Discussion Question 12A
P212, Week 12
Power in AC Circuits
An electronic device, consisting of a simple RLC circuit, is designed to be connected to an
American-standard power outlet delivering an EMF of 120 Vrms at 60 Hz frequency. However,
you have transported this device to Guyana, where standard AC power outlets deliver a blazing
240 Vrms. To overcome this problem, you connect your device to the Guyanese wall outlet
through a step-down transformer with Np turns in the primary coil and Ns turns in the secondary.
The values of the resistance, capacitance, and inductance in your device are given in the figure.
Ep
R
Np
Ns
C
L
Ep,rms = 240 V
Erms = 120 V
f = 60 Hz
R = 3.8 
L = 0.05 H
C = 177 F
(a) First, let’s design the transformer. What relation must you impose on Np and Ns to drop the
Guyanese line voltage Ep down to E = 120 Vrms?
Try thinking physically: As is typical, your transformer has an iron core which ensures that
the magnetic field produced in the primary coil is transferred exactly to the secondary coil.
Now E = -d/dt … to step down the voltage, which coil needs to have more turns?
The field in the primary will be the same as in the secondary . The flux
on the primary will be  p  BN P A. The flux on the secondary will be  s  BN s A.
By Faraday's law:  p  
We want

d
dB
dB
N
 NP A
 s  s
; s   N s A
p N p
dt
dt
dt
s 1
N
1
  s 
p 2
Np 2
(b) Let the time-dependent EMF produced by the Euro generator be Ep(t) = Ep,max sin(t) which


means   t   120 2 sin t where   t  EMF delivered to the secondary side of the
transformer . What is the time-dependent current I(t) produced in the load? Write your answer
in the form I(t) = Imax sin(t-), and determine the numerical values of Imax and .
Page 1 of 4
X L   L  377  0.05   18.85
XC 
1
1

 14.99
C 377 177  106 
X  X L  X C  3.86
3.80
18.85
5.
41
9
We begin by computing the reactances:
      rad/s

3.865
45.4º
14.99
3.80
ere is a phasor diagram for reactances. Z = 5.419   0
I max  max / Z 
120 2
 31.32 A Hence I=31.32 sin(t  45.40 ) since  lags I
5.419
(c) What is the average power <P> delivered to the load?
Your formula sheet has a couple of “cookbook” formulas you can use.
P  rms I rms cos    

 cos( 45.40 )  1.87 kW
2
(d) Obtain an algebraic expression for the <P> delivered to the load in terms of rms, R, and X 
XL - XC . Hint -write a phasor diagram for Z in terms of X and L and use trig to find the
required phase angle.

P  rms I rms cos   rms  rms
 Z
From triangle cos 
2
rms

cos

.
From
triangle
Z=R/cos

hence
P
cos 2 


R

2

R
R 2
2 
 P  rms 
 2 rms 2

R  R2  X 2 
R X
R2  X 2
R
Z

R
(e) Suppose your Guyanese AC generator has a variable frequency f, though the EMF it
supplies is fixed at 240 Vrms. What f would you select so that the maximum possible power is
delivered to your device? Hint- sketch the expression for <P> you obtained in part (d) as a
function of X. Which value of X maximizes <P>? What frequency gives you that value for X?
The formula says P monotonically decreases with X. So the maximal P
when X=0 or X L  X C   L 
1
1
 f 
 53.5 Hz
C
2 LC
is
<P>
X
Page 2 of 4
X
(f) What is the maximum average power <P>max that you can deliver to your device?
2
2
R rms
rms
P
.
the
maximum
power
is
just

R2  X 2
R
This makes sense since at resonance the load is purely resistive and this is
From our expression P 
the familar result for a resistor, . P 
2
rms
1202
 3.79 kW

3.8
R
Consider a slightly different situation. Let’s say you built your device to be purely inductive 
it contains nothing more than a coil you wound yourself, giving L = 0.05 H. However, no device
is perfect: every electronic circuit element contains some residual inductance, capacitance, and
resistance. You connect your device to the Guyanese wall outlet and measure the response in the
load. You find a current with an rms value of Irms = 8.5 A and lags the transformer EMF by 30.
Ep
R
Np
Ns
C
L
Ep,rms = 240 V
Erms = 120 V
f = 60 Hz
L = 0.05 H
Irms = 8.5 A
| = 30º
(g) What is the resistance R of your device?
You will need a couple of equations to “reverse engineer” this circuit: you measured the
voltage, current, and phase and now you have to figure out R. Big hint: obtain an equation
for R in terms of the impedance Z and the phase . (It's a useful equation, you might want to
remember it!)
From triangle Z  R / cos  or R  Z cos  . Z =
R
rms
I rms
Z

rms
 120 
0
cos  = 
 cos 30  12.23 
I rms
 8.5 
X
R
(h) What is the capacitance C of your device?
From triangle X  Z sin  =
C 
1



