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Chapter 30
Inductance, Electromagnetic
Oscillations, and AC Circuits
Copyright © 2009 Pearson Education, Inc.
Recap:
2
V
 IV
Ohm's Law: V  IR; P  I 2 R 
R
1 Q2 1
1
2
 QV
Capacitors: Q  CV ; U  CV 
2 C 2
2
RC Circuit:   RC
dQ  0 t 
 e
Q  t   Q0 1  e  ; I  t  
Charge:
R
dt
dQ  0 t 
 e
Discharge: Q  t   Q0e t  ; I  t   
R
 dt
dI
dB
NB
 L
;  t    N
Inductors: L 
dt
dt
I
t 
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30-4 LR Circuits
A circuit consisting of an inductor
and a resistor will begin with most
of the voltage drop across the
inductor, as the current is
changing rapidly. With time, the
current will increase less and less,
until all the voltage is across the
resistor.
dI  t 
0  V0  I  t  R  L
dt
dI  t  R
 I  t   V0
dt
L
I t  0  0
V0
L
t 
 I  t   1  e  ;  
R
R
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0
30-4 LR Circuits
If the circuit is then
shorted across the
battery, the current will
gradually decay away:
dI  t 
0  I t  R  L
dt
dI  t  R
 I t   0
L
dt
V0
I t  0 
R
V0 t 
L
 I t   e ;  
R
R
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0
30-4 LR Circuits
Example 30-6: An LR circuit.
At t = 0, a 12.0-V battery is
connected in series with a
220-mH inductor and a total of
30-Ω resistance, as shown. (a)
What is the current at t = 0? (b)
What is the time constant? (c)
What is the maximum current?
(d) How long will it take the
current to reach half its
maximum possible value? (e)
At this instant, at what rate is
energy being delivered by the
battery, and (f) at what rate is
energy being stored in the
inductor’s magnetic field?
Copyright © 2009 Pearson Education, Inc.
30-5 LC Circuits and
Electromagnetic Oscillations
An LC circuit is a charged capacitor
shorted through an inductor.
Q t 
dI  t 
L
C
dt
dQ  t 
I t   
 dt
d 2Q  t 
1
0
Q t  
LC
dt
0
 Q  t   Q0 cos  t    ;  
Q  t  0   Q0 ; I  t  0   0
1
LC
 Q  t   Q0 cos  t 
 I t   
dQ
 Q0 sin  t   I 0 sin  t 
dt
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30-5 LC Circuits and
Electromagnetic Oscillations
The charge and current are both
sinusoidal, but with different phases.
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30-5 LC Circuits and
Electromagnetic Oscillations
The total energy in the circuit is
constant; it oscillates between the
capacitor and the inductor:
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30-5 LC Circuits and
Electromagnetic Oscillations
Example 30-7: LC circuit.
A 1200-pF capacitor is fully charged by a
500-V dc power supply. It is
disconnected from the power supply
and is connected, at t = 0, to a 75-mH
inductor. Determine: (a) the initial
charge on the capacitor; (b) the
maximum current; (c) the frequency f
and period T of oscillation; and (d) the
total energy oscillating in the system.
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30-6 LC Oscillations with Resistance
(LRC Circuit)
Any real (nonsuperconducting) circuit will
have resistance.
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30-6 LC Oscillations with Resistance
(LRC Circuit)
Now the voltage drops around the circuit give
A current flowing through a resistor
means energy is dissipated so the solution
must die out over time.
The solutions to this equation are
damped harmonic oscillations.
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30-6 LC Oscillations with Resistance
(LRC Circuit)
The system will be underdamped for R2 < 4L/C,
and overdamped for R2 > 4L/C. Critical damping
will occur when R2 = 4L/C. This figure shows
the three cases of underdamping,
overdamping, and critical damping.
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30-6 LC Oscillations with Resistance
(LRC Circuit)
The angular frequency for critical and under damped
oscillations is given by
'
1
R2
R2
1

is imaginary in overdamped case; 2 
LC 4 L2
4L
LC
and the charge in the circuit as a function of time is
Q  Q0e

R
t
2L
cos   ' t   
The over damped case, is more complicated but the
solutions look like
Q  Q0e
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
R
t
2L
cosh   '' t  ;  '' 
R2
1

