EE2001 AC Circuits for E-learning Week

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 I_f  is equals to zero? And for v(t), the V_f 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 V_f  and i_f 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 1x1

              = 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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