SJTU2 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
SJTU3 Nodal analysis Fig. 8-28: An example node
SJTU4 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
SJTU5 EXAMPLE 8-21 Use node analysis to find the current I X in Fig Fig SOLUTION: or
7 EXAMPLE 8-24 The circuit in Fig is an equivalent circuit of an ac induction motor. The current I S is called the stator current, I R the rotor current, and I M the magnetizing current. Use the mesh-current method to solve for the branch currents I S, I R and I M.
9 EXAMPLE 8-25 Use the mesh-current method to solve for output voltage V 2 and input impedance Z IN of the circuit below. SOLUTION:
SJTU11 Example Frequency domain equivalent of the circuit
SJTU13 Find Vo/Vi, Zi See F page417
SJTU14 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.
SJTU15 EXAMPLE 8-13 Use the unit output method to find the input impedance, current I 1, output voltage V C, and current I 3 of the circuit in Fig 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,
SJTU16 Given K and Z IN, we can now calculate the required responses for an input
SJTU17 Two cases: 1.With same frequency sources. 2.With different frequency sources EXAMPLE 8-14 Use superposition to find the steady - state voltage v R (t) in Fig for R=20 , L1 = 2mH, L2 = 6mH, C = 20 F, V s1= 100cos 5000t V, and Vs2=120cos (5000t +30 )V. SUPERPOSITION
SJTU18 SOLUTION: Fig. 8-22
SJTU20 EXAMPLE 8-15 Fig Use superposition to find the steady-state current i(t) in Fig for R=10k , L=200mH, v S1 =24cos20000t V, and v S2 =8cos(60000t+30 ° ). SOLUTION: With source no. 2 off and no.1 on
SJTU21 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.
SJTU22 More examples See F page403
SJTU23 The thevenin and Norton circuits are equivalent to each other, so their circuit parameters are related as follows: THEVENIN AND NORTON EQUIVALENT CIRCUITS
SJTU24 Source transformation
SJTU25 EXAMPLE 8-17 Both sources in Fig. 8-25(a) operate at a frequency of =5000 rad/s. Find the steady-state voltage v R (t) using source transformations. SOLUTION: +
SJTU26 EXAMPLE 8-18 Use Thevenin's theorem to find the current Ix in the bridge circuit shown in Fig Fig. 8-26