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Self-Inductance and Circuits
• Inductors in circuits
• RL circuits
Inductors in Series and Parallel
LT = L1 +L2….
1/LT = 1/L1 + 1/L2…
Self-Inductance
Self-induced emf:
dI
 L  L
dt
Potential energy stored in
an inductor:
I
U L  LI
1
2
2
RL circuits: current increasing
The switch is closed at t =0;
Find I (t).
Kirchoff’s loop rule:
dI
 L
 IR  0
dt
dI   IR R  



  I
dt
L
LR

I
ε
L
R
Solution
I (t ) 


1 e 
R
t / 
Time Constant:
L

R
Note that H/Ω = seconds
(show as exercise!)
Time Constant:
 
L
R
Current Equilibrium Value:
I (t ) 
I 


R

1 e 
R
t / 
ε/R
63%
I
t
0
1τ
2τ
3τ
4τ
Example 1
Calculate the inductance in an RL circuit in which R=0.5Ω
and the current increases to one fourth of its final value
in 1.5 sec.
RL circuits: current decreasing
Assume the initial current I0 is known.
Find the differential equation for I(t)
and solve it.
I
L
R
I (t )  I e
Current decreasing:
 t /
o
Time Constant:
L

R
I
Io
0.37 I0
0τ
τ
2τ
3τ
4τ
t
Example 2:
I1
6Ω
I3
I2
12 V
50kΩ
200 mH
a) The switch has been closed for a long time.
Find the current through each component, and
the voltage across each component.
b) The switch is now opened. Find the currents
and voltages just afterwards.
Solution
LC circuits (Extra! – not on test/exam)
The switch is closed at t =0;
Find I (t).
Looking at the energy loss in each
component of the circuit gives us:
I
C
+
-
EL+EC=0
Which can be written as (remember, P=VI):
dI Q
LI
 I 0
dt C
dI Q
L

0
dt C
L
Solution
RLC circuits (Extra! – not on test/exam)
The switch is closed at t =0;
Find I (t).
Looking at the energy loss in each
component of the circuit gives us:
I
C
+
L
R
EL+ER+EC=0
Which can be written as (remember, P=VI=I2R):
dI
Q
2
LI
I R  I 0
dt
C
dI
Q
L
 IR 
0
dt
C
Solution