Discussion Question 11A
Physics 212 week 11
RLC Preview
You will learn much more about RLC circuits in the next lecture but here is a preview of their
interesting behavior. In this circuit, an initial charge of Q 0 =1C is placed on the capacitor.
At time t=0 the switch is closed.
A
L
B
+Q
C
R
I
-Q
C
Q(t=0)=1C
C = 100 nF
L = 500 H
R = 40 
The charge on the capacitor oscillates while dying out as shown below:
Q
t
Our conventions are that I is positive when flowing in the direction of the arrow, and Q is
positive when the top plate of the capacitor is positive and the bottom plate is negative as
illustrated.
(a) Immediately after the switch is thrown, what is the current I (0+)?
Right before the switch is thrown I = 0 .The inductor resists an immediate change in current
so the current will be 0 immediately after the switch is thrown.
(b) Immediately after the switch is thrown, what is the voltage drop across the inductor, and what
is the rate of change of current or dI/dt (0+) ? Hints: What is the voltage drop across the
capacitor? What must the voltage drop across the inductor be according to KVL? How does the
voltage drop across an inductor depend on dI/dt ?
Q0
1 C

 10 V.
C 0.1  F
Applying KVL (clockwise from lower plate) we have: 10 - VL - IR  0
The voltage increase across the capacitor is Vcap 
where VL is voltage drop across inductor:
or 10 - VL = 0  VL  10 V. VL  L
dI
dI
10 V


 20 kA/s
dt
dt 500  H
Discussion Question 11A
Physics 212 week 11
RLC Preview
A
L
B
+Q
C
R
I
-Q
C
Q(t=0)=1C
C = 100 nF
L = 500 H
R = 40 
(c) At a later time, the charge on the capacitor is Q = + 0.147 C and VB –VC = 3.64 V. Find the
rate of change of charge on the capacitor (dQ/dt), the voltage drop across the inductor, and dI/dt .
Is the capacitor charge decreasing or increasing at this time? Is the current decreasing or
increasing at this time?
dQ
  I since VB  VC is just the voltage drop across R  VB  VC  3.64  IR
dt
3.64
dQ
mC
I 
 91 m and hence
 91
and charge is decreasing
40
dt
s
0.147  C
Vcap 
 1.47 V By KVL from lower plate clockwise: 1.47 - VL  3.64  0
0.100  F
dI
where VL is the voltage drop. VL    3.64  1.47  = -2.17 V . Hence L =VL
dt
dI
-2.17

 4.34 kA/s . Hence current is decreasing as well.
or
dt 500  H
(d) At the time discussed in part (c), when the charge on the capacitor is Q = + 0.147 C and
VB –VC = 3.64 V. Find the total stored energy and the rate of change of stored energy.
1
1
The stored energy is U  CV 2  LI 2 . Putting in the values from part (b),
2
2
2
1
1
2
U  100 109 F  1.47 V    500 106 H  91103 A   2.18  J
2
2
The rate of change stored energy is the power loss through the resistor.
Power = I 2 R   91 103 A   40    0.331 J/s
2
(e) At a still later time, at the instant when there is no voltage drop across the resistor and the
charge on the capacitor is Q = –396 nC, find VA –VB. Is the current I increasing or decreasing?
-396 nC
 VB  VA - IR  0  VA  VB  3.96 V
100 nF
dI
dI
 0 and current
The voltage drop across the inductor is VA  VB  L 
dt
dt
is decreasing.
KVL from lower plate clockwise
Discussion Question 11B
Physics 212 week 11
LC Circuits
At right, we see the classic LC circuit, consisting of an inductor in
series with a capacitor. To be precise, this is an undriven LC circuit,
since there is no battery driving the flow of current. Nevertheless,
this simple circuit has an amazing and highly useful property: it
L
supports a resonant oscillation of current. In the ideal case shown
here where there is no resistance in the circuit, the oscillation can
continue indefinitely without any external source of EMF to drive it.
LC circuits often appear as parts of larger networks which are
designed to operate at a particular frequency. The most familiar
example is a radio, whose circuitry can be tuned to receive (i.e. respond to)
incoming electromagnetic waves of a particular frequency.
C
L = 300 mH
C = 0.025 F
(a) What is the angular frequency 0 of the current maintained by the circuit? What is
the linear frequency f0?
An undriven LC circuit can only operate in a stable way at its natural resonant frequency. As
for linear versus angular frequency, the conversion is easy if you remember the units: f is in
Hz = 1/sec,  is in rad/sec.
The natural frequency is when the net reactance is 0 or 0 L 
0 
1
0
0 C



