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Lecture 10
o Aim of the lecture
 Inductors
 Behaviour in circuits
 Energy storage
 Comparison with Capacitors
 V vs i
 Energy
 Use in Circuits
 LR time constant
 sparks
LCR Circuit
Oscillations
Filters
o Main learning outcomes
 familiarity with
 Inductors in circuits
 i and V LR curves
 Energy storage
 basic LC and LCR circuit
 Concept of filters
Reminder:
o an inductor, as used in electronics
 is typically a coil of wire wound on a core
 The core increases the inductance by mr
o Real inductors must have resistance
 because the wire has resistance
o In fact they also have ‘stray’ capacitance, but we will ignore this
inductor
Resistor
(resistance of wire in coil)
L = mrm0N2A/l
Reminder:
 For a resistor V=iR
 For a capacitor i = C dV/dt
For an inductor the relationship between
current and voltage is:
V = L di/dt
Note the ‘similarity’ between the relationships
for capacitors and inductors.
As circuit elements they can be thought of as
‘opposites’ in the sense that i and V play opposite roles
For an inductor the relationship between
current and voltage is:
V = L di/dt
So:
o If the current is constant, then V=0
 The inductor behaves simply like a (perfect) wire
o If there is current flowing, then to reduce it to zero
 requires a voltage as di/dt ≠ 0
[For a capacitor with a voltage on it, a current must
flow for the voltage to reduce to zero, as Q=CV – similar]
So:
Why?
o If the current is constant, then V=0
 The inductor behaves simply like a (perfect) wire
because
is energy
stored
into
the
inductor
o If therethere
is current
flowing,
then
reduce
it towhen
zero it has
a current
flowingathrough
 requires
voltage it,
as di/dt ≠ 0
Just like a capacitor has energy stored in it when it has a
voltage across it.
Work done = ∫ Power dt = ∫ iVdt = ∫ i(Ldi/dt)dt = ∫ iLdi = ½Li2
Energy stored, E = ½Li2
Capacitor Energy stored, E = ½CV2
Recall that there is NO energy stored in a resistor
A resistor dissipates energy, it produces heat
Inductors DO NOT dissipate any energy
Capacitors DO NOT dissipate any energy
These circuit elements can ABSORB energy
But it can come out again, it is only stored
Energy stored, E = ½Li2
Capacitor Energy stored, E = ½CV2
Where is the energy stored?
In the magnetic field!
There is an energy density associated with the field, so to create it
work must be done. In a capacitor the energy is stored in the electric
field, [but remember they are both aspects of the same e-m field]
If the current flowing decreases, then
the stored energy must decrease,
the ‘excess’ must go somewhere.
It is absorbed by the circuit that is changing the current.
The inductor does work ON the circuit
This car relies on an inductor to make the (petrol) engine run
- why? – it is used to make the spark to ignite the petrol
A petrol engine (not a diesel) needs a spark
to ignite the petrol. An inductor is used.
[in fact some modern cars do it differently, but still common]
When the switch is closed,
the voltage across and
the current through the
inductor will look like this
Because,
Kirchoff’s law says VR + VL = E
so iR + Ldi/dt = E
this is a differential equation, the solution is i = (E /R){1-e-t/(R/C)}
time
When the switch is closed,
the voltage across, and
the current through the
inductor will look like this
o But if we now open the switch again,
 there is a problem
 no circuit path
 so no current can flow
o But the stored energy is = ½Li2
 so the current MUST flow
 because there is energy stored
o To keep current flowing
 the inductor will generate a huge voltage,
 big enough to break-down the air
 cause a spark across the switch
 thousands of volts
 current will flow until energy dissipated
 it turns into heat and light
o But if we now open the switch again,
 there is a problem
 no circuit path
 so no current can flow
o But the stored energy is = ½Li2
 so the current MUST flow
o To keep current flowing
 the inductor will generate a huge voltage,
 big enough to break-down the air
 cause a spark across the switch
 thousands of volts
 current will flow until energy dissipated
o Add a second gap, smaller than the switch gap
It will break down first
The smaller gap is the ‘spark gap’
Located where the spark is needed
Engineering Detail
Because:
The ‘spark gap’ in a car engine
is in the ‘spark plug’
screwed into the engine block
at top of cylinder
gap
I or V
switch
R
L
V0
VL
IL
VL = V0e-t/(L/R) IL = (V0/R){1-e-t/(L/R)}
Time after switch closed
The sparking situation is ‘abnormal’, it is
more common to be using the components in
a ‘controlled way. Sparks are usually bad!
