# Download Circuit Analysis I with MATLAB Computing and by Steven Karris PDF

By Steven Karris

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Extra info for Circuit Analysis I with MATLAB Computing and Simulink/SimPowerSystems Modeling

Sample text

Also, the circuit constants are the same and thus the natural response has the form v Cn = k 1 e –2t + k 2 e –3t . We will find the forced (steady-state) response using phasor circuit analysis where  = 1 , jL = j20 , – j  C = – j120 , and 100 cos t  100 0 . The phasor circuit is shown below. 100  j20  VS +  – j 120  +  VC V S = 100 0 V Using the voltage division expression we obtain – j120 – j120 – 90  100 0- = 60 2 – 135 V C = ---------------------------------------- 100 0 = -------------------------- 100 0 = 120 --------------------------------------------------100 + j20 – j120 100 + j100 100 2 45 and in the t – domain v Cf = 60 2 cos  t – 135  .

2 Response of Parallel RLC Circuits with AC Excitation The total response of a parallel RLC circuit that is excited by a sinusoidal source also consists of the natural and forced response components. The natural response will be overdamped, critically damped or underdamped. The forced component will be a sinusoid of the same frequency as that of the excitation, and since it represents the AC steadystate condition, we can use phasor analysis to find the forced response. We will derive the total response of a parallel RLC circuit which is excited by an AC source with the following example.

This results in the underdamped or oscillatory natural response and has the form vn  t  = e –P t  k 1 cos  nP t + k 2 sin  nP t  = k 3 e –P t  cos  nP t +    If a second order circuit is neither series nor parallel, the natural response if found from yn = k1 e or or s1 t + k2 e s2 t yN =  k1 + k2 t  e yn = e – t s1 t  k 3 cos t + k 4 sin  t  = e – t k 5 cos  t +   depending on the roots of the characteristic equation being real and unequal, real and equal, or complex conjugates respectively.