AC analysis - many examples
The basic method for AC analysis:
1. Represent the AC sources as complex numbers:
˜ =
()
=
2. Convert resistors, capacitors, and inductors into their respective
impedances: resistor → ZR = R; capacitor → ZC = 1/(jωC);
inductor → ZL = jωL.
3. Re-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. Re-express the voltage and currents as sinusoids. (Often unnecessary.)
EE 201
AC analysis – 1
R 1 k!
Example 1
Find the series current
and the capacitor voltage
in the RC circuit at right.
i(t)
VS(t) = Vmcos ωt +
–
VM = 10 V
ω = 1000 rad/s
Transform to the complex version of
the circuit.
Vs (t)
Ṽs = Vm e jθs = Vm e j0 = Vm
1
C
ZC =
R
ZR = R
jωC
C
1 µF
+
vc(t)
–
ZR
˜
+
–
ĩ
ZC
+
ṽC
–
Use your favorite analysis method to find the required complex quantities.
Ṽs
ĩ =
ZR + ZC
Convert to polar form:
Vm
=
1
R + jωC
=
EE 201
Vm
R
j
ωC
ĩ = Im e jθi
That’s it.
AC analysis – 2
Vm
Im =
R2
θi = 0
+
1 2
ωC
arctan
= + arctan
10 V
=
2
(1000 Ω) +
= Im e
6
F)
1
ωRC
1
ωRC
1
= arctan
= 45
(1000 rad/s) (1 kΩ) (1μF)
1
1
=
jωC
j (1krad/s) (1 μF)
ṽc = ĩ · Zc
jθi
1
(1000 rad/s)(10
2
= 7.07 mA
1
·
jωC
= [7.07 mA] e j45 · [1000 Ω] e
=
j90
i (t) = [7.07 mA] cos (ωt + 45 )
j1000 Ω = [1000 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 201
AC analysis – 3
iL(t)
Example 2
Find the inductor current
and the parallel voltage in
the RL circuit at right.
IS(t) = Imsin ωt
IM = 10 mA
ω = 105 rad/s
ZR = R
L
L
10 mH
–
ĩL
Transform to the complex version of
the circuit.
Is (t)
Ĩs = Im e jθs = Im e j0 = Im
R
R
1 k!
+
vL(t)
˜
ZR
ZL = jωL
ZL
+
ṽL
–
Pick your method. Let’s try a current divider.
ĩL =
1
ZL
1
ZL
+
Ĩ
1 s
ZR
ZR
=
Ĩs
ZR + ZL
EE 201
R
=
Im
R + jωL
Convert to polar form:
ĩL = ILm e jθi
AC analysis – 4
ILm =
RIsm
R2
θi = 0
=
+ (ωL)
=
2
(1000 Ω) +
[(105
rad/s) (0.01 H)]
2
= 7.07 V
ωL
R
arctan
arctan
2
(1000Ω) (0.01 A)
ωL
R
=
arctan
105 rad/s (10mH)
=
1 kΩ
45
ṽL = ĩL · ZL
= ILm e
jθi
jωL = j 105 rad/s (10 mH) = j1000Ω
· jωL
= [7.07 mA] e
j45
iL (t) = [7.07 mA] sin (ωt
EE 201
· [1000 Ω] e j90 = [7.07 V] e j45
45 )
vL (t) = [7.07 V] sin (ωt + 45 )
AC analysis – 5
Example - equivalent impedances.
Impedances can be combined and reduced just like resistors from our
earlier work.
R 1 k!
iS
2 µF
C2
C1
C3 3 µF
VS(t) = Vmcos ωt +
–
1 µF
C4 4 µF
Vm = 10 V
ω = 1000 rad/s
ĩS
˜
+
–
ZR1
ĩS
ZC2
ZC1
ZC3
˜
+
–
Zeq
ZC4
Zeq = ZR1 + ZC1 (ZC1 + ZC2 + ZC3 )
EE 201
ṼS
ĩS =
Zeq
AC analysis – 6
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.
ZR = R = 1000 !