4 L2 LC
( don’t worry
about this)
30-6 LC Oscillations with Resistance
(LRC Circuit)
Example 30-8: Damped oscillations.
At t = 0, a 40-mH inductor is placed in
series with a resistance R = 3.0 Ω and a
charged capacitor C = 4.8 μF. (a) Show
that this circuit will oscillate. (b)
Determine the frequency. (c) What is the
time required for the charge amplitude to
drop to half its starting value? (d) What
value of R will make the circuit
nonoscillating?
Copyright © 2009 Pearson Education, Inc.
30-7 AC Circuits with AC Source
Resistors, capacitors,
and inductors have
different phase
relationships between
current and voltage
when placed in an ac
circuit.
The current through
a resistor is in phase
with the voltage.
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30-7 AC Circuits with AC Source
The voltage across the
inductor is given by
Earlier

or
.
Therefore, the current
through an inductor
lags the voltage by 90°.
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Later
30-7 AC Circuits with AC Source
The voltage across the inductor is related
to the current through it:
.
The quantity XL is called the inductive
reactance, and has units of ohms:
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30-7 AC Circuits with AC Source
Example 30-9: Reactance of a coil.
A coil has a resistance R = 1.00 Ω and
an inductance of 0.300 H. Determine
the current in the coil if (a) 120-V dc is
applied to it, and (b) 120-V ac (rms) at
60.0 Hz is applied.
Copyright © 2009 Pearson Education, Inc.
30-7 AC Circuits with AC Source
The voltage across the
capacitor is given by
Later
.
Earlier
Therefore, in a capacitor,
the current leads the
voltage by 90°.
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
30-7 AC Circuits with AC Source
The voltage across the capacitor is related
to the current through it:
.
The quantity XC is called the capacitive
reactance, and (just like the inductive
reactance) has units of ohms:
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30-7 AC Circuits with AC Source
Example 30-10: Capacitor reactance.
What is the rms current in the circuit
shown if C = 1.0 μF and Vrms = 120 V?
Calculate (a) for f = 60 Hz and then (b) for
f = 6.0 x 105 Hz.
Copyright © 2009 Pearson Education, Inc.
30-7 AC Circuits with AC Source
This figure shows a high-pass filter (allows
an ac signal to pass but blocks a dc voltage)
and a low-pass filter (allows a dc voltage to
be maintained but blocks higher-frequency
fluctuations).
XC  1
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 large for small 
 C   small for large 
30-8 LRC Series AC Circuit
Analyzing the LRC series AC circuit is
complicated, as the voltages are not in phase
– this means we cannot simply add them.
Furthermore, the reactances depend on the
frequency.
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30-8 LRC Series AC Circuit
We calculate the voltage (and current) using
phasors – these are vectors representing the
individual voltages.
Here, at t = 0, the
current and
voltage are both at
a maximum. As
time goes on, the
phasors will rotate
counterclockwise.
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30-8 LRC Series AC Circuit
Some time t later,
the phasors have
rotated.
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30-8 LRC Series AC Circuit
The voltage across
each device is given
by the x-component of
each, and the current
by its x-component.
The current is the
same throughout the
circuit.
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30-8 LRC Series AC Circuit
We find from the ratio of voltage to
current that the “effective resistance,”
called the impedance, of the circuit is
given by
BUT – only an actual resistance
dissipates energy. The inductor and
capacitor store it then release it.
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30-8 LRC Series AC Circuit
The phase angle between the voltage and
the current is given by
or
The factor cos φ is called the
power factor of the circuit.
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30-8 LRC Series AC Circuit
Example 30-11: LRC circuit.
Suppose R = 25.0 Ω, L = 30.0 mH, and
C = 12.0 μF, and they are connected in
series to a 90.0-V ac (rms) 500-Hz
source. Calculate (a) the current in the
circuit, (b) the voltmeter readings
(rms) across each element, (c) the
phase angle , and (d) the power
dissipated in the circuit.
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30-9 Resonance in AC Circuits
The rms current in an ac circuit is
Clearly, Irms depends on the frequency.
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30-9 Resonance in AC Circuits
We see that Irms will be a maximum when XC
= XL; the frequency at which this occurs is
f0 = ω0/2π is called the
resonant frequency.
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30-10 Impedance Matching
When one electrical circuit is connected to
another, maximum power is transmitted when
the output impedance of the first equals the
input impedance of the second.
The power
delivered to the
circuit will be a
maximum when
dP/dR2 = 0;
this occurs
when R1 = R2.
Copyright © 2009 Pearson Education, Inc.
Summary of Chapter 30
• LR circuit:
.
.
• Inductive reactance:
• Capacitive reactance:
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Summary of Chapter 30
• LRC series circuit:
.
• Resonance in LRC series circuit:
Copyright © 2009 Pearson Education, Inc.