INDUCTORS AND AC CIRCUITS
We begin by introducing the new circuit element provided by Faraday’s Law – the
inductor. Consider a current loop carrying a current I.
The current will produce a magnetic flux in the loop proportional to I. Hence by Faraday’s Law a
change in I will produce an EMF around the loop.
EMF  
dM
dt
We define the self-inductance of the loop by:
EMF  L
dI
dt
where L = inductance. There are two equivalent definitions which are often useful. Since
EMF 
dM
dI
L
dt
dt
we have
M  LI
(We are ignoring signs just as we did for capacitance). Finally consider the energy involved in
setting up the field:
P  IV  IL
t
t
U 

P


0
Hence:
dt 

dI
 LI
dt

dt

0
dI
dt
t
1 d
1
 L  I 2  dt  LI2
2  dt
2
0
UM 
1 2
LI
2
where UM is the magnetic field energy.
Now consider two current loops:
Loop (1) will produce a flux in loop (2) and vice-versa. Hence a change in the current in (1) will
produce an EMF around (2). We define “mutual inductance” as:
EMF2  L12
dI1
dt
Just as with capacitance, inductance is purely geometric – depends only on the shape of the
circuit or circuits.
INDUCTANCE OF SOLENOID
Consider the solenoid shown:
As we have seen there will be a field in the solenoid given by:
  N
B  0 Izˆ
L
Hence each loop has a flux:
 N
loop  0 IR 2
L
Since there are N loops, the total flux is:

  0 N 2 R 2 I  LI
L
 N 2R 2
L  0
L
AC CIRCUITS
We now turn to the use of the elements resistors, capacitors, and inductors in circuits with
an AC generator instead of a battery to provide the EMF. We begin by considering each in turn.
RESISTOR
As usual the voltages around a closed loop must add to zero. Thus:
V0 cos t  IR  0
V cos t
I  0
R
Now consider the power provided by the generator.
V02 cos 2 t
P  IV 
R
Hence the average power is:

1  V 2 cos2 t
PAV   0
dt

R
0
where τ is the period = 2π/ω. Thus
PAV 

2
2





0
V02 cos 2 t
V 2  1 2 2
V02
dt  0
cos
zdz


R
R 2  0
2R
Thus the average power is that of a battery of voltage:
VRMS 
V0
2
This “root mean square” voltage is what is specified for AC systems. The term comes from the
fact that we are taking the average value of the square and then taking the square root.
CAPACITOR
V0 cos t 
q
0
C
  V0 sin  t  
I
0
C
I   V0C sin  t 
We would like to be able to treat this like we did resistors – Ohm’s Law. This would require:
I
V
V

R eff x C
But this won’t work since V~cos(t) while I~sin(t). In other words, they are out of phase. We
can solve this problem by using complex numbers.
COMPLEX NUMBERS
Complex numbers can be thought of as points in a plane – as opposed to real numbers
which are points on a line.
We denote
1 by i. Points in the plane are then of the form:
a + ib
in Cartesian coordinates. For our purpose polar coordinates are more useful. We have:
a  r cos 
b  r sin 
We now remember that:
ix
e
 1  ix
 ix 2
2!

 ix 3
3!
 1
x2 x4
x3
x5
   ix  i  i   cos x  i sin x
2! 4!
3!
5!
Thus
a  ib  rei
where
b
  tan 1   ,
a
1/2
r   a 2  b2 
Now consider the product:
r1ei1 r2ei2  r1r2e 
i 1 2 
Hence the angles add.
CONVENTIONS
We adopt the convention that physical quantities are to be the real part of complex
numbers. Hence:
V0 cos  t   V0eit
Then in the capacitor circuit we considered above:
V eit
I 0
 V0C sin  t 
XC
Let
XC  
i
C
Then
V0eit
 iCV0eit  iCV0  cos t  i sin t 
 i 


 C 
 iCV0 cos t  CV0 sin  t   CV0 sin  t 
Hence if we take:
XC  
i
C
we can use Ohm’s Law in the form:
V = IXC
Why is XC what it is? The voltage arises from a charge build up on the capacitor. The
bigger  the less time for charge to build up before the voltage reverses and charge is removed.
The bigger C the more charge is required to produce a given voltage. Hence:
XC ~
i
C
The –i comes from the fact that voltage lags current by 90o (I first, then V). But:
e
i

2
 i
Hence
XC  
i
C
INDUCTOR
V0 cos  t   L
dI
0
dt
V
dI V0

cos  t   I  I0  0 sin  t 
L
dt L
In our case there is no constant I. Thus I0 = 0 and:
I
V0
sin  t 
L
Try
X L  i L
Then
V eit V0eit
iV
V sin t
I 0

  0  cos t  i sin t   0
XL
i L
L
L
Hence we can take:
XL = iL
Why this form? The larger  the bigger (dI/dt) and the larger voltage produced. The bigger L the
larger voltage for a given (dI/dt). The i comes from the fact that V leads I (first V, then I).
SERIES CIRCUIT
Now let’s see how this works in a series circuit.
We have
V0 cos t  IR 
 V0 sin t  R
q
dI
L 0
C
dt
dI I
d 2I
 L
0
dt C
dt 2
We solve this in the usual way:
I = IH + IP
where IH is the general solution of the homogeneous equation and IP is one solution of the
complete equation. Let
I H  Ae mt
Then
Am 2e mt 
R
1
me mt 
Ae mt  0
L
LC
 m2 
R
1
m
0
L
LC
1/2
R  R2
4 
  2 

