Physics -272 Lecture 19
LC Circuits
RLC Circuits
AC Circuit Theory
LC Circuits
•
Consider the RC and LC
series circuits shown:
++++
----
•
C R
++++
----
C
L
Suppose that the circuits are
formed at t=0 with the
capacitor charged to value Q.
There is a qualitative difference in the time development of the
currents produced in these two cases. Why??
• Consider what happens to the energy!
• In the RC circuit, any current developed will cause
energy to be dissipated in the resistor.
• In the LC circuit, there is NO mechanism for energy
dissipation; energy can be stored both in the capacitor
and the inductor!
Energy in the Electric
and Magnetic Fields
Energy stored in a capacitor ...
1
U = CV 2
2
… energy density ...
Energy stored in an inductor ….
+++ +++
--- ---
uelectric
1
= ε0 E2
2
B
1 2
U = LI
2
… energy density ...
E
umagnetic
1 B2
=
2 µ0
RC/LC Circuits
I
Q+++
---
I
Q+++
---
C R
I
I
0
0
t
L
LC:
current oscillates
RC:
current decays exponentially
0
C
0
t
LC Oscillations
(qualitative)
I =0
+ +
- -
C
I = −I0
L
⇒
C
Q=0
Q = +Q0
⇓
⇑
I =0
I = +I0
C
L
-
L
Q=0
⇐
-
+ +
C
L
Q = −Q0
Multi-part Clicker
t=0
• At t=0, the capacitor in the LC
circuit shown has a total charge
+ +
Q0. At t = t1, the capacitor is
Q = Q0
uncharged.
- C
– What is the value of Vab=Vb-Va,
2A the voltage across the inductor
at time t1?
(a) Vab < 0
(b) Vab = 0
t=t 1
a
L
Q =0
C
L
b
(c) Vab > 0
2B – What is the relation between UL1, the energy stored in the
inductor at t=t1, and UC1, the energy stored in the capacitor at
t=t1?
(a) UL1 < UC1
(b) UL1 = UC1
(c) UL1 > UC1
Multi-part clicker
t=0
• At t=0, the capacitor in the LC
circuit shown has a total charge
+ +
Q0. At t = t1, the capacitor is
Q = Q0
uncharged.
- C
– What is the value of Vab=Vb-Va,
2A the voltage across the inductor
at time t1?
(a) Vab < 0
(b) Vab = 0
t=t 1
a
L
Q =0
C
L
b
(c) Vab > 0
• Vab is the voltage across the inductor, but it is also
(minus) the voltage across the capacitor!
• Since the charge on the capacitor is zero, the voltage
across the capacitor is zero!
t=t 1
t=0
• At t=0, the capacitor in the LC
circuit shown has a total charge
+ +
Q0. At t = t1, the capacitor is
Q = Q0
uncharged.
- C
– What is the relation between UL1,
2B the energy stored in the inductor
at t=t1, and UC1, the energy stored
in the capacitor at t=t1?
(a) UL1 < UC1
(b) UL1 = UC1
• At t=t1, the charge on
the capacitor is zero.
U C1
Q12
=
=0
2C
a
L
Q =0
C
b
(c) UL1 > UC1
• At t=t1, the current is a
maximum.
U L1
L
1 2 Q02
= LI 1 =
>0
2
2C
Clicker problem:
At time t = 0 the capacitor is fully charged
with Qmax, and the current through the circuit
is 0.
2) What is the potential difference across the inductor at t = 0?
a) VL = 0
b) VL = Qmax/C
c) VL = Qmax/2C
3) What is the potential difference across the inductor when the current
is maximum?
a) VL = 0
b) VL = Qmax/C
c) VL = Qmax/2C
LC Oscillations
(mechanical analogy, for R=0)
•
What is the oscillation frequency ω0?
•
Begin with the loop rule:
I
Q
2
d Q Q
L 2 + =0
dt
C
•
+ +
- -
Q = Q0 cos(ω t + φ)
where
•
L
Guess solution: (just harmonic oscillator!)
remember:
•
C
d2x
m 2 + kx = 0
dt
φ, Q0 determined from initial conditions
Procedure: differentiate above form for Q and substitute into
loop equation to find ω.
Note: Dimensional analysis ω =
1
LC
LC Oscillations
(quantitative)
• General solution:
Q = Q0 cos(ωt + φ)
• Differentiate:
dQ
= −ω Q0 sin(ω t + φ )
dt
+ +
- -
C
L
d 2Q Q
L 2 + =0
dt
C
d 2Q
2
=
−
ω
Q0 cos(ωt + φ )
2
dt
• Substitute into loop eqn:
1
1
2
L − ω Q0 cos(ω t + φ ) + (Q0 cos(ω t + φ )) = 0 ⇒ − ω L + = 0
C
C
(
)
2
Therefore,
ω =
1
LC
which we could have determined
from the mechanical analogy to SHO:
k
1/ C
1
ω =
=
=
m
L
LC
Multi-part clicker problem
• At t=0 the capacitor has charge Q0; the resulting