 11.55 krad/sec ;
LC
0.3  0.025  106 
 =2 f 0  f 0 
 11.55 krad/sec

 1838 Hz
2
6.28318
(b) Let’s set our clock so that the current in the circuit is at its maximum value Imax at time
t = 0. Write down an expression for the time-dependence of the current I(t) in terms of Imax
and the frequency 0.
It’s either a sine or a cosine …
We are looking for a trig function that maximizes as t = 0. Hence I  I max cos 0t
1
(c) Starting from your expression for I(t), determine the voltages VL(t)
and VC(t) across the inductor and capacitor. As part of your solution, find
the peak values VL,max and VC,max in terms of Imax.
L
C
L = 300 mH
C = 0.025 F
You’ll need your familiar formulas for the voltage across R, C, and L
dQ
... and remember that, by definition, I 
and so Q   I dt .
dt
dI
d
dt  d cos t 
 L  I max cos t   LI max

   L I max sin t  VL ,max   L I max
dt
dt
dt  dt 
dQ
VC  Q / C If the current is positive in the clockwise direction, I 
dt
I
I
I
Q   dt I   dt I max cos t = max sin t  VC  max sin t  VC,max  max
C

C
VL  L
(d) To visualize what’s going on, sketch your functions I(t), VL(t), and VC(t). Don’t worry
about the amplitudes of your curves, just their shapes and phases.
t
/2
VL

/2
 
VC
I
e) Have a look at your plot  does VL lead or lag the current? How about VC?
Leading and lagging can be tricky concepts. Think of it this way: which one “gets there”
(i.e. reaches its maximum value) first, V or I? The one that “gets there first” leads the other.
VL leads. VC lags.
2
Another way of viewing leading and lagging is if
I  cos t and V  cos t    , we say V leads I by 
We can do this using cos t     cos t cos  sin t sin


VL   sin t = cos  t   hence VL leads I by 900




VC  sin t = cos  t   hence VC lags I by 900


Congratulations! You’ve just derived the master relations between current and voltage for
inductors and capacitors in an AC circuit!
Peak Values
Relative Phases
VR,max = Imax R
VL,max = Imax XL  XL = L
Across R  V in phase with I
Across L  V leads I by 90
Across C  V lags I by 90
VC,max = Imax XC  XC = 1/C
Notice how all the peak-value formulas look like good old “V = IR”  the reactances XL and XC
describe the effective resistance of inductors and capacitors in AC circuits.
We now set the circuit into oscillation by “stimulating” it with a brief pulse from some external
source of EMF. The result is that the peak voltage across the capacitor is VC,max = 120 V.
(f) What is the peak current Imax?
L = 300 mH
C = 0.025 F
VC,max = 120 V
From part (a) 0 
From part (c) VC,max
L
C

 11.5 krad/sec
LC
I
 max  I max  0C VC,max
0 C
 11.5  103 rad/sec  0.025  106 F  120 V   34.5 mA
(g) What are the maximum
3
energies UL,max and UC,max stored in the inductor and capacitor respectively?
UL,max = 1/2 L Imax2 = 0.5*(0.3)*(0.03452) = 0.18 mJ
UC,max = 1/2 C (VC,max)2 = 0.5*(0.02510-6)*(1202) = 0.18 mJ
(h) Determine time-dependent expressions UL(t) and UC(t) for the two stored energies, and
add them together to find the total stored energy U(t). Finally, plot your results for all
three functions.
Something like this: Utotal(t) = 0.18 mJ, UC is a sin2(ωt) function while UL is a cos2(ωt) function
t
/2


/2
(i) How are UL,max, UC,max, and Umax related to each other?
Knowing this relation is extremely helpful in solving LC circuit problems!
All are equal:
This is no accident but always happens in RC circuits
2
U C ,max
I2  1 
1
1
1 I 
2
 C VC2 ,max  C  X C I max   C  max   max  
2
2
2  C 
2C   
2
2
2
2
2
I max
I max
LI max
 1 