   L  rms s in  
I rms


 C  225 F
Page 3 of 4

rms
1
s in  = L 
I rms
C
Z

1
120


( 2    ( 2  0.05 
s in 30o 
8
5
.


R
X
(i) In part (c), you probably used the average-power formula <P>=Erms Irms cos() from your
formula sheet (the others, like <P>=Irms2 R, are equivalent). Let’s derive that result! First, what
is the instantaneous power P(t) delivered to the load? Give an algebraic answer (no numbers) in
terms of Erms, Irms, , , and t.
Instantaneous power is P = IV, forever and ever, under all circumstances. Remember the
time-dependent expressions you found in part (b) … and in preparation for the next question,
you should “massage” your result using sin( a  b )  sin a cos b  cos a sin b .
Let    max sin t and I  I max sin t    where  is the phase lag of
the current relative to the voltage. P =I   max I max sin t sin t   
 P   max I max sin t  sin t cos   cos t sin  
 P   max I max  cos  sin 2 t  sin  sin t cos t 
(j) Now calculate the time-average of your result for P(t) over one cycle to obtain the average
power <P> delivered to the load.
Remember, the average of sin2 and cos2 is 1/2, while the average of (sincos) is zero.
1
P   max I max  cos  sin 2 t  sin  sin t cos t 

 P   max I max cos  sin 2 t  sin  sin t cos t
0.5

As suggested by the figure sin t (solid) averages to 1/2 and
sin t cos t (dashed curve) averages to zero over one cycle.
0
0
2
Hence P 
 max I max
cos    rms I rms cos 
2
Page 4 of 4
-0.5
-1
1
2
3
4
5
6
7
Discussion Question 12B
P212, Week 12
Electromagnetic Waves
A laser beam travels through vacuum. The electric field of the plane electromagnetic wave
produced by the laser has the form given below. The wavelength of the beam is  = 514 nm, and
the amplitude of the electric field is E0 = 2.5 x 104 N/C.


E
 (x, y,z,t)  yˆ E 0 cos(kz  t  45 )
(a) In what direction is this wave propagating?
What does “propagating wave” mean anyway? It means that the shape of the wave remains
the same as a function of time, it just moves. The shape of the wave is determined by that
cosine. As time rolls forward, how do you have to change position so that the shape looks
the same?

ˆ 0 cos( 450 ). This spot obeys
Let's follow a spot on the wave corresponding to E  yE
kz  t  0 or z  -


t. Hence this spot moves along the -zˆ axis and evidently  c
k
k
(b) What are the magnitudes of the wave number k and angular frequency ?
8
2
2
2c 2  3  10 
7
-1

 1.22  10 m ;   ck 

 3.66  1015 rad/sec
k
 514  109

514 109
(c) Make a sketch of what the E field looks like at some moment in time.


At t  0, we have E y  E0 cos  kz  


which is a cosine curve that maximizes

at z = +
k
Page 1 of 2
Ey
t =0
z
(d) Write down an expression for the magnetic field B(x,y,z,t). Express your answer
algebraically (i.e. no numbers) in terms of the symbols k, , and B0 (the latter being the
magnetic field amplitude). Be sure to indicate the direction of the magnetic field.
The E and B fields of an electromagnetic wave have a fixed relationship to the direction of
propagation of the wave. You know the direction of the wave, and the direction of E, so you
can figure out the direction of B with the help of your sketch. Add the B-field to your
drawing in part (c).
The B field has the same position and time
dependence as the E field. It is directed
 
such that E  B is along the direction of
propagation (marked S in the figure)

B ( x, y, z , t )  xˆ B0 cos(kz  t  45 )
E
z
y
x
B
S
(e) What is the amplitude B0 of the magnetic field?
E0 2.5  104
B0 

 8.33 105 T
8
c
3 10
Page 2 of 2