L L
LC 
m
2
Hence IM will have a factor
et
0
and thus will go to zero at long times.
Next IP. As usual we try a solution of the form:
IP  A cos t  B sin t
Then
R
 A sin t  B cos t 
L
 V
1

 A cos t  B sin t   0 0 sin t  0
LC
L
 A2 cos t  B2 sin t 
R
A

  A2  B 
0
L
LC 

B V0 

2 R
 B  L A  LC  L   0
(Set coefficients of sin and cos each to zero so that the solution will work at all times.) Then:
 2 1 
 

LC 
B  A 
 R 


 L 
 2 1 
 

LC     2  1    A R   V0
A 


LC  
L
L
 R   


 L 
A
V0
L
 2 1 
 


R 
LC 

L
 R 


 L 
 2 1 
 

LC 
B
 R 


 L 
2

V0  R 


L  L 
2
 R   2 1 

   

 L  
LC 
V0 R
L L
2
 R   2 1 

   

LC 
 L  
2

2
 2 1  V0
 

LC  L

2
 R   2 1 

   

LC 
 L  
1/2 
I P  A cos t  B sin t   A 2  B2 
A
B

cos t 
sin t 

2
2 1/2
 A 2  B2 1/2
  A  B 

But cosθ cosφ + sinθ sinφ = cos(θ – φ). Hence
1/2
I P   A 2  B2 
cos  t   
We have
1/2
 A 2  B2 1/2 

2
2
 V 2  R 2 
1   V0  
2
0

 
   
 
 
LC   L  
 L   L  
V0
L
2
 R   2 1 

   

LC 
 L  
2 1/2
  R  2 
1 
2

   
 
LC  
 L  
2

V0
2
1/2
2

1  
2 
 R   L 
 
C  


1
B
LC
tan   
A
 R 


 L 
2 
Let
z  X R  XC  X L  R 
i
1 

 i L  R  i   L 

C
C 

1/2
2

1  

z   R 2   L 
 
C  


ei 
where
tan  
L 
1
C
R
Then
I
V0 cos  t   
1/2
2
 2 
1  
 R   L 
 
C  


exactly as above. Hence our complex solution gives the values at long times – once the transients
have died out.
We now have a general procedure for any AC circuit: solve it exactly as you would a DC
circuit, but use complex X instead of R, and take physical quantities to be the real part of the
complex numbers.
POWER
We will always end up with:
V eit V0eit V0 i t  
V


 0 cos  t   
I 0
e
z
z0
z0
z 0 ei
Then
V
V2
P  Vo cos t 0 cos  t     0 cos t cos t cos   sin t sin 
z0
z0
V2 1
V2
 PAV  0 cos   RMS cos 
z0 2
z0
This is exactly like the DC result:
PAV 
V02
R ff
except for the cos factor. This factor is called the “power factor” and means that it is possible to
have both a current and a voltage and yet no average power. This has important consequences as
we will now see.
PARALLEL CIRCUITS
Consider the circuit shown:
We solve it exactly as we would if it were DC. We first note that L and C are in parallel. They
have an effective impedance (term for effective resistance in AC circuits) given by:
1
1
1
1
1
1 
i





 i  C 
  z11 
1 
z11 X C X L iL  i  
L 



 C 

L 
 C 

Then R and z11 are in series
z  R  z11  R 
with
i
1 

 C 

L 

 z 0 ei 
1/2
1


z0   R 2 
2
1  


 C 


L  

1
1 

 C 

L 
tan    
R
Then
PAV 
cos  
V02
cos 
2z0
R
1/2
1
 2

R 
2
1  


 C 


L  

 cos  
PAV 
R
z0
V02 R
2z 0 z 0
Hence the average power will be a max or min when Z0 is. This occurs when:
C 
1
1

L
 LC 1/2
At that point Z0 becomes infinite and no power is provided by the source. At this “resonance”
frequency the energy in the parallel section is just passed back and forth between the inductor
(magnetic field) and the capacitor (electric field).
Now recall our result for the series circuit:
1/2
2

1  
2 
z 0   R   L 
 
C  


tan  
cos  
L 
1
C
R
R
2 1/2

1 

 R 2   L 
 
C  



R
z0
V2 R
PAV  0
2z z 0
Again the “resonance” frequency is:

1
(LC)1/2
but this times it leads to maximum power from the source because it makes Z0 a minimum
instead of a maximum.
A curious fact about AC circuits is that the sum of the RMS voltages around a circuit
need not be zero. The sum of the instantaneous voltages must by zero, but the sum of the
averages (RMS voltages) need not be. You should check this numerically for a couple of circuits.
This method can be used to solve any AC problem – even those where Kirchoff’s Laws
must be used. The only restriction is that we get the long term steady state solution, not the
transients. If the transients are desired we have to use the entire solution and boundary
conditions. This is not usually of interest.