Download XII. AC Circuits - Worked Examples - Mit

MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Spring 2003
XII. AC Circuits - Worked Examples
Example 1: Series RLC Circuit
A sinusoidal voltage V ( t ) = ( 40.0 V ) sin (100t ) is applied to a series RLC circuit with L =
160 mH, C = 99.0 µ F , and R = 68.0 Ω .
(a) What is the impedance of the circuit?
(b) Let the current at any instant in the circuit be I ( t ) = I 0 sin (ωt − φ ) . Find I0.
(c) What is the phase constant φ ?
Solution:
(a) The impedance of a series RLC circuit is given by
Z = R2 + ( X L − X C )
where
2
(1.1)
XL = ωL
(1.2)
1
ωC
(1.3)
and
XC =
are the inductive reactance and the capacitive reactance, respectively. Since the general
expression of the voltage source is V ( t ) = V0 sin (ωt ) , where V0 is the maximum output
voltage and ω is the angular frequency, we have V0 = 40 V and ω = 100 . Thus, the
impedance Z becomes
2
2


1 
1

2

 = 109 Ω
=
+
−
Z = R2 +  ω L −
(68)
100
.16
(
)(
)
−6


ωC 
100
99
10
×

(
)
(
)


(1.4)
(b) WithV0 = 40.0 V , the amplitude of the current is given by
1
I0 =
V0 40.0 V
=
= 0.367 A
Z 109 Ω
(1.5)
(c) The phase angle between the current and the voltage is determined by
X − XC
=
tan φ = L
R
ωL −
1
ωC
R
(1.6)
Numerically, we have
1


