(b)
v
L
(0+)
=
v
C
(0+)
=
0
But,
v
L
=
Ldi
L
/dt
and
di
L
(0+)/dt
=
v
L
(0+)/L
=
0
di
L
(0+)/dt
=
0
Again, 80
=
45i
R
+ v
C
0
=
45di
R
/dt
+ dv
C
/dt
But,
dv
C
(0+)/dt
=
i
C
(0+)/C
=
0.278 mohms/1
µ
F
=
278 V/s
Hence,
di
R
(0+)/dt
=
(1/45)dv
C
(0+)/dt
=
278/45
di
R
(0+)/dt
=
6.1778 A/s
Also, i
R
=
i
C
+ i
L
di
R
(0+)/dt
= di
C
(0+)/dt + di
L
(0+)/dt
6.1788
= di
C
(0+)/dt + 0,
or di
C
(0+)/dt
=
6.1788 A/s
(c)
As
t
approaches infinity, we have the equivalent circuit in Figure
(b).
i
R
(
∞
)
=
i
L
(
∞
)
=
80/45k
=
1.778 mA
i
C
(
∞
)
= Cdv(
∞
)/dt
=
0
.
Chapter 8, Solution 3.
At
t
=
0

, u(t)
=
0.
Consider the circuit shown in Figure (a).
i
L
(0

)
=
0, and v
R
(0

)
=
0.
But, v
R
(0

) + v
C
(0

) + 10
=
0,
or
v
C
(0

)
=
10V.
(a)
At t = 0
+
, since the inductor current and capacitor voltage cannot change abruptly,
the inductor current must still be equal to
0A
, the capacitor has a voltage equal to
–10V
.
Since it is in series with the +10V source, together they represent a direct
short at t = 0
+
.
This means that the entire 2A from the current source flows
through the capacitor and not the resistor.
Therefore,
v
R
(0
+
) =
0 V
.
(b)
At t = 0
+
,
v
L
(0+)
=
0, therefore Ldi
L
(0+)/dt
=
v
L
(0
+
)
=
0, thus, di
L
/dt
=
0A/s
,
i
C
(0
+
)
=
2 A,
this means that dv
C
(0
+
)/dt
=
2/C
=
8 V/s
.
Now for the value of
dv
R
(0
+
)/dt.
Since v
R
=
v
C
+ 10, then dv
R
(0
+
)/dt
=
dv
C
(0
+
)/dt + 0
=
8 V/s
.