Oscillating Currents
• Ch.30: Induced E Fields: Faraday’s Law
• Ch.30: RL Circuits
• Ch.31: Oscillations and AC Circuits
Review: Inductance
• If the current through a coil of wire changes,
there is an induced emf proportional to the rate of
change of the current.
•Define the proportionality constant to be the
inductance L:
ε
di
= −L
dt
• SI unit of inductance is the henry (H).
LC Circuit Oscillations
Suppose we try to discharge a capacitor, using an
inductor instead of a resistor:
At time t=0 the capacitor
has maximum charge and
the current is zero.
Later, current is increasing
and capacitor’s charge is
decreasing
Oscillations (cont’d)
What happens when q=0?
Does I=0 also?
No, because inductor does
not allow sudden changes.
In fact, q = 0 means i = maximum!
So now, charge starts to build up on C
again, but in the opposite direction!
Textbook Figure 31-1
Energy is moving back and forth between C,L
1
2
U L = U B = Li
2
1
2
UC = U E = q / C
2
Textbook Figure 31-1
Mechanical Analogy
• Looks like SHM (Ch. 15) Mass on spring.
• Variable q is like x, distortion of spring.
• Then i=dq/dt, like v=dx/dt, velocity of mass.
By analogy with SHM, we can guess that
q = Q cos(ω t )
dq
i=
= −ωQ sin(ω t )
dt
Look at Guessed Solution
q = Q cos(ω t )
q
i
dq
i=
= −ωQ sin(ω t )
dt
Mathematical description of
oscillations
Note essential terminology:
amplitude, phase, frequency, period,
angular frequency. You MUST
know what these words mean! If
necessary review Chapters 10, 15.
Get Equation by Loop Rule
If we go with the
current as shown,
the loop rule gives:
q
di
− −L =0
C
dt
Now replace i by
dq/dt to get:
d q
 1 
= −
q
2
dt
 LC 
2
Guess Satisfies Equation!
Start with
so that
q = Q cos(ω t )
dq
= −ω Q sin(ω t )
dt
d q
 1 
= −
q
2
dt
 LC 
2
And taking one more derivative gives us
2
d q
2
2
= −ω Q cos(ω t ) = −ω q
2
dt
So the solution is correct, provided our angular
frequency satisfies
 1 
2
ω =

 LC 
LC Circuit Example
Given an inductor with L = 8.0 mH and a
capacitor with C = 2.0 nF, having initial charge
q(0) = 50 nC and initial current i(0) = 0.
(a) What is the frequency of oscillations (in Hz)?
(b) What is the maximum current in the inductor?
(c) What is the capacitor’s charge at t = 30 µs?
Example: Part (a)
L = 8 mH, C = 2 nF, q(0) = 50 nC, i(0) = 0
(a) What is the frequency of the oscillations?
ω =
1
1
5 rad
=
= 2.5 × 10
−3
−9
s
LC
8 × 10 × 2 × 10
ω 2.5 × 10 rad / s
4
f =
= 4 × 10 = 40 kHz
=
2π 6.28 rad / cycle
5
Example: Part (b)
( L = 8 mH, C = 2 nF, q(0) = 50 nC, i(0) = 0 )
(b) What is the maximum current?
−9 2
Q
( 50 × 10 )
−7
E=
= 6.25 × 10 J
=
−9
2C 2 × 2.0 × 10
2
But E =
So
I=
1
2
LI
2E
=
L
2
so
2E
I =
L
2
2 × 6.25 × 10
−3
8 × 10
−7
= 12.5 mA
Example: Part (c)
( L = 8 mH, C = 2 nF, q(0) = 50 nC, i(0) = 0 )
(c) What is the charge at t = 30 µs?
q( t ) = Q cos(ωt )
−6
= ( 50nC ) cos( 2.5 × 10 × 30 × 10 )
= ( 50nC ) cos( 7.5 rad )
= ( 50nC ) cos( 430°)
5
= 50 × 0.347 = 17 nC
AC Voltage Sources
•
For an AC circuit, we need an
alternating emf, or AC power
supply.
• This is characterized by its
amplitude and its frequency.
ε =ε
amplitude
m
sin ωt
angular frequency
Notation for oscillating functions
Note that the textbook uses lower-case
letters for oscillating time-dependent
voltages and currents, with upper-case
letters for the corresponding amplitudes.
v = V sin ωt
i = I sin ωt
AC Currents and Voltages
ε =ε
m
Ohm’s Law gives:
sin ωt = vR = VR sin ωt
iR = vR / R = (ε m / R ) sin ωt
So the AC current is:
iR = I R sin ωt
So the amplitudes are related by:
VR = I R R
AC Voltage-Current Relations
•
First apply an alternating emf to a resistor, a
capacitor, and an inductor separately, before
dealing with them all at once.
• For C, L, define reactance X analogous to
resistance R for resistor. Measured in ohms.
VR = I R R
VC = I C X C
VL = I L X L
Summary for R, C, L Separately
“ELI the ICE man”
VR = I R R
{ v and i are in phase
R
{
R
1
VC = I C X C with X C =
ωC
i leads v by 90°.
