Discussion Question 10A
P212, Week 10
LR Circuits
R1
The diagram shows a classic
“LR” circuit, containing both
resistors and inductors. The
switch shown is initially
connected to neither terminal,
and is then thrown to position
“a” at time t = 0.
a
b
E
L
R2
E = 10 V
L = 15 mH
R1 = 4 
R2 = 6 
(a) At t = 0+, just after the switch is thrown to position a, what are the currents I1 and I2 across
the two resistors?
Just after the switch is thrown, what does the inductor 'look like' to the rest of the circuit?
What is the current doing? Do inductors like that? Once you know what the inductor looks
like at this point in time, it is highly advisable to redraw the circuit.
Just before the switch is thrown, there is no current in the inductor, since the inductor blocks
immediate changes in current, the current will be still be zero. Hence the same current will
. Hence
flow through R1 and R 2 . The voltage drop through both resistors will equal 
I1  I 2 
10


1 A
R1  R2 4  6
(b) After a very long time, what is the instantaneous power dissipated in the circuit?
After a very long time, what will have happened to the current? Now what will the inductor
look like to the rest of the circuit? The circuit now takes on a simple form ... what current is
flowing through the resistors?
After a long time, the rate of change of current has stabilized and hence
dI
VL  L  0. This means no current flows through R1 .
dt
2
 10 5
5

 . Power = I22 R 2    6  16.66 W
Hence   I 2 R2  I 2 
R2 6 3
3
(c) Sketch the behavior of the energy stored in the inductor as a function of time. Sure, why not?
What is the final energy stored, after a very long time? LI 2 2  15mH   25 9  2  .021 Joules
Think about the current through
the inductor: what is it at time 0?
R1
0A after a very long time? 5/3
a
E
E
max long time, the
Next, after a very
'clock' is reset to 0 and the switch
is thrown to position b.
t
E
b
L
R2
E = 10 V
L = 15 mH
R1 = 4 
R2 = 6 
(d) What is the time constant  describing the change in current through the inductor?
We have a new formula available for time constants in LR circuits:  = L/R. But the R in the
formula refers to the total resistance in series with the inductor. Redrawing your circuit will
help you to determine this R!
Essentially the inductor discharges through R1 and R 2 in parallel.
The equivalent resistance is R E 
The time constant is  
R1  R2 6  4

 2.4 
R1  R2 6  4
L 15 103

 6.25 ms
RE
2.4
(e) Sketch the time dependence of the current through the inductor.
I
Imax
The current dies away exponentially starting from the I max
computed in part (b) according to I = Imax exp  t /   and
t
resembles the figure.
(f) What is the energy stored in the inductor 8 msec after the switch goes to position b?
First you must write down an equation for the time-dependence of the current. Check that
your formula is correct: does it produce the right answer at time 0? What about at t = ∞?
The stored energy is U 
where U max
LI 2 L 2 L
LI 2
2
 2t 
 2t 
 I   I max exp(-t/    max exp     U max exp   
2
2
2
2
  
  
 2  8 ms 
is value calculated in part (c). Thus U  20.83 mJ exp    1.61 mJ
 6.25 ms 
.
Discussion Question 10B
Physics 212 Week 10
Inductors
In this question, we will explore the effective inductance of series and parallel combinations of
wire coils. In your calculations, you can assume that your coils are all very long compared to the
radii of their circular turns … which allows you to approximate them as infinite solenoids.
(a)
l
(b), (c)
(d)
L1
L1
n turns per meter
coil area A
L2
L2
(a) Calculate the self-inductance L of the coil in figure (a).
The formula is L = /I … just remember, the magnetic flux  we’re talking about
here is that caused by the coil’s own field, not some external field as in earlier
problems. (That’s what we mean by self-inductance). So run some current I
through the coil and see what flux it causes.
The field of the inductor is the field of a solenoid or B= nI .
This field pierces nL loops of area A and hence   BnA or

   n 2 I A  L    n 2 A
I
(b) Work out the equivalent inductance for two inductors in series. Write down a
relationship between the voltage and current through the pair of inductors, in terms of L1
and L2 for each coil. This relation should show that the pair of inductors is equivalent to a
single coil with inductance Ls.
Let V be the voltage across the pair, and let V1 and V2 be the voltages across the individual
inductors. Start with an equation relating these …
If the two coils are in series, one has the same I and dI/dt flowing through them.
dI
dI
The voltage drops are V1  L1
and V2  L2 . The net voltage drop across
dt
dt
dI
dI
dI
both coils is Veq  V1  V2  L1
+L2
 Lequiv
 Lequiv  L1  L2
dt
dt
dt
(d) What about two coils connected in parallel? Using the technique from part (b),
determine the effective inductance Lp of this new combination.
If the coils are in parallel, the Vequiv  V1  V2 . The voltage drops are
dI1
dI
dI
and V2  Vequiv  L2 2 and Vequiv  Lequiv
dt
dt
dt
We now write the rate of change of currents:
Vequiv dI dI1 dI 2 Vequiv Vequiv
1
1 1






 
Lequiv
dt dt
dt
L1
L2
Lequiv L1 L2
related to the currents as V1  Vequiv  L1
(e) Finally, here’s a little design challenge, to help you develop some intuition about how
inductors are constructed. Suppose you have a spool of thin wire of total length D = 5 m and
radius R = 0.2 mm. You can make an inductor out of this wire by wrapping it into tight circular
turns, thereby forming a spiral. How should you do the wrapping to achieve the maximum
possible inductance? Given that you will make circular turns, there are only two parameters you
can vary: the radius r of your turns, and the gap g between them (i.e. how tightly you pack your
turns.) Let’s try three designs and see which one gives the greatest self-inductance L.
(i) Design #1: Use circular turns of radius r = 1 cm, and pack them as tightly as possible
without causing any of the turns to overlap.
(ii) Design #2: Double the turn radius r to 2 cm.
(iii) Design #3: Go back to the 1 cm turn radius, but this time try packing the turns more
loosely: leave a gap of 1 wire diameter between each turn.
Hint- write L in terms of the winding radius r, the total wire length D, and the spacing
between the wire centers.
We begin by writing L in a form that can be used for all three cases. We will write
all quantities in the expression L= n 2 A in terms of the winding radius r , the total length of wire D , and
the spacing between the centers of adjacent wires s. A   r 2 Let N be the total number of number
D
of turns one can make with a length D of wire: N =
. Write  in terms of spacing between
2 r
adjacent wires and total number of wires: Ns   
sD
N 1
  and n  
2 r
 s
1
 sD   rD
Substitute these forms for n ,  , and A to get L = n A      r 2  

2s
s
 2 r 
2
2
Continues on next page…
Design #1 and #2 have s  2 R since the wires are spaced apart by 1 wire diameter.
 rD
Hence L = 
Thus L #2  2L #1 Since r#2  2 cm and r#1  1 cm
4R
Design #3 has r#3 =1 cm and s# 3  4 R = 2s #1 Hence L #3 =
L #1 =
 rD
2R
 4    0.01 5  314  H ;
=
2  0.1 10 
1
L . Hence the order is L #2 > L #1  L #3
2 #1

3
L #2 =628  H ; L #3  157  H