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EE2001 AC Circuits for E-learning Week

by nosper on March 22nd, 2011

Key words for  AC Circuits:

Math Background:  


AC Circuit Analysis Method:

Nodal Analysis (Same with DC)

Mesh Analysis (Same with DC)

AC superposition Principle (Similar with DC)

Power Concept:



QA section (Frequently asked questions):


1) For the series RLC circuit, is it true that for the i(t), the If  is equals to zero? And for v(t), the Vf is equals to the voltage at t=infinity across the capacitor?

For 1), the current and voltage should both be zero at the end. The R will consume the power from the C and L until there is no power left in the capacitor and inductor, which means the If and Vf are both zero.

2) Under what conditions do we need to add the Vf  and if into the v(t) or i(t) equation in the RL/RC circuit?

For 2), there is a general solution for step and natural response in the RL/RC circuit (x=xf+[x0-xf]e power(-t-t0/tao)). You can check this part in your lecture notes. The general solution works for any first-order RL/RC circuits with step or natural response.


3) I know that for apparent power, there are 2 formulas, S = V x I* and S = 1/2 x V x I* ? May i know under what situation do we need to use each other the following formula? secondly, what does the * represent in AC circuit?

For 3), First, I* is the conjugate value of I. (example 1: I=2+3j, then I* = 2-3j; example 2: I=3, then I*=3)

Base on the definition, apparent power is V_rms multiply by I_rms. As we know, I_rms is a real number, which means I_rms*=I_rms (see the above example 2).

I believe the following equation will make you understand very well.

S = |V x I*|, V=V_rms (cos + j sin) = V_p/root(2) (cos + j sin).   V_p is the peak value of V.    V_rms is the RMS value of V.

I=I_rms (cos + j sin) = I_p/root(2) (cos + j sin).   I_p is the peak value of I.  I_rms is the RMS value of I.

Then, we have

S = |V x I*| =| V_rms (cos + j sin) x( I_rms (cos + j sin) )*|

              =| V_rms x I_rms *|  |(cos + j sin) | |(cos + j sin) *|

              = |V_rms x I_rms*| x 1×1

              = V_rms x I_rms*

              = V_rms x I_rms      

S = |V x I*| =| (V_p/root(2) (cos + j sin)) x (I_p/root(2) (cos + j sin))*|

                  = …..

                        =|1/2 x V_p x I_p*|

                  =|1/2 x V_p x I_p|

                  = 1/2 x V_p x I_p

This should explain to you why there is a ½  in the second formula.  

In your question,  S = V x I*  : both of V and I should be  RMS values.

                                      S = 1/2 x V x I*  : both of V and I should be  peak values.



E-learning Week Solution:  

AC part 1: Tutorial 6

AC part 2: Tutorial 7


Other AC Circuit  Reference: Wiki

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