C
C
VL = I L X L
{ v leads i
L
L
with X L = ω L
by 90°.
Q.31-1
Which of the following is true about
the phase relation between the current
and the voltage for an inductor?
(1) The current is in phase with the voltage.
(2) The current is ahead of the voltage by 90º.
(3) The current is behind the voltage by 90º.
(4) They are out of phase by 180º.
Q.31-1
Which of the following is true about
the phase relation between the current
and the voltage for an inductor?
Remember ELI the ICE man!
Inductor: voltage leads current.
(1) The current is in phase with the voltage.
(2) The current is ahead of the voltage by 90º.
(3) The current is behind the voltage by 90º.
(4) They are out of phase by 180º.
Q.31-1
What is the phase relation between the
current and the voltage for an inductor?
Proof:
i ∝ sin(ωt )
v ∝ cos(ωt )
Let i = I sin(ωt )
di
Drop is v = L = LIω cos(ωt )
dt
v hits peak before i
Q.31-2
An inductor L carries a current with
amplitude I at angular frequency ω.
What is the amplitude V of the voltage
across this inductor?
I = 3 .0 A
ω = 200 rad / s
L = .03 H
(1) VL = 0.5 V
( 2) VL = 2.0 V
( 3) VL = 6.0 V
( 4) VL = 12 V
( 5) VL = 18 V
( 6) VL = 600 V
Q.31-2
The reactance is:
X L = ωL = 200 × .03 = 6.0 Ω
I = 3 .0 A
ω = 200 rad / s
Thus the voltage amplitude is:
L = .03 H
VL = I L X L = 3.0 A × 6.0 Ω = 18 V
(1) VL = 0.5 V
( 2) VL = 2.0 V
( 3) VL = 6.0 V
( 4) VL = 12 V
( 5) VL = 18 V
( 6) VL = 600 V
Impedance
The “AC Ohm’s Law” is:
ε m = IZ
where Z is called the impedance.
Obviously, for a resistor, Z = R
for a capacitor
Z = X C and for an inductor Z = X L
But if you have a combination of circuit
elements, Z is more complicated.
Impedance and Phase Angle
General problem:
if we are given
ε (t ) = ε m sin ω t
can we find i(t)?
We can always write
i (t ) = I sin (ω t − φ )
By definition of impedance
I = εm / Z
Given εm and ω, find Z and φ.
??
Series RLC Circuit
1. The currents
are all equal.
2. The voltage drops add
up to the applied emf as
a function of time:
= vR
ε
+ vC + vL
BUT: Because of the phase differences, the
amplitudes do not add:
≠ V +V +V
ε
So the impedance is not
just a sum:
m
R
C
L
Z ≠ R + XC + X L
Results for series circuits
ε ( t ) = ε m sin ω t
i ( t ) = I sin(ω t − φ )
It turns out that for this particular circuit
Z = R + ( X L − XC )
X L − XC
tan φ =
R
2
2
Z
φ
R
X L − XC
Series Circuit Example
Given L = 50 mH, C = 60 µF,
f=60Hz, and I=4.0A.
(a) What is the impedance?
(b) What is the resistance R?
εm=120V,
Solution to (a) is easy:
ε
Z=
120
=
= 30 Ω
I
4
m
(a) Z = 30 Ω
Example (part b)
L=50 mH, C = 60 µF, εm=120V, f=60Hz, I=4.0A
(b) What is the resistance R?
ω = 2π f = 2 × 3.14 × 60 = 377 rad / s
−3
X L = ω L = 377 × 50 × 10 = 18.8 Ω
1
1
XC =
= 44.2 Ω
=
−6
ω C 377 × 60 × 10
X C − X L = 44.2 − 18.8 = 25.4 Ω
R = Z − ( X L − X C ) = 30 − 25.4 = 16 Ω
2
2
2
2
AC Circuits
• Ch.30: Faraday’s Law
• Ch.30: Inductors and RL Circuits:
• Ch.31: AC Circuits
Review: Inductance
• If the current through a coil of wire changes,
there is an induced emf proportional to the rate of
change of the current.
•Define the proportionality constant to be the
inductance L:
ε
di
= −L
dt
• SI unit of inductance is the henry (H).
Review: LC Circuits
q = Q cos(ω t )
dq
i=
= −ωQ sin(ω t )
dt
d q
 1 
= −
q
2
dt
 LC 
2
Loop rule gives differential equation:
Solution if frequency is correct:
Natural frequency of oscillations!
 1 
ω =

 LC 
2
Review: AC Voltage Sources
•
For an AC circuit, we need an
alternating emf, or AC power
supply.
• This is characterized by its
amplitude and its frequency.
ε =ε
amplitude
m
sin ωt
angular frequency
Review: R, C, L Separately
“ELI the ICE man”
VR = I R R
{ v and i are in phase
R
{
R
1
VC = I C X C with X C =
ωC
i leads v by 90°.
C
C
VL = I L X L
{ v leads i
L
L
with X L = ω L
by 90°.
