SJTU1 Chapter 10 Sinusoidal steady-state analysis.

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Chapter 10

Sinusoidal steady-state analysis

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Steps to analyze ac circuit

1. Transform the circuit to the phasor or frequency domain

2. Solve the problem using circuit techniques(nodal analysis, mesh analysis, superposition,etc)

3. Transform the resulting phasor to the time domain

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Nodal analysis

Fig. 8-28: An example node

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Mesh analysis

planar circuits: Circuits that can be drawn on a flat surface with no crossovers

Fig. 8-29: An example mesh

the sum of voltages around mesh A is

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EXAMPLE 8-21Use node analysis to find the current IX in Fig. 8-31.

Fig. 8-31

SOLUTION:

075VC :C Node

or

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EXAMPLE 8-24 The circuit in Fig. 8-32 is an equivalent circuit of an ac induction motor. The current IS is called the stator current, IR the rotor current, and IM the magnetizing current. Use the mesh-current method to solve for the branch currents IS, IR and IM.

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EXAMPLE 8-25

Use the mesh-current method to solve for output voltage V2 and input impedance ZIN of the circuit below.

SOLUTION:

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Example

Frequency domain equivalent of the circuit

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Example

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Find Vo/Vi, Zi

See F page417

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Circuit Theorems with Phasors

PROPORTIONALITY

The proportionality property states that phasor output responses are proportional to the input phasor

where X is the input phasor, Y is the output phasor, and K is the proportionality constant.

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EXAMPLE 8-13

Use the unit output method to find the input impedance, current I1, output voltage VC, and current I3 of the circuit in Fig. 8-20 for Vs= 10 0°∠

SOLUTION:

1.Assume a unit output voltage             . 2.By Ohm's law,                        . 3.By KVL,                             4.By Ohm's law,                                  5.By KCL,                             6.By KCL,                                  

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Given K and ZIN, we can now calculate the required responses for an input

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Two cases:

1. With same frequency sources.

2. With different frequency sources

EXAMPLE 8-14

Use superposition to find the steady - state voltage vR (t) in Fig. 8 - 21 for R=20 , L1 = 2mH, L2 = 6mH, C = 20 F, V s1= 100cos 5000t V , and Vs2=120cos (5000t +30 )V.

SUPERPOSITION

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SOLUTION:

Fig. 8-22

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EXAMPLE 8-15

Fig. 8-23

 Use superposition to find the steady-state current i(t) in Fig. 8-23 for R=10k , L=200mH, vS1=24cos20000t V, and vS2=8cos(60000t+30 ° ).

SOLUTION: With source no. 2 off and no.1 on

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With source no.1 off and no.2 on

The two input sources operate at different frequencies, so that phasors responses I1 and I2 cannot be added to obtain the overall response. In this case the overall response is obtained by adding the corresponding time-domain functions.

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More examples

See F page403

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The thevenin and Norton circuits are equivalent to each other, so their circuit parameters are related as follows:

THEVENIN AND NORTON EQUIVALENT CIRCUITS

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Source transformation

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EXAMPLE 8-17

Both sources in Fig. 8-25(a) operate at a frequency of =5000 rad/s. Find the steady-state voltage vR(t) using source transformations.

SOLUTION:

+

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EXAMPLE 8-18 Use Thevenin's theorem to find the current Ix in the bridge circuit shown in Fig. 8-26.

Fig. 8-26

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SOLUTION:

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