AC Network Analysis in Software

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AC Network Analysis
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necessary to solve the circuit separately for each signal and then add the individual answers obtained for the different excitation sources Example 410 illustrates the response of a circuit with two separate AC excitations using AC superposition
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EXAMPLE 410 Example of AC Superposition
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+ vR2(t) + vR1(t) R2 R1
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+ v (t) _ S
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Compute the voltages vR1 (t) and vR2 (t) in the circuit of Figure 430
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iS(t)
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Known Quantities:
R1 = 150 , R2 = 50
iS (t) = 05 cos(2 100t) A vS (t) = 20 cos(2 1,000t) V
Find: vR1 (t) and vR2 (t) Analysis: Since the two sources are at different frequencies, we must compute a separate solution for each Consider the current source rst, with the voltage source set to zero (short circuit) as shown in Figure 431 The circuit thus obtained is a simple current divider Write the source current in phasor notation:
+ vR2(t) + vR1(t) R2 R1
iS(t)
IS (j ) = 05ej 0 = 05 0 A Then, VR1 (IS ) = IS
= 2 100 rad/s
50 R2 150 = 1875 0 V R1 = 05 0 R1 + R 2 150 + 50
= 2 100 rad/s VR2 (IS ) = IS 150 R1 R2 = 05 0 50 = 1875 0 V R1 + R 2 150 + 50
= 2 100 rad/s
+ vR2(t) + vR1(t) R2 R1
+ v (t) _ S
Next, we consider the voltage source, with the current source set to zero (open circuit), as shown in Figure 432 We rst write the source voltage in phasor notation: VS (j ) = 20ej 0 = 20 0 V = 2 1,000 rad/s
Then we apply the voltage divider law to obtain VR1 (VS ) = VS R1 150 = 20 0 R1 + R2 150 + 50 = 15 0 V
= 2 1,000 rad/s VR2 (VS ) = VS 50 R2 = 20 0 R1 + R 2 150 + 50 = 5 0 = 5 V
= 2 1,000 rad/s Now we can determine the voltage across each resistor by adding the contributions from each source and converting the phasor form to time-domain representation: VR1 = VR1 (IS ) + VR1 (VS ) vR1 (t) = 1875 cos(2 100t) + 15 cos(2 1,000t) V
Part I
Circuits
and VR2 = VR2 (IS ) + VR2 (VS ) vR2 (t) = 1875 cos(2 100t) + 5 cos(2 1,000t + ) V
Comments: Note that it is impossible to simplify the nal expression any further,
because the two components of each voltage are at different frequencies
Impedance We now analyze the i-v relationship of the three ideal circuit elements in light of the new phasor notation The result will be a new formulation in which resistors, capacitors, and inductors will be described in the same notation A direct consequence of this result will be that the circuit theorems of 3 will be extended to AC circuits In the context of AC circuits, any one of the three ideal circuit elements de ned so far will be described by a parameter called impedance, which may be viewed as a complex resistance The impedance concept is equivalent to stating that capacitors and inductors act as frequency-dependent resistors, that is, as resistors whose resistance is a function of the frequency of the sinusoidal excitation Figure 433 depicts the same circuit represented in conventional form (top) and in phasor-impedance form (bottom); the latter representation explicitly shows phasor voltages and currents and treats the circuit element as a generalized impedance It will presently be shown that each of the three ideal circuit elements may be represented by one such impedance element Let the source voltage in the circuit of Figure 433 be de ned by vS (t) = A cos t or VS (j ) = Aej 0 = A 0
vS(t)
+ ~
i(t)
(452)
vS(t)
+ ~
i(t)
without loss of generality Then the current i(t) is de ned by the i-v relationship for each circuit element Let us examine the frequency-dependent properties of the resistor, inductor, and capacitor, one at a time
vS(t)
+ ~
i(t)
The Resistor Ohm s law dictates the well-known relationship v = iR In the case of sinusoidal sources, then, the current owing through the resistor of Figure 433 may be expressed as i(t) = vS (t) A = cos( t) R R (453)
VS (j ) + I(j ) ~ AC circuits
Converting the voltage vS (t) and the current i(t) to phasor notation, we obtain the following expressions: VS (j ) = A 0 I(j ) = A 0 R (454)
Z is the impedance of each circuit element
AC circuits in phasor/impedance form
Figure 433 The impedance element
Finally, the impedance of the resistor is de ned as the ratio of the phasor voltage across the resistor to the phasor current owing through it, and the symbol ZR is
4
AC Network Analysis
used to denote it: VS (j ) =R I(j )
ZR (j ) =
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