direction, this will yield a positive current as we expect. Continuing with the other currents:

I2 = (0 - Vy) / R2 => I2 = -Vy/

Vz Vy^ Vx

I 1

I 2

I 3

I 4

common node

node Z

node Y node X

Page 2

(the common node is 0V)

I3 = (Vy - Vx) / R3 => I3 = (Vy - Vx)/

I4 = (0 - Vx) / R4 => I4 = -Vx/

Now substitute these expressions for the currents in the original equations.

-((Vy - 10)/5) + (-Vy/10) - (Vy - Vx)/10 + 2 = 0 @ node Vy

(Vy - Vx)/10 +(-Vx/10) - 2 = 0 @ node Vx

now we have two independent equations with two unknowns. Let’s solve.

from the equation for Vx, solve for Vy:

(Vy - Vx)/10 +(-Vx/10) - 2 = 0 @ node Vx Vy - Vx -Vx - 20 = 0 (multiply through by 10) Vy = 2Vx + 20 (gather terms, solve for Vy)

Taking this result for Vy and substituting into the equation at Vy:

-((Vy - 10)/5) + (-Vy/10) - (Vy - Vx)/10 + 2 = 0 @ node Vy -((2Vx + 20 - 10)/5) + (-2Vx - 20/10)

  • (2Vx + 20 - Vx)/10 + 2 = 0 (substitute Vy) -2/5Vx - 4 + 2 - 1/5Vx - 2 - 1/5Vx - 2
  • 1/10Vx + 2 = 0 (reduce fractions) -4/5Vx + 1/10Vx = 4 Vx = -40/ Vx = -5.71; Thus the voltage at node Vx is -5.71 volts.

Taking the voltage for Vx and putting it into the equation for Vx to solve for Vy yields:

(Vy - Vx)/10 +(-Vx/10) - 2 = 0 @ node Vx (Vy - -5.71)/10 + (- -5.71/10) - 2 = 0 (substitute Vx = -5.71) Vy = 2(-5.71) + 20 Vy = 8.58; Thus the voltage at node Vy is 8.58 volts.