Energy of an Inductor
• How much energy is stored in
an inductor when a current is
flowing through it?
a
• Start with loop rule:
dI
  IR  L
dt

I
I
R
b
L
• Multiply this equation by I:
εI  I 2 R  LI
•
dI
dt
From this equation, we can identify PL, the rate at which energy is
dI
dU
being stored in the inductor:
PL 
•
dt
 LI
dt
We can integrate this equation to find an expression for U, the
energy stored in the inductor when the current = I:
U
I
0
0
U   dU   LIdI

1
U  LI 2
2
Where is the Energy Stored?
• Claim: (without proof) energy is stored in the Magnetic field
itself (just as in the Capacitor / Electric field case).
• Consider the uniform field inside a long solenoid:
B  0
•
N
I
l
l
2
2
N
N
2
The inductance L is:
L  0
r   0
A
l
l
r
N turns
N2  2 1
B2
1 2 1

U  LI    0
A  I  Al
l
2
2
2
0

• Energy U:
• We can turn this into an energy density by dividing by the
volume containing the field:
uM
U
U
1 B2



Al vol 2  0
Recall for E: u E 
1
 0E
2
2
Oscillations
VC
UE
0
C
VL
LC
Circuits
0
1
L
t
0
t
UB
Physics 122 Lecture 24
0
t
Recall & Compare
Energy in the Electric Field
Work needed to add charge to capacitor...
+++ +++
q
--- --dW  dq ( V)  dq  
 C
1Q
1 Q2 1
W   qdq 
 CV 2 … total work ...
2 C 2
C0
Recall: C = 0A/d & V = Ed
u
W
1
 0E 2
volume 2
… energy density
Energy Density:
1
u electric   0 E 2
2
u magnetic
1 B2

2 0
RC/LC Circuits
i
Q+++
---
i
Q+++
---
C R
0
i
0
0
t
L
LC:
current oscillates
RC:
current decays exponentially
-i
C
0
t
LC Oscillations
(qualitative)
i0
+ +
- -
C
i  i0
L

C
Q  Q0
The image part with relationship ID rId18 was not found in the file.


i  i0
i0
C
Q0
L
L

-
-
+ +
C
L
Q  Q0
LC Oscillations
Q
Cos()
0
(qualitative)
VC
Cos()
0
i
-Sin()
0
VL
0
di
__
dt
-Cos()
1
0
-Cos()
1
t
t
How do these change if L
has a finite R?
Clicker
t=t1
t=0
At t=0, the capacitor in the LC
circuit shown has a total charge
Q0. At t = t1, the capacitor is
+ +
uncharged.
Q  Q0
L
– What is the value of Vab, the
C
voltage across the inductor
at time t1?
(a) Vab < 0
(b) Vab = 0
a
Q0
C
L
b
(c) Vab > 0
• Vab is the voltage across the inductor, but it is also the
voltage across the capacitor!
• Charge on the capacitor is zero,  VC = 0
• When Q = 0 on capacitor, I is maximum through
inductor
• and dI/dt is zero then so VL = 0 makes sense
Clicker
t=t1
t=0
At t = 0, the capacitor in the LC
circuit shown has a total charge
Q0. At t = t1, the capacitor is
+ +
uncharged.
Q  Q0
L
- -
a
Q0
C
C
L
b
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
At t = t1, the charge on the
capacitor is zero.
Q12
U C1 
0
2C
(c) UL1 > UC1
At t = t1, the current is a
maximum.
1 2 Q 20
0
U L1  LI1 
2
2C
LC Oscillations
(L with finite R)
• If L has finite Resistance, then
– energy will be dissipated in R and
– the oscillations will become damped.
Q
Q
0
0
t
R=0
t
R0
Quick checkpoint review
At time t = 0 the capacitor in the circuit below is fully charged with Qmax,
and the current through the circuit is 0.
• What is the potential difference across the
inductor at t = 0?
• Ans: VL = Qmax / C (same as capacitor)
• What is the potential difference across the
inductor when current is maximum?
• Ans: 0
• How much energy is stored in C when I is max?
• Ans: U = 0 (it’s all in the inductor)
LC Oscillations
(quantitative)
i
•
Begin with the loop rule:
Q
d 2Q
Q
L 2  0
C
dt
•
+ +
- -
C
L
Guess solution: (harmonic oscillator)
Q  Q 0 cos( 0 t  )
where:
In mechanics
d 2x
 kx  m 2
dt
•  determined from equation
• , Q0 determined from initial conditions
•
Procedure: differentiate above form for Q and substitute into
loop equation to find 
LC Oscillations
(quantitative)
• General solution:
+ +
- -
Q  Q 0 cos( 0 t  )
• Differentiate twice:
dQ
  0Q 0 sin( 0 t  )
dt
L
d 2Q
C
d 2Q
dt 2