 (100 )( 0.160 ) − 100 99.0 × 10−6 
( )(
)  = −51.3°
φ = tan −1 


68.0




(1.7)
2
Example 2: Series RLC Circuit
Suppose an AC generator with V ( t ) = (150 V ) sin (100π t ) is connected to a series RLC
circuit where R=40.0 Ω, L=185 mH, and C=65.0 µF.
Calculate the following:
(a) VR0, VL0 and VC0, the maximum voltage drops across each circuit element, and
(b) the maximum voltage drop across points b and d shown in the figure.
Solution:
(a) The inductive reactance, capacitive reactance and the impedance of the circuit are
given by
XC =
1
1
=
= 49.0 Ω
ωC (100π ) ( 65.0 ×10−6 )
(2.1)
X L = ω L = (100π ) (185 ×10−3 ) = 58.1 Ω
(2.2)
and
Z = R2 + ( X L − X C ) =
2
( 40.0 ) + ( 58.1 − 49.0 )
2
2
= 41.0 Ω
(2.3)
respectively. Therefore, the corresponding maximum current amplitude is
I0 =
V0 150
=
= 3.66 A
Z 41.0
(2.4)
3
The maximum voltage across the resistance would be just the product of maximum
current and the resistance:
VR 0 = I 0 R = ( 3.66 )( 40 ) = 146 V
(2.5)
Similarly, the maximum voltage across the inductor is
VL 0 = I 0 X L = ( 3.66 )( 58.1) = 212 V
(2.6)
and the maximum voltage across the capacitor is
VC 0 = I 0 X C = ( 3.66 )( 49.0 ) = 179 V
(2.7)
(b) The maximum input voltage V0 is related to VR0, VL0 and VC0 by
V0 = VR 0 2 + (VL 0 − VC 0 ) 2
(2.8)
Thus, from b to d, the maximum voltage would be the difference between VL 0 and VC 0 :
Vbd = VL 0 − VC 0 = 212.5 − 179.1 = 33.4 V
(2.9)
4
Example 3: Resonance
A sinusoidal voltage V ( t ) = (100 V ) sin ωt is applied to a series RLC circuit with L =
20.0 mH, C = 100 nF and R = 20.0 Ω . Find the following quantities:
(a) the resonant frequency,
(b) the amplitude of the current at the resonant frequency,
(c) the quality factor Q of the circuit, and
(d) the amplitude of the voltage across the inductor at resonance.
Solution:
(a) The resonant frequency for the circuit is given by
f =
ω0
1
=
2π 2π
1
1
=
LC 2π
1
( 20.0 ×10 )(100 ×10 )
−3
−9
= 3560 Hz
(3.1)
(b) At resonance, the current is
I0 =
V0 100
=
= 5.00 A
R 20.0
(3.2)
(c) The quality factor Q of the circuit is given by
Q=
ω0 L
R
=
2π ( 3560 ) ( 20.0 × 10−3 )
( 20.0 )
= 22.4
(3.3)
(d) At resonance, the amplitude of the voltage across the inductor is
VL 0 = I 0 X L = I 0ω 0 L = ( 5.00 )( 2π × 3560 ) ( 20.0 × 10−3 ) = 2.24 × 103 V
(3.4)
5
Example 4: High-pass RL filter
A high-pass RL filter can be represented by the circuit in the figure below, with r being
the internal resistance of the inductor.
(a) Find Vout ,0 / Vin ,0 , the ratio of the maximum output voltage Vout ,0 to the maximum input
voltage Vin ,0 .
(b) Let r =20.0 Ω, R=5.0 Ω, and L=250 mH. What is the frequency if
Vout ,0
Vin ,0
=
1
?
2
Solution:
(R + r)
(a) The impedance for the input circuit is Z in =
2
+ X L2 where X L = ω L and
Z out = R 2 + X L2 for the output circuit. The maximum current is given by
I0 =
Vin ,0
Z in
V0
=
(R + r)
2
+ X L2
(4.1)
Similarly, the maximum output voltage is related to the output impedance by
Vout ,0 = I 0 Z out = I 0 R 2 + X L2
(4.2)
This implies
Vout ,0
Vin ,0
(b) For
Vout ,0
Vin ,0
=
=
R 2 + X L2
( R + r ) + X L2
2
(4.3)
1
, we have
2
R 2 + X L2
(R + r)
2
+X
2
L
=
1
4
(4.4)
6
Rearranging the terms, we have
XL =
(r + R)
2
− 4R2
(4.5)
3
Since X L = ω L = 2π fL , we have
1
X
f = L =
2π L 2π ( 0.250 )
( 25.0 )
2
− 4 ( 5.00 )
= 8.42 Hz
3
2
(4.6)
Example 5: RLC Circuit
Consider the circuit shown below, assuming that R, L, V0 and ω are known. If both
switches are closed initially, find the following:
(a) the current as a function of time,
(b) the average power delivered to the circuit,
(c) the current as a function of time after only switch 1 is opened.
(d) the capacitance C after switch 2 is also opened, with the current and voltage in phase,
(e) the impedance of the circuit when both switches are open,
(f) the maximum energy stored in the capacitor during oscillations,
(g) the maximum energy stored in the inductor during oscillations.
(h) the phase difference between the current and the voltage if the frequency of the
voltage source is doubled, and
7
(i) the frequency that makes the inductive reactance one-half the capacitive reactance.
Solution:
(a) When both switches are closed, the current goes through only the generator and the
resistor, so the total impedance of the circuit is R and the current is
I (t ) =
V0
cos ωt
R
(5.1)
(b) The average power is given by
< P > = < I (t )V (t ) >=
V0 2
V2
< cos 2 ωt >= 0
R
2R
(5.2)
(c) If only switch 1 is opened, the current will pass through the generator, the resistor and
the inductor. For this RL circuit, the impedance becomes
Z=
1
R 2 + X L2
=
1
R 2 + ω 2 L2
(5.3)
and the phase angle φ is
 ωL 

