Ch. 32
Self Inductance
B
0 NI
 B1 
0 NIA
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
2
d
dB
 0 N 2 IA 
d

N
A dI
0
 N
 B1   
EL  
 
dt
dt
dt 
dt

dI
E  L
L  0 N 2 A
dt
Warmup 18
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)
L
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
What is Kirchoff’s law for the loop shown?
A) E + L (dI /dt) = 0 B) E – L (dI /dt) = 0
C) None of the above
D) I don’t know Kirchoff’s law for switches
L
dI
0  E L
dt
dI E
E

I t
dt L
L
JIT Quick Quiz 32.1
A coil with zero resistance has its ends labeled a and b. The potential at a is higher than at b.
Which of the following could be consistent with this situation?
(a) The current is constant and is directed from a to b
(b) The current is constant and is directed from b to a.
(c) The current is increasing and is directed from a to b.
(d) The current is decreasing and is directed from a to b
(e) The current is increasing and is directed from b to a
(f) The current is decreasing and is directed from b to a
Ans c, f
Ex- (Serway 32-9) The current in a 90.0 mH inductor changes with time as I = t2 - 6t (in SI units). Find
the magnitude of the induced emf at (a) t = 1.00 s and (b) t = 4.00 s. (c) At what time is the emf zero.
Solve on
Board
Warmup 18
Energy in Inductors
•Is the battery doing work on the inductor?
P  I  V 
dI
 IL
dt
L
+
E–
•Integral of power is work done on the inductor
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
L  0 N 2 A
U  12 LI 2
B
0 NI
•Energy density inside a solenoid?
U
0 N 2 AI 2
2
2
2 2
U
B

N
I
u

 0 2
A
2 0
2
•Just like with electric fields, we can associate the energy
with the magnetic fields, not the current carrying wires
B2
u
2 0
6A
6A
6A
R2 = 4 
L
+
–
•Circuits with resistors (R) and inductors (L)
In the steady state, with the switch closed,
how much current flows through R2? How
much current flows through R2 the moment
after we open the switch?
A) 0 A
B) 6 A
C) 3 A
D) 2 A
E) None of the above
R1 = 2 
RL Circuits
E = 12 V
•In the steady state, the inductor is like a wire
•Both ends of R2 are at the same potential: no current through R2
•The remaining structure had current I = E/R1 = 6 A running through it
I = E /R1 = 6 A
•Now open the switch – what happens?
•Inductors resist changes in current, so the current
instantaneously is unchanged in inductor
•It must pass through R2
I=6A
I
R = 4 
L
+
–
•What happens after you open the switch?
•Initial current I0
•Use Kirchoff’s Law on loop
•Integrate both sides of the equation
R1 = 2 
RL Circuits (2)
dI
dI
RI
E = 12 V
0  L
 RI


dt
dt
L
dI
R
Rt
dI
R
  dt
ln I  
 constant


dt
I
L
I

L
L
I  e  Rt L
 L R
I  I 0 e t 
RL Circuits (3)
•Where did the energy in the inductor go?
•How much power was fed to the resistor?
I  I 0 e t 
 L R
P  RI 2  RI 02 e  2tR L
•Integrate to get total energy dissipated

U   P dt  RI
0
U  12 LI 02

2
0
e
 2 Rt L
0
L  2 Rt L 
2 L
dt   RI
  RI 0
e
0
2R
2R
2
0
•It went to the resistor
•Powering up an inductor:
•Similar calculation
R
L
–
E
+
E
I  1  e t  
R
 L R
I  I 0 e t 
U 0  12 LI 02
I 2f  12 I 02
L = 4.0 mH
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
Sample Problem
U  12 LI 2  12 U 0  14 LI 02
I
1