L
Q
0
C
2



0 Q 0 cos( 0 t  )
2
dt
• Substitute into loop eqn:
1
L   20 Q 0 cos( 0 t  )   Q 0 cos( 0 t  )   0
C

0 
1
LC
1
C
   20 L   0
Clicker
At t = 0 the capacitor has charge Q0; the resulting
oscillations have frequency 0. The maximum
current in the circuit during these oscillations has
value I .
– What is the relation between 0 and 2 , the
frequency of oscillations when the initial
charge = 2Q0 ?
(a) 2 = 1/2 0
(b) 2 = 0
t=0
+ +
Q  Q0
- -
C
L
(c) 2 = 2 0
• Q0 determines the amplitude of the oscillations (initial condition)
• The frequency is determined by the circuit parameters (L,C) only
Clicker
t=0
• At t = 0 the capacitor has charge Q0; the resulting
oscillations have frequency 0. The maximum
current in the circuit during these oscillations has
value I .
+ +
Q  Q0
- -
C
What is the relation between I and I , the maximum current in
the circuit when the initial charge = 2Q0 ?
(a) I = I
(b) I = 2 I
(c) I = 4 I
• The initial charge determines the total energy : U0 = Q02/2C
• Maximum current occurs when Q = 0
• When all the energy is in the inductor: U = 1/2 LIo2
• Doubling initial charge quadruples total energy.
• Implies maximum current must double
L
LC Oscillations
Does solution conserve energy?
Energy in E field:
1 Q 2 (t )
1 2
U E (t ) 

Q0 cos 2 (0t   )
2 C
2C
Energy in B field:
U B (t ) 
1 2
1
Li ( t )  L 02Q02 sin 2 (0 t   )
2
2
0 
1
LC
1
U B (t ) 
Q 02 sin 2 ( 0 t   )
2C
Q 02
U E (t )  U B (t ) 
2C
YES !!
Quick checkpoint review
The capacitor charged such that the top plate has a charge +Q0 and the
bottom plate -Q0. At time t=0, the switch is closed …
• What is the value of the capacitor C?
• Ans: (500)2 x L = 1 / C
• C = 10-3 F
• Which of the following plots best represents the energy in the
inductor as a function of time starting just after the switch is closed?
Energy is always POSITIVE (proportional to Square of current)
Energy Plotted vs Time
UE
Q 20
UE (t)  UB (t) 
2C
0
t
UB
0
t
Clicker
At t = 0 the current flowing through the circuit is 1/2
of its maximum value.
– Which of the following is a possible value for
the phase , when the charge on the capacitor
is described by: Q(t) = Q0cos(t + ).
(a)  = 30
(b)  = 45
i
Q
+ +
- -
C
L
(c)  = 60
• We are given a form for the charge on the capacitor as a function of
time, but we need to know the current as a function of time.
I(t) 
dQ
  ω0Q 0sin(ω0 t  φ)
dt
• At t = 0, the current is given by: I(0)   ω0 Q 0sinφ
• The maximum value of the current is: I max  ω 0 Q 0
• Therefore, the phase angle must be given by: sinφ  
1
2
 φ  30
i
Clicker
At t = 0 the current flowing through the circuit is 1/2 of
+ +
its maximum value.
Q
Which of the following plots best represents UB, the - energy stored in the inductor as a function of time?
UB
UB
0
0
UB
0
time
L
(c)
(b)
(a)
C
0
0
time
0
time
Energy stored in the inductor proportional to the CURRENT SQUARED.
If the current at t = 0 is 1/2 its maximum value, the energy stored in the
inductor will be 1/4 of its maximum value.
LCR Damping
For your interest, we do not derive here, but only illustrate the
following behavior
R  R0
R
+ +
- -
Q
C
L
Q  Q 0e t cos( 'o t  )

R
2L
 1
R2 
'o  
 2
 LC 4 L 
0
t
In a LRC circuit,  depends also on R !
Q
R
R0
4
0
t