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A sample circuit for AC analysis
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Figure 442 An AC circuit
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The rst step in the analysis of an AC circuit is to note the frequency of the sinusoidal excitation Next, all sources are converted to phasor form, and each circuit element to impedance form This is illustrated in the phasor circuit of Figure 442 At this point, if the excitation frequency, , is known numerically, it will be possible to express each impedance in terms of a known amplitude and phase, and a numerical answer to the problem will be found It does often happen, however, that one is interested in a more general circuit solution, valid for an arbitrary excitation frequency In this latter case, the solution becomes a function of This point will be developed further in 6, where the concept of sinusoidal frequency response is discussed With the problem formulated in phasor notation, the resulting solution will be in phasor form and will need to be converted to time-domain form In effect, the use of phasor notation is but an intermediate step that greatly facilitates the computation of the nal answer In summary, here is the procedure that will be followed to solve an AC circuit analysis problem Example 415 illustrates the various aspects of this method
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F O C U S O N M E T H O D O L O G Y
AC Circuit Analysis Identify the sinusoidal source(s) and note the excitation frequency Convert the source(s) to phasor form Represent each circuit element by its impedance Solve the resulting phasor circuit, using appropriate network analysis tools 5 Convert the (phasor-form) answer to its time-domain equivalent, using equation 446 1 2 3 4
4
AC Network Analysis
EXAMPLE 415 Phasor Analysis of AC Circuit
Problem
50 iS(t) + v (t) ~ S
Apply the phasor analysis method just described to the circuit of Figure 443 to determine the source current
200 100 F
Solution
vS(t) = 10 cos(100t)
Known Quantities: = 100 rad/s; R1 = 50 Find: The source current iS (t)
; R2 = 200
, C = 100 F
Analysis: De ne the voltage v at the top node and use nodal analysis to determine v Then observe that vS (t) v(t) iS (t) = R1
Next, we follow the steps outlined in the Methodology Box: AC Circuit Analysis Step 1: vS (t) = 10 cos(100t) V; = 100 rad/s Step 2: VS (j ) = 10 0 V Step 3: ZR1 = 50 , ZR2 = 200 , ZC = 1/(j 100 10 4 ) = j 100 The resulting phasor circuit is shown in Figure 444 Step 4: Next, we solve for the source current using nodal analysis First we nd V: VS V V = ZR1 ZR2 ||ZC 1 1 VS =V + ZR1 ZR2 ||ZC ZR1 V= 1 1 + ZR2 ||ZC ZR1
VS = ZR1
1 1 + 40 j 80 50
VS 50
= 7428 0381 V Then we compute IS : IS = VS V = 0083 0727 A ZR1
Step 5: Finally, we convert the phasor answer to time domain notation: is (t) = 0083 cos(100t + 0727) A
Z1 = 50 IS
VS = 10e j0
+ ~
Z2 = 200
Z3 = j100
Figure 444 Focus on Computer-Aided Tools: You will nd the solution to the same example
computed by MathCad in the electronic les that accompany this book An EWB solution is also enclosed
Part I
Circuits
EXAMPLE 416 AC Circuit Solution for Arbitrary Sinusoidal Input
Problem
Determine the general solution of Example 415 for any sinusoidal source, A cos( t + )
Solution
Known Quantities: R1 = 50
; R2 = 200
, C = 100 F
Find: The phasor source current IS (j ) Analysis: Since the radian frequency is arbitrary, it will be impossible to determine a
numerical answer The answer will be a function of The source in phasor form is represented by the expression VS (j ) = A The impedances will be ZR1 = 50 ; ZR2 = 200 ; ZC = j 104 / Note that the impedance of the capacitor is a function of Taking a different approach from Example 415, we observe that the source current is given by the expression IS = VS ZR1 + ZR2 ||ZC 2 106 ZR2 ZC 200 104 /j = 4 = 4 /j ZR2 + ZC 200 + 10 10 + j 200 25 106 + j 104 2 106 = + j 200 104 + j 200
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