An important note on units (for solving
problems):
f has units of Hz = 1/s
ω has units of rad/s
To convert between the two:
ω=2πf
ALTERNATING CURRENT
CIRCUITS I
Purdue University – Physics 241 – Lecture 19
Brendan Sullivan
Overview

Room Mean Square (rms) quantities

Resistors in AC circuits

Inductors in AC circuits

Capacitors in AC circuits

LC and RLC circuits
Root Mean Square Values are a way
to average periodic functions



When talking about AC circuits we wish to discuss
average quantities since instantaneous would be
meaningless
The average of a sine wave is zero, so we gain no
information by analyzing its average
To get the rms of any periodic function, average the
square of the function, then take the square root
1.5
1
X rms
2
( X ) avg
0.5
0
-0.5 0
-1
-1.5
1
2
3
4
5
6
7
RMS of Sine and Cosine

Average sin2 over a whole period:
2
(sin ( ))avg

1
2
2
1
2
2
sin ( )d
0
2
1
[1 cos(2 )]d
2
0
1
2
To get the rms, we take the root of the average of
the square
(sin 2 ( ))avg
(sin( ))rms
1
2
1
2
0.707
1.5
1
0.5
0
-0.5
-1
-1.5
0
1
2
3
4
5
6
7
Sine (blue) and sine squared
(orange). Sine’s rms value is shown
by the red line.
RMS of a Sawtooth Function


X max
)t
T
Sawtooth wave:
0
We’ll get the average of X2:
X (t ) (
T
( X (t ))

2
avg
x T
T
1
1 X max 2
X (t )dt
(
t ) dt
T0
T0 T
1 2
X max
3
Now, the rms is just the square root of that:
X max
( X (t ) 2 ) avg
( X (t ))rms
3
X m ax
X
T
T
T
T
t
X max
3
Quiz Question 1
In the United States, wall sockets are regulated at
120V rms at 60Hz. Determine the peak voltage
out of a wall socket.
a) 2V b) 42.4V c) 85V d) 120V e) 169V
Alternating current generators are like
batteries with sinusoidal voltage



ac generator: spin a coil
through a magnetic field
at a constant frequency
This induces a sinusoidal
potential difference in
time
In circuits, we use these
like a battery with a time
varying voltage
The flux (blue ~sine) through the
loop and the induced current
(orange ~cosine)
Resistors in AC circuits

--
+
At all times, the potential difference
across the resistor must be the same as
that across the battery (for this circuit)
 For
an AC generator, this means a
sinusoidal voltage drop
I(t)
By Ohm’s Law, this means we must have
(t ) I (t ) R VR (t ) a sinusoidal current

I rms
I peak
 That
rms
R
peak
R

varies only by the constant R
We can discuss both peak and rms
values for current and voltage
Finding the average power delivered
by the generator

Earlier in the semester, we found the power output
by a battery. We add time dependence:
P IV


peak
cos( t )) I peak
peak
cos2 ( t )
Now, we’ll take the average
Pavg

( I peak cos( t ))(
( I peak
2
peak cos ( t ))avg
I peak
2
peak
I peak
2
peak
2
I rms
rms
So, the power is just the rms voltage times the rms
current!
You can verify that this is also the power dissipated
in the resistor
Inductors in AC circuits


Loop Rule:
dI
L
max cos( t )
dt
0 P
Differential Equation Soln:
I (t )
VL , peak
L
sin( t )
I peak sin( t )
With VL,peak=εmax (at some
point, all voltage must be
dropped across the
inductor)
For a driven inductor, voltage leads
current by 90° (quarter cycle)!
There is no net power delivered to
this circuit (an inductive load):
IV
1
( I peak sin( t ))( max cos( t ))
I peak
2
Pavg (sin( 2 t )) avg 0
max
sin(2 t )
Inductive Reactance

Let’s consider the rms current for an
inductor/generator: I V L
This looks almost like Ohm’s Law: I = V/R
We can make this behave like Ohm’s Law if we
define the reactance of an inductor: X L
L , rms
rms


L
I rms

VL,rms
L
VL,rms
XL
Reactance is a frequency dependent, effective
resistance for an inductor. It has units of Ohms (Ω)
Quiz Question 2
An inductor of inductance
5mH is being driven at
200Hz (ω = 1257 rad/s).
If the generator has an rms
voltage of ε =12V, find the
rms current.
a) 1.3 A
b) 1.9 A
c) 2.7 A
d) 8.5 A
e) 12 A
Capacitors in AC circuits



Now, current leads voltage by
a quarter of a cycle
 It was the opposite for an
inductive load
Again, the average power
delivered by the circuit is zero
We can easily relate the
max(rms) current and charge:
Loop Rule:
max cos( t )
q(t )
C
Solve for q(t):
q(t ) VcC cos( t )
Q peak cos( t )
Use this to get current:
I peak
Q peak
I rms
Qrms
I (t )
dq(t )
dt
q peak sin( t )
0
Capacitive Reactance

Just as we defined an effective, frequency
dependent resistance for inductors, we can do the
same for capacitors
 Allows
I peak
us to nicely recover Ohm’s Law
Qpeak
XC
CVC , peak
VC , peak
XC
1
C
Like inductive reactance, capacitive
reactance has units of Ohms!
Undriven LC Circuits
We start with a
charged capacitor
and no current.
LC Oscillations
2
d Q
dt 2
f0
1
Q
LC
0
0
1
LC
0
2
The capacitor acts like
a battery, so a current
starts running.
We went from zero
current to some
current, so an emf is
induced in the
inductor.
The cycle repeats
itself.
UE
Q2
, UB
2C
1 2
LI , I
2
dQ
dt
Undriven LC Circuits

Loop Rule:
dI
L
dt
d 2Q
L 2
dt
Q
C
0
Q
C
0
This is the “mass on a
spring” equation:
Q (t )
I (t )

The circuit oscillates back and forth
between charge stored on the
capacitor and charge moving around
the circuit
It oscillates back and forth with a
1
natural frequency
LC
Qpeak
t
Q peak cos( t )
I peak sin( t )
1
LC
Ipeak
Charge (blue)
and current (orange)
Energetics of an Undriven LC Circuit

Energy stored in a capacitor:
UC

Q2
2C
2
Q peak
cos 2 ( t )
Energy stored in an inductor:
UL
1 2
LI
2
1 2
LI peak sin 2 ( t )
2
U
Q (t )
I (t )
Orange:
Capacitor
Energy
Q peak cos( t )
Blue: Inductor
Energy
I peak sin( t )
Green: Sum
(notice it’s
constant)
t
Undriven RLC Circuit


Now, we’ll add a resistor to the circuit
Apply the loop rule: d 2Q dQ Q
L


R
dt
C
0
This is like the equation of motion for a
damped mass on a spring


dt
2
The resistance term acts as a damping
force; it slowly drains energy from the
system
Large R: Energy is drained too quickly
to oscillate
Small R: Gradually diminishing
oscillation
Comparing (R)LC circuits to a harmonic
oscillator
Charge behaves like position (q,x)
Current behaves like velocity (I=dq/dt, v = dx/dt)
U
Inductance behaves like mass (L,m)
Capacitance behaves like inverse spring constant
Energy stored in the capacitor
is comparable to kinetic energy
t
Energy stored in the inductor is
comparable to kinetic energy
Quiz Question 3
Suppose we have an undriven RLC circuit oscillating at
frequency ω0. We then double the capacitance.
Determine the new frequency of oscillations, ω.
a) ω = 4 ω0
b) ω = 2 ω0
c) ω = ω0
d) ω = ½ ω0
e) ω = ¼ ω0