 U L ,max
LC 
2C  1 / LC 
2C
2
4
Discussion Question 11C
Physics 212 week 11
Phasors
Next, we encounter the rich subject of driven RLC circuits.
The most basic example is shown in the diagram: a resistor,
a capacitor, and inductor, and an AC generator all connected
in series. The generator is just a fancy type of battery that
produces not a constant “DC” voltage as we have
encountered so far, but an oscillating “AC” voltage. The
voltage from the generator oscillates sinusoidally at some
angular frequency , and has a maximum magnitude Emax
(also called the “amplitude” or “peak voltage”).
R
L
C
E
If the AC generator weren’t present, our circuit would be a plain old undriven LC circuit with
some resistance R thrown in. It would support a nice oscillating current of frequency
0 = 1/LC… except the resistor would damp out the oscillations over time. To keep the circuit
going, we attach the generator to drive the oscillations. But … the AC generator is driving the
circuit at its own frequency , which need not be the same as the circuit’s resonant frequency
0. The result is something of a mess.  The generator forces the current to oscillate at
frequency , but if  doesn’t match 0, the driving voltage will be out of phase with the current.
To help us visualize what is going on, we use phasors. A phasor is just a way of graphically
representing the time-dependence of something which oscillates. Specifically, a phasor is a
vector in the xy plane. We then imagine that this vector rotates around the origin in a
counterclockwise direction, with angular velocity  radians/second. The phasor describes the
behavior of something oscillating with frequency  in the following way: the value of that
something, at any given point in time, is equal to the projection of the phasor onto the vertical
axis. All clear? OK. Phasors are just a graphical trick to help you figure out how oscillating
things behave. Now let’s see how to apply them to a driven RLC circuit.
(a) Let’s practice by drawing a few phasors. Below are expressions describing three timedependent currents. Draw the phasors for these currents at two different times:
t = 0 and t = /2. Note that the length of the phasor never changes. What does it
represent?

1(t)

I2(t) = -I2,max cos(t)

I3(t) = I3,max sin(t-/3)
= I1,max sin(t)
Phasors at
t = 
Phasors at
t = /2
I1
I2
I3
I1
I3
I2
R
Now consider the RLC circuit shown. The values of R, L, and C are
all known. We also know that the generator is driving the circuit at
frequency f = 50 Hz and that the peak current is Imax = 0.5 A. We will
set our clock so that the current is zero at time t = 0 … thus:
I(t)
= Imax sin(t)
L
C
E
(b) Using the “master relations” for peak current and voltage,
calculate VR,max, VC,max, and VL,max.
L = 40 mH
C = 2.5 mF
R = 15 
=2 rad/sec
f = 50 Hz
Imax = 0.5 A
VR,max = Imax R = (0.5)(15)=7.5 V
VL,max = Imax ωL = (0.5)(314.16)(0.04)= 6.28 V
VC,max = Imax / ωC = 0.5/(314.16*0.0025)=0.637 V
(c) Draw a diagram showing the phasors for I(t), VR(t), VC(t),
and VL(t) at t = 0.
You must use the “master relations” for the phase between
current and voltage to determine the angles of the phasors.
Also, be sure to draw the lengths of your voltage phasors to
scale.
Phasors at
t=0
tan φ = (XL – XC)/R
φ = 37˚
VR
6.28-0.637=5.64 V
VL
37°
VC
7.5 V
(d) The phasor diagram makes it easy to answer questions like these:

True or false? Whenever the voltage across the capacitor hits is maximum value, so does
the voltage across the inductor. False. When the capacitor voltage hits maximum the
inductor voltage hits minimum since they are 180 degrees out of phase.

True or false? The voltage across the capacitor is always zero when the current through
the inductor is zero. False. The current is the same through the inductor, capacitor, or
resistor since all elements are in series. The voltage across the capacitor lags the voltage
across the resistor (or the current) by 90 degrees and hence the voltage across the
capacitor is either a minimum or maximum when the current is zero.

