PROBLEM # 7.7
In the circuit shown, the switch makes contact with position b just before breaking
contact with position a. As already mentioned, this is known as a make – before
– break switch and is designed so that the switch does not interrupt the current in
an inductive circuit. The interval of time between “making” and “breaking” is
assumed to be negligible. The switch has been in the a position for a long time.
At t = 0, the switch is thrown to position b.
a) Determine the initial current in the inductor.
The circuit for t< 0.
i (0)  i (0) 
24
 2A
12
b) Find i, v1, and v2 for t>=0.
Circuit for t>0.
i ( )  0
L 1.6
 
 0.02 s
R 80
1
1

 50
 0.02
i (t )  i ()  [i (0)  i ()]e
R
 t
L
i (t )  2e 50t , t  0
di (t )
v1 (t )  L
dt
d

v1 (t )  1.6   (2e 50t )   1.6  (100e 50t )
 dt

v1 (t )  160e 50tV , t  0
v2 (t )  i (t )  R  2  72  144V , t  0
c) What percentage of the initial energy stored in the inductor is dissipated in the
72 Ω resistor 15 ms after the switch is thrown from position a to position b?
w(0) 
1 2
1
Li (0)   1.6  22  3.2 J
2
2
t
t
w(t )   i ( x) Rdx   (2e
2
0
0
t
w(t )  288 e 100 x dx  
0

t
) (72) dx   (4e 100 x )(72) dx
50 x 2
0
288 100 x t
e
  2.88 1  e 100t  J
0
100
w(15ms )  2.88 1  e  (100)(1510
3
)
  2.88 1  0.223
w(15ms )  2.24 J
%dissipated 
2.24
 100%  70%
3.2