4th Week: "Nodal Analysis - AC"

"Nodal Analysis"


The basis of nodal analysis is Kirchhoff’s current law. Since KCL is validfor phasors, we can analyze ac circuits by nodal analysis.

Example:

Solution,

1. Convert the circuit to phasor domain:

Given a frequency of 1MHz, the radian frequency is 1MHz*2*pi = 6.28Mrad/s. Thus we have:

2. Solve for the desired value in the phasor domain. Let's use node voltage to find the voltage over the capacitor. Choose the bottom node as the reference and the top node as "V", giving us the following circuit:

The single KCL equation we get is:
(V - 12<0)/1k + V/-j795.8 - 20m<0 = 0

Solve for V. First, factor out V and put the constants on the right side.
V(1/1k +j/795.8) = 20m + 12/1k

Now add the constants and rearrange:
V ( 1E-3 +j1.26E-3) = 32m
V = 32m/(1E-3 + j1.26E-3)

Next, to find the final answer, we could either (a) multiply numerator and denominator by the conjugate of the denominator or (b) convert the numerator and denominator to polar format and then divide. I choose to do the latter:

V = 32m<0 / 1.6m<51.6
V = 20 < -51.6

3. Convert answer to time-domain. Simply copy over the form (cos/sin) and the frequency (1MHz) and fill in our solution for magnitude and phase:
Vc(t) = 20 sin(wt - 51.6 degrees)



Reflection:

• I learned that all the concepts and process in solving the unknowns in Ac Analysis are the same with Dc Analysis. It only involves complex numbers.

Videos:

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That's all. Thank You for visiting my blog.

GOD Bless! :)

By:

AYALA, ARNY  S.   BSECE -3

ECE 321

Professor:

ENGR. JAY S. VILLAN, MEP - EE