8.022 (E&M) – Lecture 16
Topics:
Inductors in circuits
RL circuits
LC circuits
RCL circuits
Last time
Our second lecture on electromagnetic inductance
3 ways of creating emf using Faraday’s law:
Change area of circuit S(t)
Change angle between B and S
AC generators
Change B magnitude
Self and mutual inductance
Energy stored in inductor
Applications: transformers
Today is our 3rd lecture on inductance: inductors in circuits
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8.022 – Lecture 16
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RL circuits: intuitive description
At t=0, close S1 :
Lentz’s law opposes change in ΦB through L
Since ΦB (t=0)= 0, L will impede current flow
I(0)=0
As time passes, I will start flowing saturating at I=V/R
After a long time, simultaneously open S1 and close S2:
Lentz’s law opposes change in ΦB through L
Back emf will keep current flowing for a while
R dissipates power
the current will die exponentially
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8.022 – Lecture 16
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RL circuits: quantitative description
At t=0: close S1
Kirchoff’s rule #2: V − IR − L
Rewrite as: − I +
⇒ ln
V
L dI
=
R R dt
I −V / R
R
=− t
−V / R
L
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⇒
⇒ I −
dI
=0
dt
dI
R
= − dt
V
L
I −
R
V
V −R t
=− e L
R
R
8.022 – Lecture 16
⇒
I =
R
− t
V
(1 − e L )
R
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RL circuits: quantitative description(2)
At t=t’: open S1 and close S2
Kirchoff’s rule #2:
−IR − L
Rewrite as: − I =
⇒
ln
Graphically:
R
I
=− t
L
I0
L dI
R dt
⇒
dI
=0
dt
I = I (t )
∫
⇒
I =I 0
I =
t
dI
R
dt
=−∫
I
t =0 L
V − RL t
e
R
I(t)
R
− t
V
(1 − e L )
R
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V − RL t
e
R
t
8.022 – Lecture 16
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RL circuits: interpretation of results
I(t)
R
− t
V
(1 − e L )
R
V − RL t
e
R
t
How do we interpret these results?
Inductors cause currents to have an “inertia”
If no current flowing: L forces I to build up gradually
If current is flowing: L will do what it takes to make it continue (backemf)
Asymptotic behavior when “charging” L
At t=0, I=0, as if L were an open circuit
At t=infinity, I=V/R, as if L did not exist
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8.022 – Lecture 16
⎧t=0: L → open circuit
⎨
⎩t=∞: L → short circuit
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RL circuits: time constant
I(t)
V − RL t
e
R
R
− t
V
(1 − e L )
R
t
Results of RL circuit are exponentials, as in RC circuits
RC circuit: time constant τ=RC
RL circuits: time constant τ=L/R
NB: time constant is the time it takes the exponential function to
decrease (increase) to 1/e (1-1/e) of its original (final) value
Check units
cgs: [L]/[R]=(sec2/cm)/(sec/cm)=sec
SI: [L]/[R]= H/Ω = (V sec/A)/(V/A) = sec
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8.022 – Lecture 16
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LR time constant
Consider the following
circuit
On the oscilloscope:
Vinput , VL, VR, I in the circuit
Vin
L
75 Hz
t
I
R
VR
t
VL
t
VL = L dI/dt
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8.022 – Lecture 16
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t
4
LC circuits
C
L
Start with charged capacitor and close switch at t=0:
Q
dI
Kirchoff’s second rule:
−L
=0
C
dt
2
dQ
d Q
Q
+
=0
:
Since I=2
dt
LC
dt
How to solve this? Educated guess: Q (t ) = A cos ω0t + B sin ω0t
⇒
d 2Q
= −ω 2 A cos ω0t − ω 2B sin ω0t = −ω 2Q (t ) ⇒
dt 2
0
0
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ω =
0
0
8.022 – Lecture 16
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LC
9
LC circuits: solution
Plug this in the differential equation:
d 2Q (t )
1
1
Q (t ) ⇒
Q (t ) ⇒ − ω 2Q (t ) = −
=−
LC
dt 2
LC
0
ω =
0
1
LC
Determine constants A and B from initial conditions:
Q(t=0)=Q0= A cos(0) + B sin(0)
I(t=0)=0 = -ω0A sin(0) + ω0B cos(0)
A=Q0
B=0
Complete solution:
Q (t ) = Q 0 cos ω0t ⇒ VC (t ) =
Q0
Q (t )
=
cos ω0t
C
C
Q0
dQ
sinω0 t
=
dt
LC
NB: current and voltages are off by 90 degrees
I(t) = -
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8.022 – Lecture 16
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5
LC circuits: solution
Graphical representation of the solution:
I(t)
V(t)
t
Q0
⎧
⎪⎪VC (t ) = C cos ω0t
⎨
⎪I(t) = Q0 sinω t
0
⎪⎩
LC
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NB: Q and I have a phase of 90 deg
8.022 – Lecture 16
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Energy conservation
Energy stored in the capacitor over time:
Q 2 (t ) Q 0 2 (t )
cos 2 ω0t
U C (t ) =
=
2C
2C
Energy stored in the inductor:
U L (t ) =
Q2
1
1 Q2
LI (t )2 = L 0 sin2 ω0t = 0 sin2 ω0t
2
2 LC
2C
Total energy:
Q 02
Q2
(sin2 ω0t + cos 2 ω0t ) = 0
2C
2C
What is happening over time?