True or false? When the magnetic energy stored in the inductor hits is maximum value,
so does the electric energy stored in the capacitor. False. The magnetic energy is proportional to
the square of current. The electrical energy is proportional to VC . The capacitor voltage lags
the current by 90 degrees and hence when the current hits a maximum the voltage across the
capacitor is zero.
We now know the voltages across the capacitor, resistor, and inductor. The only thing left to
calculate is the EMF E(t) across the generator. From Kirchoff’s Voltage Law, we see
immediately that E(t) = VR(t) + VC(t) + VL(t). To perform the sum, we need one more principle:
Phasors add like vectors.
(e) What is the voltage E across the generator? Copy your
previous phasor diagram, and add a new phasor representing E.
What is the length of this phasor? What is the angle  that this
phasor makes with respect to the current phasor?
VL
E = 9.39 V, φ = 37˚
=9.39 V
VC
6.28-0.637=5.64 V
E
VR
37°
7.5 V
If you avoided plugging in numbers until the end of your calculation, you just found the last of
the “master relations”:
Peak Values
Emax  I max Z  Z  R 2   X L  X C 
Relative Phases
2
Across E  V leads I by  , with
tan    X L  X C  R
These are the current/voltage relationships for the entire RLC circuit, as seen by the generator.
The impedance Z is particularly important: it is the effective resistance of the entire circuit.
(f) What condition would you have to place on the frequency  to make the EMF E across
the generator lead the current?
Look at your phasor diagram, and remember that adjusting the frequency allows you to
change the effective resistances of the inductor and capacitor. The one with the highest
reactance will determine the phase of the circuit.
As shown in the two figures, if X L  X C ,  will lead I whereas
if X L < X C ,  will lag I. X L  X C implies  L 
1
 
 
LC
1
or
C
1
or   
LC

I XL
IR
I XC
IR

(g) What frequency  would you need to make the peak EMF Emax as small as possible for
a fixed Imax ?
From that last “master relation”, it’s clear that you need to minimize the impedance =
effective total resistance of the circuit … what is the minimum possible impedance Z?
Since the reactance leads the resistance by  900 by right triangles
we have Z= R 2   X L  X C  . The minimal impedance occurs when
2
X L  XC  0  L 
1
 
C
1

LC
1
 40 10  2.5 10 
3
3
 100 rad/sec
(h) The frequency you found in (f) is the magic resonant frequency 0 of the circuit. Draw
the phasor diagram for the circuit when it is in resonance.
Note how that nasty phase  disappears at resonance. 
VL
VR E
VC
I XL
I XC
Discussion Question 11D
P212, Week 11
RLC Circuits
(a) Calculate the maximum EMF Em and the maximum current Im in
the RLC circuit described at right.
The “rms” = root-mean-square value of anything oscillating
L
sinusoidally is its peak value divided by 2.
Em = Erms √2 = 170 V
Im = Em/Z = 170/223.6
= 0.76 A
C
E
The figure shows R, X L and
X C . The net reactance is 100
R = 200 
L = 40 mH
C = 0.20 F
Erms = 120 V
 = 104 rad/sec
in the direction of capacitance.
The right triangle gives
Z = 223.6 .
X L  L
 400 
R  200 
1
C
 500 
XC 
200 
100 
Z=223.6 
(b) Find the magnitude and sign of the phase  by which the driving EMF leads the
current.
A negative phase means that the driving EMF lags the current … which is the case here?
Does your answer make sense given the reactances you calculated earlier?
200 
26.60
Z=223.6 
100 
Here is our reactance triangle from before. Z    lags R
  I  by   Tan -1 
100 
0
  26.6
 200 
(c) Draw the phasor diagram for this circuit, giving numerical values for the lengths of
each phasor (E, VR, VC, VL).
Be sure to draw your diagram carefully: use longer phasors for larger peak voltages.
VR = Im R = 152 V
VC = Im / (ωC) = 380 V
VL = Im ω L = 304 V
VL  400  0.76
VR  200  0.76
VC  500  0.76
(d) What is the resonant frequency 0 of this circuit?
L
C
E
ωo = 1/√(LC) = 11.2 krad/sev
R = 200 
L = 40 mH
C = 0.20 F
Erms = 120 V
 = 104 rad/sec
(e) Calculate the maximum energies UL,max and UC,max stored in the inductor and
capacitor.
UL,max = ½ L Im2 . Inserting Im= 0.76 A from part (a) we have UL,max = 11.6 mJ
UC,max = ½ C (VC,max)2 = 14.4 mJ . Inserting VC,max = 380 V from part (c) gives us
UC,max = 14.4 mJ
(f) Assume that the angular frequency  of the generator is variable. For what  is the
total impedance Z equal to 2R? Hint- you will get a quadratic equation.
R 2   X L  X C   2 R  R 2   X L  X C  =4 R 2   X L  X C   3R 2
2
  X L  XC 
Let 0 
2
2
2
2
2
1 
1 