oscillations have frequency ω0. The maximum
current in the circuit during these oscillations has
value I0.
3A – What is the relation between ω0 and ω2, the
frequency of oscillations when the initial charge =
2Q0?
(a) ω2 = 1/2 ω0
(b) ω2 = ω0
t=0
+ +
Q = Q0
- -
C
(c) ω2 = 2ω0
3B – What is the relation between I0 and I2, the maximum current in
the circuit when the initial charge = 2Q0?
(a) I2 = I0
(b) I2 = 2I0
L
(c) I2 = 4I0
Clicker problem
t=0
• At t=0 the capacitor has charge Q0; the resulting
oscillations have frequency ω0. The maximum
current in the circuit during these oscillations has
value I0.
3A – What is the relation between ω0 and ω2, the
frequency of oscillations when the initial charge =
2Q0?
(a) ω2 = 1/2 ω0
(b) ω2 = ω0
+ +
Q = Q0
- -
C
L
(c) ω2 = 2ω0
• Q0 determines the amplitude of the oscillations (initial condition)
• The frequency of the oscillations is determined by the circuit
parameters (L, C), just as the frequency of oscillations of a mass
on a spring was determined by the physical parameters (k, m)!
Clicker problem
• At t=0 the capacitor has charge Q0; the resulting
oscillations have frequency ω0. The maximum
current in the circuit during these oscillations has
value I0.
– What is the relation between I0 and I2, the
3B
maximum current in the circuit when the initial
charge = 2Q0?
(a) I2 = I0
(b) I2 = 2I0
t=0
+ +
Q = Q0
- -
C
L
(c) I2 = 4I0
• The initial charge determines the total energy in the circuit:
U0 = Q02/2C
• The maximum current occurs when Q=0!
• At this time, all the energy is in the inductor: U = 1/2 LIo2
• Therefore, doubling the initial charge quadruples the total
energy.
• To quadruple the total energy, the max current must double!
Clicker question:
The current in a LC circuit is a
sinusoidal oscillation, with
frequency ω.
5) If the inductance of the circuit is increased, what will happen
to the frequency ω?
a) increase
b) decrease
c) doesn’t change
6) If the capacitance of the circuit is increased, what will
happen to the frequency?
a) increase
b) decrease
c) doesn’t change
LC Oscillations
Energy Check
1
• Oscillation frequency ω =
LC
loop equation.
has been found from the
• The other unknowns ( Q0, φ ) are found from the initial
conditions. E.g., in our original example we assumed initial
values for the charge (Qi) and current (0). For these values:
Q0 = Qi, φ = 0.
•
Question: Does this solution conserve energy?
1 Q 2 (t )
1 2
U E (t ) =
=
Q0 cos 2 (ω t + φ )
2 C
2C
1 2
1
U B (t ) = Li (t ) = Lω 2 Q02 sin 2 (ω t + φ )
2
2
Energy Check
Energy in Capacitor
UE
1 2
U E (t ) =
Q0 cos 2 (ωt + φ )
2C
Energy in Inductor
1 2 2 2
U B (t) = Lω Q0 sin (ω t + φ)
2
1
LC
⇒
ω =
1 2
U B (t ) =
Q 0 sin 2 (ω t + φ )
2C
Therefore,
Q02
UE (t) +UB (t) =
2C
0
t
UB
0
t
Inductor-Capacitor Circuits
Solving a LC circuit problem; Suppose ω=1/sqrt(LC)=3 and
given the initial conditions,
Q(t = 0) = 5C
I (t = 0) = 15 A
Solve find Q0 and φ0, to get complete solution using,
Q(t = 0) = 5 = Q0 cos(0 + φ0 )
I (t = 0) = 15 = −Q0ω sin (0 + φ0 ) = −3Q0 sin (0 + φ0 )
and we find,
2
 15 
2
(5) +  −  = Q0 2 sin 2 (φ0 ) + cos 2 (φ0 ) = Q0 2 , Q0 = 5 2
 3
[
 15 
o
, φ0 = −45
 5⋅3
φ0 = inv. tan −
]
Remember harmonic oscillators !!
The following are all equally valid solutions
Q(t ) = Q0 cos(ωt + φ0 )
Q(t ) = Q0 sin(ωt + φ1 )
Q(t ) = Q0 (cos(ωt ) cos(φ0 ) − sin(ωt ) sin(φ0 ))
Q(t ) = A cos(ωt ) + B sin(ωt )
Inductor-Capacitor-Resistor Circuit
Q
d 2Q
0 = + RI + L
C
dt 2
d 2Q
dQ Q
0=L
+R
+
dt C
dt 2
Solution will have form of
Q(t ) = Ae
If,
−αt
2
cos(ω ' t + φ )
1
R
>
LC 4 L2
Inductor-Capacitor-Resistor Circuit
3 solutions, depending on L,R,C values
Very important !!
2
1
R
>
LC 4 L2
2
1
R
=
LC 4 L2
2
1
R
<
LC 4 L2
Inductor-Capacitor-Resistor Circuit
Solving for all the terms
Q(t ) = Ae −αt cos(ω ' t + φ )
R