U (t ) = U L (t ) + U C (t ) =
Energy swings back and forth between C and L but at any moment in time
the total energy is equal to the energy initially stored in the capacitor:
Energy is conserved!
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8.022 – Lecture 16
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RCL circuits
LC circuits don’t belong to this world:
R is never exactly 0!
So let’s concentrate on RCLs
Start with a charged C
Intuitively:
LC
R
oscillatory part: sin and cos solution
dissipative part: exponential damping
Rigorous solution:
Use Kirchoff:
Since I(t) = -
dI
Q
− IR − L
=0
dt
C
dQ
dt
⇒
d 2Q
R dQ
1
Q =0
+
+
dt 2
L dt
LC
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8.022 – Lecture 16
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RCL circuits: solution
d 2Q
R dQ
1
Q =0
+
+
How to solve this equation?
2
dt
L dt
LC
Educated guess!
Intuition tells us that the solution must have an oscillatory term
and a damping term
Strategy #1: exponential * sin/cos functions:
Q (t ) = e -t / τ ( A cos ω0t + B sin ω0t )
Very heavy on algebra!!!
Strategy #2: complex exponentials
Idea: the solution is the real part of a complex solution
Q (t ) = A e i φ e i αt ⇒ Q (t ) = Re ⎣⎡Q (t ) ⎦⎤
0
Much easier algebra!!!
NB: a can be complex!
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8.022 – Lecture 16
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See handout on complex number
+ sections next week
Complex number notation
Complex number: number with both a real and an imaginary part
z = x +i y
y
with i= -1
x
z = x +i y
y
Complex plane representation z=(x,y)
x
Another useful representation
y
Set magnitude r= x +y and phase θ =arctg ⇒ z = r (cos θ + i sin θ )
x
2
2
e i θ = cos θ + i sin θ
Given Euler’s relation:
Prove it using Maclaurin expansion (see handout)
⇒ z = re i θ
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(Phasor representation)
8.022 – Lecture 16
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RCL circuits: solution (cont)
Plug expected solution Q (t ) = e i φ0 e i αt into the differential equation
1 d 2Q
R dQ
Q =0
+
+
2
dt
L dt
LC
dQ
d 2Q
⎛ -α 2 + i α R + 1 ⎞ = 0
= i αQ ;
= −α 2Q ⇒ Q
⎜
dt
dt 2
L LC ⎟⎠
⎝
Simple quadratic equation: - α 2 + i α
R2
1
R
1
R
+
=0 ⇒ α =i
±
− 2
LC 4L
L LC
2L
1 R
R
⎧
i
t
−
− t
⎪⎪Q+ (t ) = Ae i φ e 2 L e LC 4 L
This gives us 2 complex solutions for Q(t): ⎨
1 R
R
t
−i
−
⎪
− t
iφ
LC 4 L
⎪⎩Q − (t ) = Ae e 2 L e
2
0
2
2
0
⇒ real part: Q (t ) = Ae
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−
R
2L
t
cos( ±ωt + φ0 ) w ith ω =
8.022 – Lecture 16
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LC
−
R2
4L2
2
16
8
The weak damping limit
Weak damping limit: small R the damping is small
several
oscillations occur before amplitude start decreasing in sizable way
I (t ) = −
R
− t
dQ
R
cos(ω t + φ 0 ) ]
= Q 0e 2 L [ω si n(ω t + φ 0 ) +
2L
dt
W hen ω > > R/(2L) (dam ping l im i t), the second term can be i gnored and
I (t ) ~ Ae
−
R
2L
t
ω si n(ω t ) w i th ω =
1
LC
−
R2
~
4 L2
1
LC
= ω0
R
− t
⎧
2L
cos(ω 0t + φ 0 )
⎪Q (t ) ~ Q 0e
⎪
R
− t
⎪
⇒ final solution for "w eak dam ping": ⎨ I (t ) ~ ω 0Q 0e 2 L si n(ω 0t + φ 0 )
⎪
1
⎪ω 0 =
⎪⎩
LC
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8.022 – Lecture 16
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RCL in weak damping limit
Initial conditions: Q(0)=Q0 =Acos(φ0 ) and I(0)=0=Aω0 sinφ0 ⇒ A = Q 0 ; φ0 = 0
R
− t
⎧
2L
cos(ω 0t )
⎪Q (t ) ~ Q 0e
⇒ ⎨
R
− t
⎪ I (t ) ~ ω Q e 2 L si n(ω t )
0 0
0
⎩
Graphical representation of solution:
I(t)
t
Demo L2: Dumped RCL
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8.022 – Lecture 16
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Summary and outlook
Today:
What happens when we put L in circuits?
RL circuits: exponential solutions
LC circuits: oscillatory solution
RCL circuits: damped oscillation
Next Tuesday:
Quiz # 2: good luck!!!
G. Sciolla – MIT
8.022 – Lecture 16
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