Transcript Inductors - Stephen F. Austin State University

Inductors
Energy Storage

Current passing through a coil causes a
magnetic field


Energy is stored in the field
Similar to the energy stored by capacitors


We saw a charging time for a capacitor
An inductor takes time to store energy also
Simple RL Circuit
I
V  IR  L
t
L must have units of Ohmsseconds
From the Construction

Inductance
N A
L

2
N = number of turns on the coil
 = permeability of the core (henrys/m)
A = cross sectional area (m2)
l = length of core (m)
L = inductance in henrys
Relative Permeability

Many texts and handbooks publish Km, where
 = Km o
o = permeability of free space = 4p X 10-7 Wb/A

Ex: Compute L for the following coil:


N = 100 turns
l = 25 X 10-3 m
A = 1.3 X 10-4 m2
Km = 400 (steel)
N 2 K m o A
L

(100)2 ( 400 )( 4pX 107 )( 1.3 X 104 )

25 X 103
 0.0261H  26.1m H
Time Dependence
E
t / 
I  (1 e
)
R
,
 = L/R
This is the same way that voltage varied
in the capacitor
Try it!
Notes

The final current (E/R) doesn’t depend on L




There is no voltage drop across the inductor after the full
current has been established
The coil then acts as a short circuit (as if it weren’t there)
The inductance depends on the change of current
(once I is established, I/t → 0 and V=IR)
At first I = 0, so V = IR = 0

As current rises the voltage drop across the resistor (IR)
gets greater, leaving less voltage to be dropped through the
coil.
Voltage
V  Ee
t / 
Inductors in Series
Kirchhoff’s Voltage Law
E  VL1  VL2  VL3
 L1
I
( L1  L2  L3 )
t
I

LT
t

LT = L1 + L2 + L3
I
I
I
 L2
 L3
t
t
t
Inductors in Parallel
E  VL1  VL2  VL3
The analysis is difficult in a dc circuit since the
voltage drains to zero, but the result is…
1
1 1 1
  
LT L1 L2 L3
Real Inductors

Inductors have…



Internal Resistance
Internal Capacitance between windings
So a real inductor in a circuit looks like…
Example
The equivalent circuit is
1
1
1


RT 4 k 8 k
RT  8 k
3
Continued
LT
20 X 103 H


 7.5 s
3
RT 2.66 X 10 
E
t / 
t / 7.5 s
I
(1 e
)  3.76m A( 1  e
)
RT
Comparing inductors to capacitors
After about 5, the current has reached a
maximum for the coil and zero for a capacitor.
The coil acts as a short, while the capacitor acts
like an open circuit.
Sample RLC Circuit
After about 5 , the equivalent circuit is
No current
flows
through C1
and L1 acts
as if it’s not
there
Solve Circuit

R1 and R2 are in series, so…
E
10 V
I

 2A
R1  R2 5 

For the path ABCD
IR1 + IR2 = E
VR1  VR2  E
Notice that R2 and C1 are
in parallel, so VR2 is the
voltage drop across the
capacitor also.
VR2  E  VR1  10V  IR1
 10 V - (2A)(2)  6V
Stored Energy

Capacitor


WC = ½CV2
Inductor

WL = ½LI2