Download RL Circuits

RL Circuits
AP Physics C
Montwood High School
R. Casao
• A circuit that contains a coil, such as a solenoid,
has a self-inductance that prevents the current
from increasing or decreasing instantaneously.
• A circuit element that has a large inductance is
called an inductor.
• The circuit symbol for an inductor is
• We will always assume that the self-inductance
of the remainder of the circuit is negligible
compared to that of the inductance.
• Keep in mind that even a circuit without a coil
has some self-inductance that can affect the
behavior of the circuit.
• Because the inductance of the inductor esults in a
back EMF, an inductor in a circuit opposes
changes in the current in that circuit.
– The inductor attempts to keep the current the same as
it was before the change occurred.
– If the battery voltage in the circuit is increased so that
the current rises, the inductor opposes this change and
the rise is not instantaneous.
– If the battery voltage is decreased, the presence of the
inductor results in a slow drop in the current rather
than an immediate drop.
– The inductor causes the circuit to be “sluggish” as it
reacts to changes in the voltage.
• Consider a circuit consisting of a resistor, an
inductor, and a battery.
– The internal resistance of the battery will be neglected.
• The switch is closed at t = 0 and the current
begins to increase.
• Due to the increasing current, the inductor will
produce an EMF
(the back EMF)
that opposes the
increasing current.
• The inductor acts like a battery whose polarity is
opposite that of the battery in the circuit.
• The back EMF produced by the inductor is:
dI
EMFL  L 
dt
• Since the current is
increasing, dI/dt is
positive; therefore,
EMFL is negative.
• This corresponds to the fact that there is a
potential drop in going from a to b across the
inductor.
• Point a is at a higher potential than point b.
• Apply Kirchhoff’s loop rule to the circuit:
dI
EMF  I  R  L 
0
dt
• Divide through by R:
dI
dI
EMF  I  R  L 
0
EMF  I  R  L 
dt
dt
EMF I  R L dI
EMF
L dI

 
I  
R
R
R dt
R
R dt
• Let
; EMF and R are constant;
EMF
x
I
I is variable
R
• Take the derivative of x:
EMF
 EMF

dx  d 
Id
 dI  0  dI
R
 R

dx  dI
dI  dx
• Replace the variables in the equation:
L dx L dx
x 


R dt
R dt
EMF
L dI
I  
R
R dt
• Combine like terms and differentiate the differential
equation:
L dx L dx
R
1
x 


 dt   dx
R dt
R dt
L
x
R
1
R
1
 dt 
 dx
 dt 
 dx
L
x
L
x
R
R
x
 t  ln x x
 t  ln x  ln xo
o
L
L
R
x
 t  ln
L
xo




• Exponentiate both sides of the equation:
R
x
 t  ln
L
xo
x  xo
R
t
e L
e
ln
R
t
e L
• At t = 0 s, I = 0 A, so:
EMF
EMF
xo 
I 
0
R
R
EMF
xo 
R
x
xo
R
t
e L
x

xo
• Replacing the substituted variables:
x  xo
R
t
e L
EMF
x
I
R
R
t
e L
EMF
EMF
I 
R
R
R 

t
EMF
I 
 1  e L 


R


EMF
xo 
R
EMF EMF
I 

R
R
• The current I as a function of time:
R 

t
EMF
I t  
 1  e L 


R


R
t
e L
• The equation shows how the inductor effects the current.
• The current does not increase instantly to its final
equilibrium value when the switch is closed but instead
increases according to an exponential function.
• If the inductance approaches zero, the exponential term
becomes zero and there is no time dependence for the
current – the current increases instantaneously to its
final equilibrium value in the absence of the inductance.
• Time constant for an RL circuit:
L
 
R
so
R 1

L 
• The current I as a function of time:
t
R 



t
EMF
EMF
L

I t  
 1  e
 1  e 



R
R







• The time constant τ is the time it takes for the current to
reach (1 – e-1) = 0.63 of the final value, EMF/R.
• The graph of current versus time, where I = 0 A at t = 0
s; the final equilibrium value of the current occurs at
t =  and is equal to EMF/R.
• The current rises very fast initially
and then gradually approaches the
equilibrium value EMF/R as
t  .
• The rate of increase of current dI/dt is a maximum
(equal to EMF/L) at t = 0 s and decreases exponentially
to zero as t  .
• Consider the RL circuit shown
in the figure below:
• The curved lines on the switch S represent a switch that
is connected either to a or b at all times. If the switch is
connected to neither a nor b, the current in the circuit
suddenly stops.
• Suppose that the switch has been set at position a long
enough to allow the current to reach its equilibrium
value EMF/R. The circuit is described by the outer loop
only.
• If the switch is thrown from
a to b, the circuit is described
by just the right hand loop
and we have a circuit with
no battery (EMF = 0 V).
• Applying Kirchhoff’s loop rule to the right-hand loop at
the instant the switch is thrown from a to b:
dI
I  R  L 
0
dt
• Divide through by R:
dI
I  R  L 
0
dt
I  R L dI


R
R dt
dI
I  R  L 
dt
L dI
I 

R dt
• Combine the current I terms and differentiate the
differential equation:
L dI
1
R
I 

 dI 
 dt
R dt
I
L
1
R
 dI 
 dt
I
L
R
R
I
ln I I 
 dt
ln I  ln I o 
t
o
L
L
I
R
ln

t
Io
L



• Exponentiate both sides of the equation:
I
R
ln

t
Io
L
I

Io
R
t
e L
e
I
ln
Io
I  Io

R
t
e L
R
t
e L
• The current I as a function of time for an RL circuit in
which the current is decaying over time:
I (t )  I o
R
t
e L
• When t = 0 s, Io = EMF/R and τ = L/R:
t
t
EMF
I (t )  I o  e 
e 
R
• The current is continually decreasing with time.