 0.707
I0
2
 0.707
 t   ln  0.707   0.347
I
 e t 
I0
3
1.2

10
s
t

 0.00346 s
 
0.347
0.347
L
.00400 H
R
R

.00346 s
R  1.16 
R
I=1A
R2 = 1 k
L
+
–
The circuit at right is in a steady state.
What will the voltmeter read as soon as
the switch is opened?
A) 0.l V
B) 1 V
C) 10 V
D) 100 V
E) 1000 V
R1 = 10 
Concept Question
E = 10 V
•The current remains constant at 1 A
•It must pass through resistor R2
V  IR  1 A 1000  
•The voltage is given by V = IR
V  1000 V
+
–
•Note that inductors can produce very high
voltages
•Inductance causes sparks to jump when you
turn a switch off
Loop has unintended inductance
V
JIT Quick Quiz 32.2
The circuit in the figure consists of a resistor, an inductor, and an ideal battery with no internal
resistance.
At the instant just after the switch is closed, across which circuit element is the voltage equal to
the emf of the battery?
(a) the resistor
(b) the inductor
(c) both the inductor and resistor
After a very long time, across which circuit element is the voltage equal to the emf of the
battery?
(d) the resistor
(e) the inductor
(f) both the inductor and resistor
Ans b, d
CT-1- When the switch is closed, the current through the circuit exponentially approaches a
value I=E/R. If we repeat this experiment with an inductor having twice the number of turns per
unit length, the time it takes for the current to reach a value of I/2
A increases.
B. decreases.
C. is the same.
Ans A
CT-2 When the switch in the circuit below is closed, the brightness of the bulb

+
R
-
Assume inductor has no
resistance
Bulb
L
A.
B.
C.
D.
Starts off at its brightest and then dims.
Slowly reaches its maximum brightness.
Immediately reaches it maximum, constant brightness.
Something else.
Ans A
Inductors in series and parallel
L1
L2
•For inductors in series, the
inductors have the same current
dI
dI
dI
•Their EMF’s add
E   L1
 L2
   L1  L2 
L  L1  L2
dt
dt
dI1
E   L1
•For inductors in parallel, the
dt
inductors have the same EMF
dI 2
but different currents
E   L2
dt
E
E E
dI dI1 dI 2

 


L
L1 L2
dt
dt
dt
1
1
1


L L1 L2
dt
L1
L2
Parallel and Series - Formulas
Capacitor
Series
Resistor
Inductor
1
1
1


R  R1  R2
C C1 C2
L  L1  L2
1
1
1


R R1 R2
1
1
1


L L1 L2
V  IR
dI
EL   L
dt
Parallel
C  C1  C2
Fundamental
Formula
Q
V 
C
Warmup 19
CT - 3- The primary coil of a transformer is connected to a battery, a resistor, and a switch.
The secondary coil is connected to an ammeter. When the switch is thrown closed, the
ammeter shows
A.
B.
C.
D.
zero current.
a nonzero current for a short instant.
a steady current.
Something else.
Ans B
LC Circuits
Q0  C V  CE
•Inductor (L) and Capacitor (C)
I
•Let the battery charge up the capacitor
Now flip the switch
C Q
+
•Current flows from capacitor through inductor E –
•Kirchoff’s Loop law gives:
Q
dI
L
0

L
•Extra equation for capacitors:
C
dt
dQ
 I
dt
d 2Q
1

Q
2
dt
CL
dI
d  dQ 
Q  CL
 CL 

dt
dt  dt 
Q cos  t 
Q  Q0 cos t 
Q sin t 
•What function, when you take two derivatives, gives the same things with a minus sign?
•This problem is identical to harmonic
oscillator problem
Q  Q0 cos t 
LC Circuits (2)
•Substitute it in, see if it works
d 2Q
1

Q
2
dt
CL
dQ
 Q0 sin t 
dt
d 2Q
2




Q0 cos t 
2
dt
1
2
 Q0 cos t   
Q0 cos t 
CL
•Let’s find the energy in the capacitor and
2
Q
the inductor
UC 
dQ
2C
 Q0 sin t 
I 
dt 2
U L  12 LI  12 LQ02 2 sin 2 t 
Q02
UL 
sin 2 t 
2C
Energy sloshes
back and forth
1
CL
I
C Q
L
Q02
UC 
cos 2 t 
2C
U C  U L  Q02 2C
Warmup 19
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
Q  Q0 cos t 

1
CL
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?
T
T  2
f 1 T
  2 f
  2 f  2  88.5 106 s 1   5.56 108 s 1 C  3.23 pF
 2 LC  1
1
1
C 2 
8
1 2
6
 L
5.56

10
s
10

  H
Ex – Serway (32-49) A 1.00 µF capacitor is charged by a 40 V power supply. The fully-charged
capacitor is then discharged through a 10.0 mH inductor. Find the maximum current in the resulting
oscillations.
Solve on
Board
RLC Circuits
•Resistor (R), Inductor (L), and Capacitor (C)
•Let the battery charge up the capacitor
Now flip the switch
•Current flows from capacitor through inductor E
Q
dI
•Kirchoff’s Loop law gives:
0 L
 RI
•Extra equation for capacitors: C
dt
dQ
 I
dt
Q
dQ
d 2Q
0 R
L 2
C
dt
dt
I
+
–
•This equation is hard to solve, but not impossible  
•It is identical to damped, harmonic oscillator
Q  Q0e Rt 2 L cos t 
C Q
L
R
1
R2
 2
LC 4 L