Review: Simple series circuit
ε ( t ) = ε m sin ω t
i ( t ) = I sin(ω t − φ )
It turns out that for this particular circuit
Z = R + ( X L − XC )
X L − XC
tan φ =
R
2
2
Z
φ
R
X L − XC
Series Circuit Example
Given L = 50 mH, C = 60 µF,
f=60Hz, and I=4.0A.
(a) What is the impedance?
(b) What is the resistance R?
εm=120V,
Solution to (a) is easy:
ε
Z=
120
=
= 30 Ω
I
4
m
(a) Z = 30 Ω
Example (part b)
L=50 mH, C = 60 µF, εm=120V, f=60Hz, I=4.0A
(b) What is the resistance R?
ω = 2π f = 2 × 3.14 × 60 = 377 rad / s
−3
X L = ω L = 377 × 50 × 10 = 18.8 Ω
1
1
XC =
= 44.2 Ω
=
−6
ω C 377 × 60 × 10
X C − X L = 44.2 − 18.8 = 25.4 Ω
R = Z − ( X L − X C ) = 30 − 25.4 = 16 Ω
2
2
2
2
Q.31-3
A certain series RLC circuit is driven by an
applied emf with angular frequency 20 radians
per second. If the maximum charge on the
capacitor is 0.03 coulomb, what is the
maximum current in the circuit?
(1) 6 A
(2) 1.5 A
(3) 0.6 A
(5) Not enough information
(4) 0.15 A
Q.31-3
A certain series RLC circuit is driven by an
applied emf with angular frequency 20 radians
per second. If the maximum charge on the
capacitor is 0.03 coulomb, what is the
maximum current in the circuit?
dq
q = Q sin(ωt )
i=
= ωQ cos(ωt )
dt
I = ωQ = 20 × 0.03 = 0.6 A
(1) 6 A (2) 1.5 A (3) 0.6 A
(5) Not enough information
(4) 0.15 A
Q.31-4
In a series RLC circuit, find the resistance, given the impedance
and phase constant:
Z = 500 Ω
(1) 400 Ω
φ = 60°
(2) 350 Ω
R=?
(3) 300 Ω
(4) 250 Ω
In an RLC circuit:
Z = 500 Ω
φ = 60°
Z
φ
R=?
X L − XC
R = Z cos φ
R
(1) 400 Ω
Q.31-4
1
cos(60° ) = sin(30° ) =
2
R = Z / 2 = 250 Ω
(2) 350 Ω
(3) 300 Ω
(4) 250 Ω
Resonance
For a given series RLC circuit, what applied
frequency will give the biggest current?
I=
ε
m
R + ( X L − XC )
2
2
I is biggest when denominator is smallest,
which is when reactances cancel:
1
ωL=
ωC
1
Which gives ω =
LC
2
Same as natural frequency for oscillations!
Resonance Plot: I vs ω
Peak at ω = 1 / LC
Becomes sharper as R → 0
Power in AC Circuits
To know how much power will be consumed
in an AC circuit we need more than the
impedance Z; we also need the phase angle φ.
AC Power
What is the power provided by an AC source?
• Only interested in the time average!
• Time average power to L and C is zero.
• So Pave(from source) = Pave(to resistor).
Pave = vR iR = (VR sin ωt )( I R sin ωt )
1
= VR I R sin ωt = VR I R
2
2
RMS Values
VRMS =
v
2
=
Likewise
1
V sin ωt = V
=V / 2
2
2
2
I RMS =
i
2
=I/ 2
So for a resistor:
Pave = VRMS I RMS
V I
1
=
= VI
2 2 2
Power and Phase Angle
Power to resistor
P = iR R = I R2 R sin (ω t − φ )
2
2
2
Pavg = I R R sin
2
avg
1 2
= IR R
2
But we want result in terms of impedance and phase angle:
R = Z cos φ
So
Pavg
1 2
= I Z cos φ
2
Power factor
What about more complicated circuits?
Method of Phasors
•Useful mathematical trick for working with oscillating
functions having different phase relationships.
•We’ll use it to derive the known result for the series
circuit, to show how it works. Good for general case.
iR = I R sin(ω t )
I R = Length of phasor
iR =
Vertical component
of phasor
Series RLC Circuit
1. The currents
are all equal.
2. The voltage
drops add up to the
applied emf:
ε =v
R
+ vC + vL
Because of the phase differences, the amplitudes
do not add, but in the phasor diagram, this means
that the vertical components do add, so we get the
right answer if we add the phasors as vectors.
Phasors for Series RLC Circuit
ε = vR + vC + vL = ε m sin ω t
iR = iC = iL = I sin(ω t − φ )
i
I
(ω t − φ )
Adding the Voltage Phasors
ε = vR + vC + vL = ε m sin ω t
ε
I
Adding the Voltage Phasors
ε = vR + vC + vL = ε m sin ω t
ε
2
m
= VR + (VL − VC )
2
VL − VC
tan φ =
VR
Impedance and Phase Angle
ε
2
m
= VR + (VL − VC )
2
VL − VC
tan φ =
VR
Divide by the common current amplitude I:
ε
m
= IZ , VR = IR , VL = IX L , VC = IX C
Z = R + ( X L − XC )
2
X L − XC
tan φ =
R
2