R
2
 L 
  3R    
  3 
C 
 LC 


L
2
1
R
=11.18 kHz and 1   5 kHz
L
LC
2
2

2 
Then    0   32   2  02   32 2
 

 4  202 2  04  32 2   4   202  3   2  04  0
Using quadratic equation:  
2
325 106 
 325 10 
6 2
 2
2
0
 3  
 2
2
0
 3   404
2
2
 4 11.18  10

3 4
325 106  207 106
2
2
2
6
6
  266 10 and 59 10    16.3 krad/sec and 7.66 krad/sec
 
2

2
Discussion Question 11E
P212, Week 11
Low-Pass and High-Pass Filters
R
The circuit shown at right has an AC generator of peak EMF Emax and
frequency  connected in series with a resistor R and a capacitor C. Let’s
analyze the behavior of this circuit at extreme values of the frequency .
E
C
Important note: Since there is no inductor present, you may think that we’re back to the material of
“RC Circuits” that we studied a month ago. But not so! Those “RC circuits” were connected to
DC batteries of constant voltage! This one is being driven by an AC generator, which makes it
behave in a totally different way. This is an RLC circuit like any other … the fact that L = 0 just
simplifies things a little bit.
(a) Draw a phasor diagram for this circuit, showing the relative phases of the voltages VR, VC, and
E.
VR
VC

(b) Now draw phasor diagrams for the cases when the driving frequency  is extremely small, and
when it is extremely large. In both cases, please answer the following questions:
 How does the peak voltage VR,max across the resistor compare with the generator Emax?
(Are they about the same? Or is one very-much smaller than the other?)

How does the peak voltage VC,max across the capacitor compare with the generator Emax?
(Are they about the same? Or is one very-much smaller than the other?)
Super-high frequency case
VR
VC

Super-low frequency case
VR

VC
For the resistor, which is is true?
a) VR,max   b) VR,max   c)VR,max  
For the resistor, which is is true?
a) VR,max   b) VR,max   c)VR,max  
For the capacitor, which is is true?
For the capacitor, which is is true?
a) VC,max   b) VC,max  
a) VC,max   b) VC,max  
c)VC,max  
c)VC,max  
VC ,max
VR ,max

VR ,max
I max X C
R

0
I max R C
VC ,max

I max R
 RC  0
I max X C
as   . From figure   VR
as   0. From figure   VC
and VC ,max  VR  
and VR ,max  VC ,max  
A common application of the circuit we’ve been considering is in audio amplifiers, which process
music signals containing many different frequencies. In this application, the generator EMF is
produced by something like a CD player, or an old-fashioned record player. This input signal Vin is
sent through the resistor and capacitor, and an output signal Vout is produced by connecting an
output device (like a speaker) in parallel across either R or C. Here are the two configurations:
C
Circuit A
Vin
Vin
R
R
C
Vout
Vout
Circuit B
Suppose our music input comes from an old record-player, and is subject to hiss-and-pop scratch
noise. These distortions of the music are very short-lived, and so are of very high frequency. Thus,
we would like our amplifier to remove signals of very high frequency. This is called a low-pass
filter (or “scratch filter”)  when the input signal is of low frequency, the input signal Vin is passed
through to Vout without much change … but high-frequency input signals are blocked.
(c) Based on your phasor diagrams, which of the circuits shown would provide a low-pass filter?
The loud speaker in Circuit B records the voltage across the capacitor. In the high  limit , VC,max
approaches zero and hence the high frequency components are blocked. Circuit B is then the low
pass filter.
(d) Would the other one provide a high-pass filter (also called a “rumble filter”) which deletes
very-low frequency bass noise?
The loud speaker in Circuit A records the voltage across the resistor. In the low  limit, VR,max  0
and hence the low frequency components are blocked. Circuit A is thus a high pass filter.
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(e) Low-pass and high-pass filters can also be constructed using inductors rather than capacitors.
Use your expertise with phasor diagrams to determine which of these two circuits would provide a
low-pass filter and which would provide a high-pass filter.
Circuit A
L
Vin
R
Vout
Circuit B
R
Vin
VL ,max
VR ,max

Vout
L
X L L

so as    VL ,max  VR ,max
R
R
and as   0 VL ,max  VR ,max . The phases look
like this
High limit
Low limit


VL
VR
VR << 
VL
VR
VL << 
In Circuit A, the loud speaker measures the voltage across R which
is very low when    Thus Circuit A is a low pass filter.
In Circuit B, the loud speaker measures the voltage across L which
is very low when   0. Thus Circuit B is a high pass filter.
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