Blog #32 - Calculation of a DC circuit using nodal analysis

As part of my problem series in this video, I will be analyzing the two loop circuit with multiple resistors and 2 power supplies using nodal analysis. Once I am finished with this series of problems I will be posting them in my Stan store, at this web address…https://stan.store/GVB

Calculate the current through each of the resistors in this DC circuit using Nodal Analysis or the branch-current method of solution. 

I’m going to re-draw the circuit slightly just to make the node more obvious.

Step #1 Label the Circuit … all nodes. One of the nodes…node A, is chosen as the reference node. It can be thought of as a circuit ground, which is at zero voltage or ground potential. 

Node B and node D are already known to be at the potential of the source voltages. The voltage at node C the voltage VC is unknown.

Let’s assume that VC > the voltage at node B and VC > the voltage at node D when all three currents are drawn arbitrarily… remember, these directions are arbitrary and may change depending on the outcome of the mathematics.

The direction of I1, I2, and I3 is assumed to be emanating from node C, and toward the reference node A.

Step #2 Write Kerckhoff's current law at Node C…I1 + I2 + I3 = 0

Step #3 Express Currents in Terms of Circuit Voltages Using Ohm’s Law

I1 = V1/R1 = (VC - 8)/2, 

I2 = V2/R2 = (VC - 24)/1, and 

I3 = V3/R3 = VC/4.

Substituting the current equations obtained in Step 3 into Kerckhoff’s Current Law of Step 2, we find I1 + I2 + I3 = 0 becomes 

(VC - 8)/2 + (VC - 24)/1 + VC/4 = 0. removing the denominators by multiplying the equation by 4…

and removing the brackets…gives us this 2VC - 16 + 4VC - 96 + VC = 0

Bringing all of the unknowns to the left-hand side of the equation gives us…2VC + 4VC + VC = 112 which reduces to…this 

7VC = 112 and this simple equation can be solved to obtain VC = 16 Volts.

Solving for the current is very simple… All we have to do is substitute 16 for the voltage VC…

in our equation for I1…we get 4 Amps and in our equation for I2…we get -8 Amps  and in our equation for I3…we get 4 Amps 

And not surprisingly we get the same answers that we have previously found for the currents. Noticed that for I2 we obtained an answer of -8 Amps which means we assume the wrong direction in the beginning and this means that this current is actually 8 Amps flowing in the other direction.

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