166
Practical MATLAB
® Applications for Engineers
and
v t
V i t R
L ( )
( ) *
ϭ Ϫ 1
1
eq
where
R
R
R R
R
R R
R
R
eq1
1
2
3
1
2
3
2
3
8
6 3
6 3
8 2 10
ϭ ϩ
ϭ ϩ ϩ
ϭ ϩ ϩ
ϭ ϩ ϭ
( | | )
*
*
and
1 = L/R eq1 = 10 H/10 = 1 s
At t = t 1 = 2 s, the current i 1 (t 1 ) is at its maximum as given by
I
i t
V
R
e
1
1
1
2
2
1
1
max
/
(
)
* (
)
ϭ
ϭ ϭ
0
eq
Ϫ
Ϫ
At t = 2 s, the switch moves back to position a, where it remains indefi nitely. The branch
currents would then be given by
i t
I
e
t t
1
1
1
2
( )
max
(
)/
ϭ
Ϫ Ϫ
i t
R
R
R
i t
2
3
2
3
1
( )
* ( )
ϭ
ϩ
i t
R
R
R
i t
3
2
2
3
1
( )
* ( )
ϭ
ϩ
and
v t
i t R
L ( )
( ) *
ϭ 1
2
eq
where
R
R
R R
R
R R
R
R
eq2
4
2
3
4
2
3
2
3
18
6 3
6 3
18 2 20
ϭ
ϩ
ϭ
ϩ
ϩ
ϭ ϩ ϩ
ϭ ϩ ϭ
( | | )
*
*
and
2 = L/R eq2 = 10 H/20 = 0.5 s
MATLAB Solution
% Script file: analysis _ exam
V=100; R1 = 8; R2 =3; R3 =6; R4 =18; L=10;
% circuit elements.
Req1 = R1+(R2*R3)/(R2+R3);
CRC_47760_CH002.indd 166
CRC_47760_CH002.indd 166
7/23/2008 1:38:52 PM
7/23/2008 1:38:52 PM
Practical MATLAB
® Applications for Engineers
and
v t
V i t R
L ( )
( ) *
ϭ Ϫ 1
1
eq
where
R
R
R R
R
R R
R
R
eq1
1
2
3
1
2
3
2
3
8
6 3
6 3
8 2 10
ϭ ϩ
ϭ ϩ ϩ
ϭ ϩ ϩ
ϭ ϩ ϭ
( | | )
*
*
and
1 = L/R eq1 = 10 H/10 = 1 s
At t = t 1 = 2 s, the current i 1 (t 1 ) is at its maximum as given by
I
i t
V
R
e
1
1
1
2
2
1
1
max
/
(
)
* (
)
ϭ
ϭ ϭ
0
eq
Ϫ
Ϫ
At t = 2 s, the switch moves back to position a, where it remains indefi nitely. The branch
currents would then be given by
i t
I
e
t t
1
1
1
2
( )
max
(
)/
ϭ
Ϫ Ϫ
i t
R
R
R
i t
2
3
2
3
1
( )
* ( )
ϭ
ϩ
i t
R
R
R
i t
3
2
2
3
1
( )
* ( )
ϭ
ϩ
and
v t
i t R
L ( )
( ) *
ϭ 1
2
eq
where
R
R
R R
R
R R
R
R
eq2
4
2
3
4
2
3
2
3
18
6 3
6 3
18 2 20
ϭ
ϩ
ϭ
ϩ
ϩ
ϭ ϩ ϩ
ϭ ϩ ϭ
( | | )
*
*
and
2 = L/R eq2 = 10 H/20 = 0.5 s
MATLAB Solution
% Script file: analysis _ exam
V=100; R1 = 8; R2 =3; R3 =6; R4 =18; L=10;
% circuit elements.
Req1 = R1+(R2*R3)/(R2+R3);
CRC_47760_CH002.indd 166
CRC_47760_CH002.indd 166
7/23/2008 1:38:52 PM
7/23/2008 1:38:52 PM