I or V
switch
R
C
V0
VC
IC
Vc = V0e-t/CR IC = (V0/R){1-e-t/CR}
Time after switch closed
Provided that L/R = RC then the shape of the
I and V curves simply swap
note ‘similarly’ in behaviour of L and C
I or V
switch
R
L
V0
VL
IL
VL = V0e-t/(L/R) IL = (V0/R){1-e-t/(L/R)}
Time after switch closed
Provided that L/R = RC then the shape of the
I and V curves simply swap
note ‘similarly’ in behaviour of L and C
I or V
switch
R
C
V0
VC
IC
Vc = V0e-t/CR IC = (V0/R){1-e-t/CR}
Time after switch closed
Provided that L/R = RC then the shape of the
I and V curves simply swap
note ‘similarly’ in behaviour of L and C
Capacitors and Inductors are complementary devices (Often)
o They can be ‘energised’ and ‘emptied of stored energy’
o This is called charging and discharging
o (but recall this is NOT a statement about net electric charge)
The relationship between voltage and current is:
o
o
o
V=iR
resistors
V=Ldi/dt inductors
i=CdV/dt capacitors
o Note that only resistors can ‘use’ energy
 Dissipate it as heat
Now consider using inductors and capacitors in together:
o In the circuit below, as shown,
 the capacitor will energise (charge)
 after a ‘long’ time VC = V0.
 ILC = 0
switch
L
C
V0
VC
ILC
So the Energy stored in the circuit is E= ½CVC2
o Now change the switch position, as shown below
 the capacitor is energised, E = ½ CVC2
 VC = V0.
 ILC = L dILC/dt ≠ 0
switch
L
C
V0
VC
ILC
o But now there is an inductor connected across the capacitor
o There is a voltage VC across L
 Current must flow
o As current grows, the voltage across capacitor will decrease
 Because Q=CV and I = dQ/dt
o After some time, the capacitor will be fully de-energied
 the capacitor has no energy, E = ½ CVC2 = ½ C 02 = 0
 VC = 0.
 ILC = maximum = ILCmax
switch
L
C
V0
VC
ILC
o As there is current flowing through an inductor
o There is energy stored in its magnetic field
 EL = ½ L(ILCmax)2
o As no energy can be lost in this circuit (no resistance!)
 EL = ½ L(ILCmax)2 = ½ CV02
 so ILCmax = V0 √(C/L)
o All the energy is now stored in the inductor
 EL = ½ CV02
 VC = 0 = VL
 ILC =V0 √(C/L)
switch
L
C
V0
VC
ILC
o As there is current flowing through the capacitor
o It will energise (charge up)
 The current will decrease
 as the energy is transferred back to the capacitor
o Eventually the current reaches zero and
 EL = ½ L(ILC)2 = ½ L02 =0
 EC = ½ CV02
o The energy oscillates back and forth between
 The inductor, stored as energy in magnetic field
 The capacitor, stored as energy in the electric field
 (and between the extremes, also shared between L and C)
switch
L
C
V0
ILC
VC
V0 √(C/L)
V0
ILC
VC
The final answer
is that the voltages
and currents in this
circuit oscillate like
a sine wave
But phase shifted
time
switch
L
Write Kirchoff’s voltage law round the loop
VC
C
V = 0 V=0LdILC/dt + Q/C
ILC
take derivative wrt to time
V = 0 = Ld2ILC/dt + (1/C)dQ/dt
= Ld2ILC/dt + ILC/C
You will recognise
this asrecognise
SHM withit!!w0 = 1/LC
Or you should
switch
L
C
V0
VC
ILC
0 = Ld2ILC/dt + ILC/C
ILC = {V0√(C/L)} sin(w0t+f)
w0 = 1/√(LC)
f really matters when it is used to show the difference between the
f is
found from
initialand
boundary
different
voltages
currentsconditions:
in the circuit, soBut
thef shift
is notbetween
very important
Current and voltage in capacitor is p/2 for this case, it just gives
As ILC Similarly
= minimumfor
=0
when
t=0, but it is –p/2
the
inductor,
a definition of t=0
= 0 Capacitor and inductor
The voltages onf the
are p different.
In practice, all circuits have some resistance, so consider:
.
2
dQ Q
d Q
L 2 +R
+ =0
dt capacitor
C
dtto charge
 Close S1
 Open S1 to disconnect battery
-(t/)sin(w't + ).
Q=
 Close
S2Qto
circuit with L,R,C
0e make
1
1  R 
ω= ω2 - 2=
- 
LC  2 L 
τ
2
ILC = dQ/dt
The solutions are similar,
This
is simply oscillation,
with
a sinusoidal
but Damped
now alsoSHM
with an exponential
Identicalfactor
equations to
damping
mechanics
VC
Sinusoidal voltage source
VC
Filters:
The combination of LC can be
used to select frequencies
(eg to choose a channel in a radio)
You
know everything
but will
to take
another course
All don’t
the analysis
discussedyet!
has–been
‘in need
the time
domain’
toAC
findsignals
out all (not
the clever
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This new kind of application, with
batteries)
is ayou
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thefrequency
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new topic area, and called ‘analysis
is ininthe
- engineering -
Some real filters
Band-stop filter
Vout/Vin
Enjoy!
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