Today, we discuss how to apply the circuit analysis technique we have studied before and the phasor techniques to find the steady-state response of circuits with sinusoidal sources.
Part 1:
Nodal Analysis
Nodal analysis can apply to the sinusoidal sources circuit the same way we did before. Only difference is we may calculate with complex number. This
write down nodal equations at two essential nodes.
do the algebra work to find the voltage at two essential nodes.
Mesh Analysis
write down KVL for each loop. I3 is given, then solve the system to find each current.
Part 2: Phasor Lab
The cut of frequency wc=R/L, 10*wc , (1/10)*wc.
In RL circuit, VL leads I, f increases, then the phase shift increases.
The cut off frequency, phase shift between Vl and I is about 45, and the Vl leads the current.
The results from experiments.
wc (w-cut-off)
phase shift (i-vl) = -45
10*wc
phase shift = -84.28
wc/10
phase shift= -5.7
Superposition
Since ac circuits are linear, the superposition theorem
applies to ac circuits the same way it applies to dc circuits. The theorem becomes important if the circuit
has sources operating at different frequencies. In this case, since the
impedances depend on frequency, we must have a different frequency-domain
circuit for each frequency. The total
response must be obtained by adding the individual responses in the time
domain. It is incorrect to try to add the responses in the phasor or frequency
domain.
the review of source transformation
Thevernin and Norton Equivalent circuits
We transform the circuit to the impedance form.
To find V Th ,
we apply KCL at node 1 to find I0. Then apply KVL to the right hand loop. Vth = 55, -90.
To find Zth, remove the independent source and connect an arbitrary fixed current source (In this case 3A since it makes the math easy) to terminals a and b,
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