AC analysis - many examples
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1 AC analysis - many examples The basic method for AC analysis:. epresent the AC sources as complex numbers: ( ). Convert resistors, capacitors, and inductors into their respective impedances: resistor Z ; capacitor Z C /(jωc); inductor Z L jωl. 3. e-draw the circuit using the complex sources and impedances. 4. Use your favorite method to find expressions for the complex currents and voltages (phasors) in terms of the sources and impedances 5. Do whatever complex math is needed. (This is the longest part!) 6. Express the answer in magnitude/phase form (usually). 7. e-express the voltage and currents as sinusoids. (Often unnecessary.) EE 0 AC analysis
2 k! Example Find the series current and the capacitor voltage in the C circuit at right. i(t) VS(t) Vmcos ωt VM 0 V ω 000 rad/s Transform to the complex version of the circuit. Vs (t) V s Vm e jθs Vm e j0 Vm C ZC Z jωc C µf vc(t) Z i ZC v C Use your favorite analysis method to find the required complex quantities. V s i Z ZC Convert to polar form: Vm jωc EE 0 Vm j ωc i Im e jθi That s it. AC analysis
3 Vm Im θi 0 ωc arctan arctan 0 V (000 Ω) Im e 6 F) ωc ωc arctan 45 (000 rad/s) ( kω) (μf) jωc j (krad/s) ( μf) v c i Zc jθi (000 rad/s)( ma jωc [7.07 ma] e j45 [000 Ω] e j90 i (t) [7.07 ma] cos (ωt 45 ) j000 Ω [000 Ω] e [7.07 V] e j90 j45 vc (t) [7.07 V] cos (ωt 45 ) Now try it using a voltage divider to vc first and then find the current. EE 0 AC analysis 3
4 il(t) Example Find the inductor current and the parallel voltage in the L circuit at right. IS(t) Imsin ωt IM 0 ma ω 05 rad/s Z L L 0 mh i L Transform to the complex version of the circuit. Is (t) I s Im e jθs Im e j0 Im k! vl(t) Z ZL jωl ZL v L Pick your method. Let s try a current divider. i L ZL ZL I s Z Z I s Z ZL EE 0 Im jωl Convert to polar form: i L ILm e jθi AC analysis 4
5 ILm Ism θi 0 (ωl) (000 Ω) [(05 rad/s) (0.0 H)] 7.07 V ωl arctan arctan (000Ω) (0.0 A) ωl arctan 05 rad/s (0mH) kω 45 v L i L ZL ILm e jθi jωl j 05 rad/s (0 mh) j000ω jωl [7.07 ma] e j45 il (t) [7.07 ma] sin (ωt EE 0 [000 Ω] e j90 [7.07 V] e j45 45 ) vl (t) [7.07 V] sin (ωt 45 ) AC analysis 5
6 Example - equivalent impedances. Impedances can be combined and reduced just like resistors from our earlier work. i k! S µf V S (t) V m cos ωt V m 0 V ω 000 rad/s µf C C C 3 C 4 4 µf 3 µf ĩ S ĩ S Z Z C Z C Z C3 Z eq Z C4 Z eq Z Z C (Z C Z C Z C3 ) ĩ S ṼS Z eq EE 0 AC analysis 6
7 Oftentimes, it is useful to carry through with symbols for as long as possible, but in this case we may as well go to numbers now. Z 000! ZC ZC j j jωc ωc (000 rad/s) (0 jωc j500 Ω ZC3 jωc3 6 F) j333 Ω j000 Ω ZC4 jωc4 Z34 ZC ZC3 ZC4 ( j50 Ω) ( j50 Ω) ( j50 Ω) ZC Z34 j000 Ω j083 Ω Zeq Z ZC Z Ω j50 Ω j083 Ω j50 Ω j50 Ω Zeq (7 Ω) exp ( j7.5 ) V S 0 V i S (8.87 ma) exp (j7.5 ) Zeq (7 Ω) exp ( j7.5 ) EE 0 AC analysis 7
8 The complex current tells us that the current sinusoid has an amplitude of 8.87 ma with a phase of 6.5 relative to the source. i S (t) I m cos (ωt θ i ) I m 8.87 ma θ i 6.5 With this circuit, we might find the equivalent capacitance of the small network capacitors before switching over to impedance. In finding equivalent capacitance, we must us the rules for combining capacitors. (As we recall, series capacitors add like resistors in parallel, and parallel capacitors add like resistors in series.) C eq C C C 3 C 4.9 μf Then Z eq 000 Ω jωc eq j50 Ω same as the other way EE 0 AC analysis 8
9 Example - voltage divider For the circuit at right, find v C for sinusoidal frequencies of 00,,000, and 0,000 rad/s. V S (t) V m cos ωt V m 5 V k! k! C µf v C Make the complex version of the circuit. Z Z Z C ṽ C Combine the parallel combination into a single impedance. Z Z C ṽ C Z C Z Z C ( ) jωc jωc EE 0 AC analysis 9 jω C
10 Z Now use a voltage divider: v C ZC V S ZC Z ZC jω C Vm jω C v C Vm jω C In magnitude and phase form: v C Vm ( ) (ω C) exp (jθ) θ arctan ω C at ω 00 rad/s: v C (.49 V) exp ( j5.7 ) vc(t) (.49 V)cos( ωt 5.7 ) ω,000 rad/s: v C (.77 V) exp ( j45.0 ) etc. ω 0,000 rad/s: v C (0.49 V) exp ( j84.3 ) EE 0 AC analysis 0
11 Example - source transformation For the circuit below, use a source transformation to find the resistor current. 50! V S (t) V m cos ωt V m 5 V ω 5000 rad/s i L I S (t) I m cos(ωtθ i ) 50 mh I m 50 ma θ i 45 ("/4 rad) Form the complex version of the circuit. Note that there is a phase difference between the the two source and that difference must be show up in the complex numbers (phasors) describing the sources. Z Ṽ S ĩ Z L V m I m e j45 EE 0 AC analysis
12 Z Then do the source transformation. Since we are finding the current in V S the resistor, we cannot transform it and so we must transform the currentsource / inductor pair. V S ZL I S (jωl) Im ejθi ωlej90 ZL V S i Im ejθi (ωlim ) ej(θi 90 ) V S (.5 V) e j V j8.84 V Vm e jθv The circuit analysis is trivial use KVL around the loop. V S i Z i ZL V S 0 V S V S Vm Vm ejθv i Z ZL jωl i 5 V (8.4 V) e j8.7 (8.84 V j8.84 V) (5. ma) e j45 (353.6 Ω) e 50Ω j50ω j73.7 i(t) (5. ma)cos( ωt 73.7 ) EE 0 AC analysis
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