Transcript Chapter 7-Ex 7.2-Ex-7.4

Since the current through an inductor is given as
t

i  1 v ( )d  i (t 0 )
L
t0
Therefor to find i1,i 2
We need to Find
v (t)
6W
Finding
v (t)
The circuit reduce to

v (t )  8i (t )  8 12e 2t

v (t )
 v (t )
10
415 10 
 v (t )  15
2t
i3  
1015  5.7e A
10


t  0+
i 3  5.7e 2t A
t  0+
v 2 (t )
p (t ) 
=
8
The voltage across the capacitor =
For t >= 0
KCL on the upper junction
Similar to the previous example 7.2 , the voltage across a capacitor is given as
t
v (t )  1
C

i ( )d v (t 0 )
t0
Question: how to find i(t) ?
Therefor to find v 1(t ),v 2 (t )
We need to Find
i(t)
t
v (t )  1
C

i ( )d v (t 0 )
t0
Therefor to find v 1(t ),v 2 (t )
We need to Find
i(t)
Question: how to find i(t) ?
Do we find i(t) through the capacitor voltage
i (t )   C1
The answer is no of course because we do not know
However we can find i(t) through the resistor 250 kW
as
we need to find v(t) as will be shown next
dv 1(t )
dv (t )
or i (t )   C 2 2
dt
dt
v 1(t ), v 2 (t )
i (t ) 
v (t )
(250X103 )
The circuit reduce to


(5)(20)
(5  20)
The solution of the voltage v(t) is given as
v (t ) V 0e
t / 
t 0
were
  RC
 (250X 103 )(4X 106 )  1 s
V 0  20 V
v (t )  20e t
t 0
The circuit reduce to
v (t )  20e t
t 0
Note the direction of the current i (t) and the polarity of v 1(t ), v 2 (t )
and the polarity of v 1(t ), v 2 (t ) with respect to the initial voltages
24 V and 4 V
The circuit reduce to
v (t )  20e t
t 0
v (t )  20e t
t 0
v (t )  20e t
t 0
(a)
(b)
(c)
d) Show that the total energy delivered to the 250 kW resistor is the difference between
the results in (b) and (c)
Comparing the results obtained in (b)
and in (c)
Solving for t