− t

= Ae  2 L  cos

1
R
−
t +φ
 LC 4 L2



2
R
1
R2
α = and ω ' =
−
2L
LC 4 L2
Solution for underdamped circuit;
1
R
>
LC 4 L2
For other solutions, use starting form, solve for λ and λ′,
Q(t ) = Ae −λt + Be −λ 't
2
Application of magnetic induction: “smart”
traffic lights
Traffic light in California
Another version with
two loops
Application of magnetic induction
Mechanical
ignition in a
car
Magnetic energy from ignition coil is used to fire
the automotive spark plug.
2 –part Clicker question:
The current in a LC circuit is a
sinusoidal oscillation, with
frequency ω.
I) If the inductance of the circuit is increased, what will happen
to the frequency ω?
a) increases
b) decreases
c) doesn’t change
II) If the capacitance of the circuit is increased, what will
happen to the frequency?
a) increases
b) decreases
ω=
1
LC
c) doesn’t change
Energy Check for LC circuits
Energy in Capacitor
UE
1 2
U E (t ) =
Q0 cos 2 (ωt + φ )
2C
Energy in Inductor
1 2 2 2
U B (t) = Lω Q0 sin (ω t + φ)
2
1
LC
⇒
ω =
1 2
U B (t ) =
Q 0 sin 2 (ω t + φ )
2C
Therefore,
Q02
UE (t) +UB (t) =
2C
0
t
UB
0
t
Inductor-Capacitor (LC) Circuit Example
Solving a LC circuit problem; Suppose ω=1/sqrt(LC)=3 and
given the initial conditions,
Q(t = 0) = 5C
I (t = 0) = 15 A
Solve find Q0 and φ0, to get complete solution using,
Q(t = 0) = 5 = Q0 cos(0 + φ0 )
I (t = 0) = 15 = −Q0ω sin (0 + φ0 ) = −3Q0 sin (0 + φ0 )
and we find,
2
 15 
2
(5) +  −  = Q0 2 sin 2 (φ0 ) + cos 2 (φ0 ) = Q0 2 , Q0 = 5 2
 3
[
 15 
o
, φ0 = −45
 5⋅3
φ0 = inv. tan −
]
Remember harmonic oscillators !!
The following are all equally valid solutions
Q(t ) = Q0 cos(ωt + φ0 )
Q(t ) = Q0 sin(ωt + φ1 )
Q(t ) = Q0 (cos(ωt ) cos(φ0 ) − sin(ωt ) sin(φ0 ))
Q(t ) = A cos(ωt ) + B sin(ωt )
Resistor-Inductor-Capacitor (RLC) Circuit
Q
d 2Q
0 = + RI + L
C
dt 2
d 2Q
dQ Q
0=L
+R
+
dt C
dt 2
Solution will have form of
Q(t ) = Ae
If,
−αt
2
cos(ω ' t + φ )
1
R
>
LC 4 L2
Inductor-Capacitor-Resistor Circuit
3 types of solutions, depending on L,R,C values
Very important !! This is just like
the damped SHO
2
1
R
>
LC 4 L2
2
1
R
=
LC 4 L2
2
1
R
<
LC 4 L2
Inductor-Capacitor-Resistor Circuit
Q(t ) = Ae −αt cos(ω ' t + φ )
R