ZC1
ZC2
1
j
j
=
=
=
jωC1
ωC1
(1000 rad/s) (10
1
=
=
jωC2
j500 Ω
ZC3
1
=
=
jωC3
6
F)
j333 Ω
=
j1000 Ω
ZC4
1
=
=
jωC4
Z234 = ZC2 + ZC3 + ZC4 = ( j250 Ω) + ( j250 Ω) + ( j250 Ω) =
ZC1 Z234 =
1
+
j1000 Ω
1
j1083 Ω
Zeq = ZR + ZC1 Z234 = 1000 Ω
j250 Ω
j1083 Ω
1
=
j520 Ω
j520 Ω
Zeq = (1127 Ω) exp ( j27.5 )
ṼS
10 V
ĩS =
=
= (8.87 mA) exp (+j27.5 )
Zeq
(1127 Ω) exp ( j27.5 )
EE 201
AC analysis – 7
The complex current tells us
that the current sinusoid has
an amplitude of 8.87 mA
with a phase of 26.5° relative
to the source.
iS (t) = Im cos (ωt + θi )
Im = 8.87 mA
θi = 26.5°
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.)
1
EE 201
1
1
1
Ceq = C1 +
+
+
= 1.92 μF
C2
C3
C4
1
= 1000 Ω j520 Ω same as the other way
Then Zeq = R +
jωCeq
AC analysis – 8
Example - voltage divider
R1
For the circuit at right, find
VS(t) = Vmcos ωt
vRC for sinusoidal
frequencies of 200, 2,000,
Vm = 5 V
and 20,000 rad/s.
˜
+
–
ZR2
ZR1
Combine the parallel
combination into a single
impedance.
EE 201
R2
1 k!
C
1 µF
ZR1
Make the complex
version of the circuit.
ZRC = ZR2 ZC =
+
–
1 k!
˜
(R 2 )
R2 +
1
jωC
1
jωC
+
–
ZRC
ZC
+
vRC
–
+
ṽRC
–
+
ṽRC
–
R2
=
1 + jωR2 C
AC analysis – 9
ZR1
Now use a voltage divider:
ṽRC
ZRC
=
ṼS
ZRC + ZR1
˜
=
+
–
ZRC
R2
1+jωR2 C
Vm
R2
1+jωR2 C + R1
+
ṽRC
–
R2
=
Vm
R1 + R2 + jωR1 R2 C
In magnitude and phase form:
ṽRC =
R2 Vm
2
(R1 + R2 ) + (ωR1 R2 C)
2
exp (jθ)
θ=
arctan
ωR1 R2 C
R1 + R 2
at ω = 200 rad/s: ṽRC = (2.49 V) exp ( j5.7 ) → vRC(t) = (2.49 V)cos( ωt – 5.7°)
ω = 2,000 rad/s: ṽRC = (1.77 V) exp ( j45.0 )
etc.
ω = 20,000 rad/s: ṽRC = (0.249 V) exp ( j84.3 )
EE 201
AC analysis – 10
Example - source transformation
For the circuit below, use a source transformation to find the resistor
current.
R 250 !
VS(t) = Vmcos ωt +
–
Vm = 25 V
ω = 5000 rad/s
iR
IS(t) = Imcos(ωt+θi)
Im = 50 mA
θi = –45° ("/4 rad)
L
50 mH
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.
ZR
ṼS
= Vm
EE 201
+
–
ĩR
ZL
˜
= Imej45°
AC analysis – 11
ZR
Then do the source transformation.
Since we are finding the current in
+
ṼS –
the resistor, we cannot transform it
and so we must transform the currentsource / inductor pair.
ṼS1 = ZL ĨS = (jωL) Im ejθi = ωLej90
ZL
+ ṼS1
–
ĩR
Im ejθi = (ωLIm ) ej(θi +90
)
ṼS1 = (12.5 V) e j45 = 8.84 V + j8.84 V = Vm1 e jθv
The circuit analysis is trivial – use KVL around the loop.
ṼS
ĩR ZR
ĩR ZL
ṼS1 = 0
ṼS ṼS1
Vm Vm1 ejθv
ĩR =
=
ZR + ZL
R + jωL
ĩR =
25 V
(18.4 V) e j28.7
(8.84 V + j8.84 V)
=
= (52.1 mA) e
j45
(353.6 Ω) e
250Ω + j250Ω
j73.7
iR(t) = (52.1 mA)cos( ωt – 73.7°)
EE 201
AC analysis – 12