 R 
(5.4)
ωL 

cos  ωt + tan −1

R 

R +ω L
(5.5)
φ = tan −1 
Thus, the current as a function of time is
I (t ) =
V0
2
2 2
(d) If both switches are opened, then this would be a driven RLC circuit, with the phase
angle φ given by
1
ωL −
X L − XC
ωC
=
(5.6)
tan φ =
R
R
If the current and the voltage are in phase, then φ = 0 , implying tan φ = 0 . Let the
corresponding angular frequency be ω0 , we then obtain
ω0 L =
1
ω0C
(5.7)
8
and the capacitance is
C=
1
(5.8)
ω0 2 L
(e) From (d), we see that when both switches are opened, the circuit is at resonance
with X L = X C . Thus, the impedance of the circuit becomes
Z = R 2 + ( X L − X C )2 = R
(5.9)
(f) The energy stored in the capacitor is
1
1
U C = CVC2 = CI 2 X C2
2
2
(5.10)
It attains maximum when the current is at its maximum I 0 :
2
V02 L
1
1 V 
1
U C ,max = CI 02 X C2 = C  0 
=
2
2  R  ω 0 2C 2 2 R 2
(5.11)
where we have used ω 02 = 1 / LC .
(g) The maximum energy stored in the inductor is given by
U L ,max =
1 2 LV02
LI 0 =
2
2R2
(5.12)
(h) If the frequency of the voltage source is doubled, i.e., ω = 2ω0 = 1/ LC , then the
phase angle is given by
(
) (
 2 / LC L −
 ω L − 1/ ωC 
−1 
=
tan


R
R



φ = tan −1 
)
LC / 2C 


 = tan −1  3 L 



 2R C 

(5.13)
(i) If inductive reactance is one-half the capacitive reactance, i.e.,
1 1 
ωL = 

2  ωC 
then
ω=
ω
1
= 0
2 LC
2
(5.14)
(5.15)
9
Example 6: Parallel RLC Circuit
The figures below illustrate a parallel RLC circuit and its corresponding phasor diagram.
The instantaneous voltages and rms voltages across the three circuit elements are the
same, and each is in phase with the current through the resistor. The currents in C and L
lead or lag behind the current in the resistor.
(a) Show that the rms current delivered by the source is
1 
1 
+ ωC −
2
R 
ω L 
I rms = Vrms
2
(6.1)
(b) Find the phase angle φ between Vrms and I rms .
Solution:
Denote I R , I L and I C as the currents that pass through the resistor, the inductor and the
capacitor, respectively. Since the instantaneous voltages and rms voltages across the three
circuit elements are the same, we then have
Vrms
R
(6.2)
Vrms Vrms
=
X L ωL
(6.3)
Vrms
= ωCVrms
XC
(6.4)
IR =
IL =
and
IC =
From the phasor diagram, we see that the rms current is given by
10
I rms = I R2 + ( I C − I L )
2
(6.5)
or
2
I rms
2
V 
1 
1 
V  
=  rms  +  ωCVrms − rms  = Vrms
+  ωC −
2
R 
ωL 
ω L 
 R  
2
(6.6)
(b) From the phasor diagram, we see that the phase angle can be obtained as
Vrms Vrms
−
 1
 IC − I L  X C X L
1 
= R
−
tan φ = 

=
Vrms
 IR 
 XC X L 
R
(6.7)
or
  1
1 
−

  X C X L 
φ = tan −1  R 
(6.8)
11
Example 7: RL low-pass filter
The circuit below represents an RL low-pass filter.
Suppose the input voltage is V (t ) = (10.0 V ) sin 200t with L = 500 mH, find
(a) the value of R such that the output voltage lags behind the input voltage by 30.0° ,
(b) the amplitude of the output voltage.
Solution:
(a) The phase relationship between VL and VR is given by
tan φ =
Thus, we have
VL IX L ω L
=
=
VR IX R
R
(7.1)
ω L ( 200s ) ( 0.500 H )
R=
=
= 173 Ω
tan φ
tan 30.0°
(7.2)
Vout VR
=
= cos φ
Vin Vin
(7.3)
Vout = Vin cos φ = (10.0 V ) cos 30.0° = 8.66 V
(7.4)
−1
(b) Since
we have
12