− t

= Ae  2 L  cos

1
R
−
t +φ
 LC 4 L2



2
R
1
R2
α = and ω ' =
−
2L
LC 4 L2
Solution for underdamped circuit;
1
R
>
LC 4 L2
For other solutions, use starting form, solve for λ and λ′,
Q(t ) = Ae −λt + Be −λ 't
2
Alternating Currents (Chap 31)
We next study circuits where the battery is replaced by
a sinusoidal voltage or current source.
v(t ) = V0 cos(ωt )
or
i(t ) = I 0 cos(ωt )
The circuit symbol is,
An example of an LRC circuit connected to sinusoidal source is,
Important:
I(t) is same throughout – just
like the DC case.
Alternating Currents (Chap 31.1)
Since the currents & voltages are sinusoidal, their values change
over time and their average values are zero.
A more useful description of sinusoidal currents and voltages are
given by considering the average of the square of this quantities.
We define the RMS (root mean square), which is the
square root of the average of ,
i
2
2
(
(
)
)
(t ) = I 0 cos ωt
2
I
1
2
i 2 (t ) = (I 0 cos(ωt )) = I 02 (1 + cos(2ωt ) ) = 0
2
2
I0
2
I RMS = i (t ) =
2
V0
2
Similarly:
VRMS = v (t ) =
2
Alternating Currents ; Phasors
A convenient method to describe currents and voltages in AC
circuits is “Phasors”. Since currents and voltages in circuits with
capacitors & inductors have different phase relations, we introduce
a phasor diagram. For a current,
i = I cos(ωt )
We can represent this by a vector
rotating about the origin. The angle
of the vector is given by ωt and the
magnitude of the current is its
projection on the X-axis.
If we plot simultaneously currents &
voltages of different components we
can display relative phases .
Note this method is equivalent to imaginary numbers approach
where we take the real part (x-axis projection) for the magnitude
Alternating Currents: Resistor in AC circuit
A resistor connected to an AC source will have the voltage, vR, and
the current across the resistor has the same phase. We can represent the
current phasor and the voltage phasor with the same angle.
v R = VR cos(ωt ) = iR = I cos(ωt )R
and
VR = IR
(just like DC case)
Phasors are rotating 2 dimensional vectors
Resistor in AC circuit; I & V versus ω t
I(t)=Icos(ωt)
V(t)=RIcos(ωt)
ωt →
a
b
c
d
e
NB: for a resistor voltage is in phase with current
f
Alternating Currents: Capacitor in AC circuit
In a capacitor connected to an AC current source the voltage
lags behind the current by 90 degree. We can draw the current
phasor and the voltage phasor behind the current by 90 degrees.
dq
i=
= I cos(ωt )
dt
Find voltage:
q 1
I
v = = ∫ idt =
sin (ωt )
ωC
C C
1 
VMAX = I 

ω
C


q=
I
ω
sin(ωt )
Alternating Currents ; Capacitor in AC circuit
We define the capacitive reactance, XC, as
MAX
cap
V
I
 1 
=
= I
 = I XC
ωC
 ωC 
XC =
1
ωC
Like: VR = IR
But frequency dependent
We stated that voltage lags by 90 deg., so equivalent solution is
I
I
[cos ωt cos 90 + sin ωt sin 90]
cos(ωt − 90) =
v=
ωC
ωC
I
=
sin ωt
ωC
Capacitor in AC circuit; I & V versus ω t
i(t)=Icos(ωt)
v(t)=(I/ωC)sin(ωt)
a
b
c
d
e
f
ωt →
Note voltage lags 90 deg. Behind current
V(t)=(I/ωC)sin(ωt)= (I/ωC)cos(ωt-π/2)
Alternating Currents: Inductor in AC circuit
In an inductor connected to an AC current source, the voltage will
lead the current by 90 degrees. We can draw the current
phasor and the voltage phasor ahead of the current by 90 degrees.
i = I cos(ωt )
di
and V = L = − ILω sin (ωt )
dt
VMAX = ILω
Define inductive reactance, XL, as
MAX
Vind
= IωL = I (ωL ) = I X L
X L = ωL
Like: VR = IR
But frequency dependnt
Inductor in AC circuit; I & V versus ω t
i(t)=Icos(ωt)
v(t)= - ILωsin(ωt)
a
b
c
d
e
f
ωt →
Draw phasor diagram for each point
Note voltage is 90 deg. ahead of current
v(t)=ILω sin(ωt)= ILω cos(ωt + π/2)
AC summary
i(t)
resistor
v(t)
I cosωt VR cos ωt
capacitor I cosωt VC sin ωt
inductor
I cosωt -VLsin ωt
X
V
Phasor
R
IR
in phase
1/ωC
IXC
lags
ωL
IXL
leads