– The slope, dI/dt, is always negative and has its maximum
value at t = 0 s.
dI
– The negative slope indicates that the EMFL  L 
dt
is now positive; that is, point a is at a lower
potential than point b.
Time Constant of an RL Circuit
• The circuit shown in the figure consists of a 30 mH
inductor, a 6  resistor, and a 12 V battery. The switch
is closed at t = 0 s.
• Find the time constant of the circuit.
L 0.03 H
 
 0.005 s
R
6
• Determine the current in the circuit at t = 0.002 s.
t
I (t )  I o  e

t
EMF
12V


e 
e
R
6
0.002 s
0.005 s
 0.659 A
Kirchhoff’s Rule and Inductors
• For the circuit shown in the figure:
A. find the currents I1, I2, and I3 immediately after switch
S is closed.
• Answer: The current in an inductor cannot change
instantaneously from 0 A to the steady state current,
therefore, the current in the inductor must be zero after
the switch is closed because it is 0 A before the switch is
closed.
• I3 = 0 A
• The current in the left loop equals the EMF divided by
the equivalent resistance of the two resistors in series.
• Req = 10  + 20  = 30 
EMF 150V
I1  I 2 

 5A
Re q
30 
B. find the currents I1, I2, and I3 a long time after switch S
has been closed.
• Answer: when the current reaches its steady-state value,
dI/dt = 0, so there is no potential drop across the
inductor. The inductor acts like a short circuit (a wire
with zero resistance).
Parallel resistances: R   20  
p

1
1 
  20   

1
Total resistance = 10  + 10  = 20 
Total current:
EMF 150V
I T  I1 

 7.5 A
RT
20 
 10 
The 7.5 A current will split in half to travel through the
two equal 20  resistors; I2 = I3 = 3.75 A
C. find the currents I1, I2, and I3 immediately after switch
S has been opened.
• Answer: The current in an inductor cannot change
instantaneously from the steady state current to 0 A,
therefore, the current in the inductor must be the same
after the switch is opened as it was just before the switch
is opened.
• I3 = 3.75 A
• When the switch is opened, I1 drops to 0 A.
• To oppose the change in the direction of the magnetic
flux through the inductor, the direction of the induced
current in the loop changes direction;
I2 = I3 = 3.75 A.
D. find the currents I1, I2, and I3 a long time after switch
S has been opened.
Answer: all currents must be 0 A a long time after the
switch is opened.
I1 = I2 = I3 = 0 A
RL Circuits Summary
I 
EMF
R
• The inductor acts like a wire when t is large:
• The inductor acts like an open circuit when t = 0 s;
i=0A
• The current starts from 0 A and increases up to a
EMF
I

maximum of
with a time constant given by L
R
L 
• As the current increases from 0 A
to its steady-state value:
– Current:
t

EMF
i
 1  e  L

R





R
– Voltage across the resistor:
t

VR  i  R  EMF  1  e  L






– Voltage across the inductor:
VL  EMF  VR
t

VL  EMF  EMF  1  e  L


t
VL  EMF  e  L




• The potential difference across a resistor depends on the
current.
• The potential difference across an inductor depends on
the rate of change of the current.
• The potential difference across an inductor depends on
the rate of change of the current.
• When the EMF source is removed from the circuit and
the current begins to decay from the steady-state current
to 0 A:
– Current decay:
t
EMF  L
i
e
R
t
VR  i  R  EMF  e
VR (V)
– Voltage across resistor:
L
– Voltage across inductor:
t R
L
di
EMF de
VL  L 
L

dt
R
dt
 EMF  e
t R
L
What is Happening During Current Decay?
• When the battery is removed and the RL series circuit is
shorted, the current keeps flowing in the same direction
it was for awhile. How can this be?
• What is happening is that the current tries to drop
suddenly, but this induces an EMF to oppose the
change, causing the current to keep flowing for awhile.
• Another way of thinking about it is that the magnetic
field that was stored in the inductor is “collapsing.”
• There is energy stored in the magnetic field, and when
the source of current is removed, the energy flows from
the magnetic field back into the circuit.
•
•
•
•
•
•
•
The switch in a circuit like the one at right has to be a special
kind, called a “make before break” switch.
The switch has to make the connection to b before breaking
the connection with a.
If the circuit is allowed to be in the state like this…even
momentarily, midway between a and b, then a big problem
results.
Recall that for a capacitor, when we disconnect the circuit the
charge will merrily stay on the capacitor indefinitely.
Not so on an inductor. The inductor needs current, i.e.
flowing charge. It CANNOT go immediately to zero.
The collapsing magnetic field in the inductor will force the
current to flow, even when it has no where to go.
The current will flow in this case by jumping the air gap.
Make Before Break Switches
November 7, 2007
Link to video
You have probably
seen this when
unplugging something
with a motor—a spark
that jumps from the
plug to the socket.
Start Your Engines: The
Ignition Coil
• The gap between the spark plug in
a combustion engine needs an
electric field of ~107 V/m in order to
ignite the air-fuel mixture. For a
typical spark plug gap, one needs
to generate a potential difference >
104 V.
spark
12V
•Breaking the circuit changes the current
through the “primary coil”
• Result: LARGE change in flux thru
secondary -- large induced EMF
• But, the typical EMF of a car
battery is 12 V. So, how does a
spark plug work?
The “ignition coil” is a double layer solenoid:
• Primary: small number of turns , connected to
12 V battery
• Secondary: MANY turns -- connected to spark
plug
Transformer: P=i·V