Circuits with Sources I
Concept Test
R
+
+
u(t)
C
–
+
v1
–
L
i2
y(t)
–
For the circuit above, take v1 and i2 to be the
state variables. Find the differential equation
for the state i2 .
My confidence in my answer is:
1. 0%
2. 20%
3. 40%
4. 60%
5. 80%
6. 100%
Circuits with Sources I
Concept Test
R
+
+
u(t)
C
–
+
v1
–
L
i2
y(t)
–
For the circuit above, take v1 and i2 to be the
state variables. The differential equation for
the state i2 is
di2
dt
=
1
L
v1
My answer was
1. Completely correct.
2. Had one or two small errors.
3. Incorrect.
4. I didn’t understand the problem.
Circuits with Sources I
Solution
Lecture S13: Circuits with Sources I (Confidence)
1
Answer
2
3
4
5
6
0
5
10
Number of Students
15
Lecture S13: Circuits with Sources I (Correctness)
Answer
1
2
3
4
0
5
10
15
Number of Students
20
25
The differential equation for i2 is
di2
1
= v2
dt
L
But v2 = v1 , since the inductor and capacitor are in parallel,
and the voltage across the inductor is referenced the same
way as the capactior. Therefore,
di2
1
= v1
dt
L
Note that differential equation is in terms of states and
parameters of the circuit only.
The class had some difficulty with this problem. There should
be plenty of practice on the problem set.
Circuits with Sources II
Concept Test
R
+
+
u(t)
C
–
+
v1
–
L
i2
y(t)
–
For the circuit above, take v1 and i2 to be the
state variables. Find the differential equation
for the state v1 .
My confidence in my answer is:
1. 0%
2. 20%
3. 40%
4. 60%
5. 80%
6. 100%
Circuits with Sources II
Concept Test
R
+
+
u(t)
+
v1
–
C
–
L
y(t)
i2
–
For the circuit above, take v1 and i2 to be the
state variables. The differential equation for
the state v1 is
dv1
dt
=−
1
RC
v1 −
1
C
i2 (t) +
1
RC
u(t)
My answer was
1. Completely correct.
2. Had one or two small errors.
3. Incorrect.
4. I didn’t understand the problem.
Circuits with Sources II
Solution
Lecture S13: Circuits with Sources II (Confidence)
1
Answer
2
3
4
5
6
0
5
10
15
20
Number of Students
25
30
Lecture S13: Circuits with Sources II (Correctness)
Answer
1
2
3
4
0
5
10
15
Number of Students
20
25
The differential equation for v1 is
1
dv1
=
i1
dt
C
So we need to solve for i1 . We can apply KCL to the node at
the top of the capacitor to obtain
i1 +
v1 − u
+ i2 = 0
R
Solving for i1 ,
i1 = −
1
1
v1 − i2 (t) + u(t)
R
R
Therefore,
1
dv1
1
1
=−
v1 −
i2 (t) +
u(t)
RC
C
RC
dt