Alexander-Sadiku Fundamentals of Electric Circuits Chapter 8 ...

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8.4 Step-Response SeriesRLC Circuits (3)Example 4Having been in position for a long time, theswitch in the circuit below is moved to position bat t = 0. Find v(t) and v R (t) for t > 0.Please refer to lecture or textbook for more detail elaboration.Answer: v(t) = {10 + [( 2cos3.464tv R (t)= [2.31sin3.464t]e 2t V1.1547sin3.464t)e 2t]} V158.5 Step-Response ParallelRLC Circuits (1)The step responseis obtained by thesudden applicationof a dc source.The 2ndorder ofexpression2d i2dt1RCdidtiLCIsLCIt has the same form as the equation for source-free parallelRLC circuit.The same coefficients (important in determining thefrequency parameters).Different circuit variable in the equation.168
8.5 Step-Response ParallelRLC Circuits (2)The solution of the equation should have two components:the transient response v t(t) & the steady-state response v ss(t):i( t)i ( t)i ( t)tThe transient response i tis the same as that for source-free cases1ts2titt)A1e A2e( (over-damped)i )tt( t)( A1 A2te(critical damped)tit( t)e ( A1 cosdtA2sindt)ss(under-damped)The steady-state response is the final value of i(t).i ss(t) = i( ) = I sThe values of A 1and A 2are obtained from the initial conditions:i(0) and di(0)/dt.178.5 Step-Response ParallelRLC Circuits (3)Example 5Find i(t) and v(t) for t > 0 in the circuit shown incircuit shown below:Please refer to lecture or textbook for more detail elaboration.Answer: v(t) = Ldi/dt = 5x20sint = 100sint V189
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Alexander-Sadiku Fundamentals of Electric Circuits Chapter 8 ...

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