Inductance
Self-Inductance
A
l
+
–
•Consider a solenoid L, connect it to a battery
•Area A, length l, N turns
•What happens as you close the switch?
•Lenz’s law – loop resists change in magnetic field
•Magnetic field is caused by the current
•“Inductor” resists change in current
E
 NI
 NIA
B 0
 B1  0
d  0 N 2 IA 
dB
0 N 2 A dI
d
EL  
  N  B1   

dt 
dt
dt
dt

E  L
dI
dt
L
0 N 2 A
Inductors
•An inductor in a circuit is denoted by this symbol:
•An inductor satisfies the formula:
dI
E  L
•L is the inductance
dt
•Measured in Henrys (H)
1 H  1 V  s/A
Kirchoff’s rules for Inductors:
I
•Assign currents to every path, as usual
•Kirchoff’s first law is unchanged
•The voltage change for an inductor is L (dI/dt)
+
•Negative if with the current
E–
•Positive if against the current
•In steady state (dI/dt = 0) an inductor is a wire
L
L
Energy in Inductors
•Is the battery doing work on the inductor?
dI
P  I  V   IL
dt
L
+
•Integral of power is work done on the inductor
E–
dI
U   P dt   IL dt  L  IdI  12 LI 2  k
dt
•It makes sense to say there is no energy in inductor with no current
U  12 LI 2
L  0 N 2 A
•Energy density inside a solenoid?
U
0 N 2 I 2
0 N 2 AI 2
u

U
A
2 2
2
B
0 NI
•Just like with electric fields, we can associate the energy
with the magnetic fields, not the current carrying wires
B2
u
2 0
RL Circuits
 L R
I  I 0 e t 
Rt
ln I    constant
L
I
–
I  e  Rt L
dI
R
 I   L  dt
R
dI
R
  dt
I
L
L
+
•An RL circuit has resistors and inductors
•Suppose initial current I0 before you open the switch
•What happens after you open the switch?
•Use Kirchoff’s Law on loop
•Integrate both sides of the equation
dI
RI
dI

0  L
 RI
dt
L
dt
RL Circuits (2)
•Where did the energy in the inductor go?
•How much power was fed to the resistor?
P  RI 2  RI 02 e  2tR L
I  I 0 e t 
 L R
•Integrate to get total energy dissipated
U R   P dt  RI
0

2
0
e
 2 Rt L
0
•It went to the resistor
•Powering up an inductor:
•Similar calculation
R
L
E
+
E
I  1  e t  
R
L  Rt L  1 2
e
dt   RI
 2 I0 L
0
2R
U L  12 LI 02
2
0
–

An inductor with inductance 4.0 mH is
discharging through a resistor of resistance R. If,
in 1.2 ms, it dissipates half its energy, what is R?
I  I 0 e t 
U 0  12 LI 02
U  12 LI 2  12 U 0  14 LI 02
I
1

 0.707
I0
2
I 2f  12 I 02
I
 e t   0.707
I0
 t   ln  0.707   0.347
t
1.2 103 s


 0.00346 s
0.347
0.347
.00400 H
R 
.00346 s

L
R  1.16 
I
 L R
L = 4.0 mH
Sample Problem
R
Inductors in Series
L1
L2
•For inductors in series, the
inductors have the same current
•Their EMF’s add
dI
dI
dI
E


L

L


L

L
 1 2
1
2
L  L1  L2
dt
dt
dt
Inductors in Parallel
•For inductors in parallel, the
inductors have the same EMF
but different currents
L1
dI1
E   L1
dt
dI 2
E   L2
dt
L2
E
dI dI1 dI 2   E  E



L1 L2
L
dt dt
dt
1 1 1
 
L L1 L2
Parallel and Series - Formulas
Capacitor
Resistor
Inductor
1 1
1
 
C C1 C2
R  R1  R2
L  L1  L2
Parallel
C  C1  C2
1 1
1
 
R R1 R2
1 1 1
 
L L1 L2
Fundamental
Formula
Q
V 
C
V  IR
dI
EL   L
dt
Series
Materials Inside Inductors
•For capacitors, we gained significantly by putting materials inside
•Can we gain any benefit by putting something in inductors?
•It gives you something to wrap around
Can materials increase the inductance?
•Most materials have negligible magnetic properties
•A few materials, like iron are ferromagnetic
•They can enhance inductance enormously
•Many inductors (and similar devices, like
transformers) have iron cores
•We will ignore this
Symbol for iron core inductor:
•We won’t make this distinction
Mutual Inductance
+
–
•Consider two solenoids sharing the same volume
What happens as you close the switch?
•Current flows in one coil
•But Lenz’s Law wants mag. flux constant
•Compensating current flows in other coil
•Allows you to transfer power without circuits
E
being actually connected
•It works even better if source is AC from
generator
I1
I2
LC Circuits
Q0  C V  CE
•Inductor (L) and Capacitor (C)
•Let the battery charge up the capacitor
Now flip the switch
•Current flows from capacitor through inductor
•Kirchoff’s Loop law gives:
Q
dI
0
L
•Extra equation for capacitors:
dt
C
dQ
dI
d  dQ 
 I
Q  CL  CL 

dt
dt
dt  dt 
d 2Q
1

Q
2
dt
CL
Q
cos t 
Q
sin t 
I
E
Q  Q0 cos t 
•What function, when you take two derivatives, gives the same things with a minus sign?
•This problem is identical to harmonic
oscillator problem
+
–
C Q
L
LC Circuits (2)
•Substitute it in, see if it works
Q  Q0 cos t 
dQ
 Q0 sin t 
1
dt

2
CL
d Q
2
  Q0 cos t 
2
dt
1
 2Q0 cos t   
Q0 cos t 
CL
Q02
•Let’s find the energy in the
Q2
UC 
UC 
cos 2 t 
capacitor and the inductor
2C
2C
dQ
I 
 Q0 sin t 
dt
U L  12 LI 2  12 LQ02 2 sin 2 t 
Q02
UL 
sin 2 t 
2C
Energy sloshes
back and forth
d 2Q
1

Q
2
dt
CL
I
C Q
L
Q02
UC  U L 
2C
Frequencies and Angular Frequencies
•The quantity  is called the angular frequency
•The period is the time T you have to wait for it to repeat
•The frequency f is how many times per second it repeats
1
Q  Q0 cos t 

CL
T  2
f 1 T
  2 f
T
WFDD broadcasts at 88.5 FM, that is, at a
frequency of 88.5 MHz. If they generate
this with an inductor with L = 1.00H,
what capacitance should they use?
8 1
  2 f  2  88.5 106 s 1   5.56 10 s
 LC  1
2
C
1
 L
2

 5.56 10
1
8
s
 10
1 2
6
H
C  3.23 pF
RLC Circuits
•Resistor (R), Inductor (L), and Capacitor (C)
•Let the battery charge up the capacitor
Now flip the switch
E
•Current flows from capacitor through inductor
•Kirchoff’s Loop law gives:
Q
dI
0


L
 RI
•Extra equation for capacitors:
C
dt
2
dQ
Q
dQ
d
Q
 I 0   R
L 2
dt
C
dt
dt
•This equation is hard to solve, but not impossible
•It is identical to damped, harmonic oscillator
Q  Q0e
 Rt 2 L
cos t 

1
R2
 2
LC 4 L
I
+
–
C Q
L
R