# 1. Suppose that the switch has been closed for a long time and is

```NCTU 2008 Course: Electric Circuit(I)
Midterm-2 (Cap7 to Chap10)
1. Suppose that the switch has been closed for a long time and is opened at t=0. Determine the
inductor current i(t) for t&gt;0.
24V
2Ω
3Ω
i(t)
t=0
+
−
2H
2i(t)
Sol:
24V
2Ω
3Ω
i(t)
+
−
2H
2i(t)
For t&lt;0, we have
( ( ) ( ))
( )
( )
( )
24 = 3 2 i 0 − + i 0 − + 2 i 0 − = 11i 0 − ⇒ i 0 − =
24V
2Ω
3Ω
+
−
i(t)
2H
For t&gt;0, we have
24 = (3 + 2)i (t ) + 2
5
− t
2
di (t )
di(t ) 5
⇒
+ i (t ) = 12
dt
dt
2
( )
24
24
. Since i (0 ) = i 0 − =
, we have
5
11
24 24
24 24 24 &times; (− 6)
144
=−
A+
=
⇒ A=
−
=
5 11
11 5
55
55
Hence, i (t ) = Ae
Therefore, i (t ) = −
+
5
144 − 2 t 24
e +
A.
55
5
1
24
11
(12%)
NCTU 2008 Course: Electric Circuit(I)
Midterm-2 (Cap7 to Chap10)
2. Suppose that the switch has been opened for a long time and is closed at t=0. Find the output
voltage vo(t) for t&gt;0.
80V
+
−
(15%)
4Ω
4Ω
+
vo(t)
1H
t=0
−
10mF
Sol:
4Ω
4Ω
80V
iL(t)
+
vC(t)
+
−
−
For t&lt;0, we have
( )
iL 0 − =
1H
( )
80
= 10 and vC 0 − = 0
4+4
+
vC(t)
−
di L (t )
= vC (t )
dt
(1)
iL(t)
+
−
For t&gt;0, we have
−
10mF
4Ω
80V
and
( )
+
vo(t)
1H
10mF
+
vo(t)
−
dvC (t )
 80 − vC (t )

= 100
− i L (t )
4
dt


( )
Therefore, from (1) we obtain vC 0 − = 0 and v&amp;C 0 − = 25 &times; 80 − 100 &times; 10 = 1000 . Besides,
di L (t )
= −25 v&amp;C (t ) − 100 vC (t )
dt
i.e., v&amp;&amp;C (t ) + 25 v&amp;C (t ) + 100 vC (t ) = 0 . The eigenvalues satisfy λ2 + 25λ + 100 = 0 or λ = −5 , − 20 ,
v&amp;&amp;C (t ) = −25 v&amp;C (t ) − 100
which results in vC (t ) = Ae −5 t + Be −20 t and v&amp;C (t ) = −5 Ae −5 t − 20 Be −20 t . Thus, A+ B = 0 and
200
200
− 5 A− 20 B = 1000 . Then we have A =
.
and B = −
3
3
200 −5 t 200 − 20 t
Hence, vo (t ) = vC (t ) =
e −
e .
3
3
2
NCTU 2008 Course: Electric Circuit(I)
Midterm-2 (Cap7 to Chap10)
3. Based on phasor method, determine the inductor current iL(t).
iL(t)
(15%)
2mH
2sin1000t
50&micro;F
20Ω
2Ω
Sol:
IL
2∠−90&deg;
j2Ω
−j20Ω
20Ω
2Ω
Based on phasor method, we have
1




− j2
20 + j 2

=
I L = 2∠ − 90&deg; &times;
 j 1

1
 j 1
+ +

 (20 + j 2) +  + 1
 20 2 
 20 2 20 + j 2 
− j 40
=
= −0.0326 − j 0.1775 = 0.1805∠ − 100.4&deg;
(20 + j 2)( j + 10) + 20
Hence, we obtain
i L (t ) = 0.1805 cos (1000 t − 100.4&deg;)
3
NCTU 2008 Course: Electric Circuit(I)
Midterm-2 (Cap7 to Chap10)
4. Solve the phasor current I.
5Ω
5Ω
j2Ω
(15%)
10∠90&deg; V
I
−jΩ
+
−
+
−
3Ι
2∠0&deg; V
Sol:
5Ω
j10 V
V1
I
+
−
5Ω
j2Ω
V2
−jΩ
+
−
3Ι
2V
From nodal analysis, we have
V1 − j 10 V1 V1 − V2
+
+
= 0 ⇒ (2 + j 5)V1 − V2 = j 10
5
−j
5
V2 − 2
V
V − V1
+3 1 + 2
= 0 ⇒ (− 2 + j 30)V1 + (2 − j 5)V 2 = − j 10
j2
−j
5
Hence, V1=1.0129 − j0.7551 and V2=5.8011− j 6.4457. Therefore, we have
I=
V1 1.0129 − j 0.7551
=
= 0.7551 + 1.0129 j = 1.2634∠53.3&deg;
−j
−j
4
NCTU 2008 Course: Electric Circuit(I)
Midterm-2 (Cap7 to Chap10)
5. For the circuit with an ideal OP-amp, if the capacitor voltage is initially charged to be vC(0)=1V,
determine the output voltage vo(t) for t&gt;0.
100&micro;F
10kΩ
+ vC(t) −
(12%)
5kΩ
5kΩ
t=0
10V
+
−
−
+
+
vo(t)
−
Sol:
C
10kΩ
i
10V
+
−
100&micro;F
+ vC(t) −
R
−
+
5kΩ
5kΩ
i
+
vo(t)
−
For t&gt;0, we have
i(t ) =
10
= 1 mA
10k
and from the KCL, we obtain
dvC (t ) vC (t )
+
dt
R
dv (t ) v (t ) i (t )
⇒ C + C =
dt
RC
C
⇒ v&amp;C (t ) + 2 vC (t ) = 10
i(t ) = C
⇒ vC (t ) = 5 + Ae − 2 t
Since vC (0 ) = 1 , we have A=−4.
Furthermore, vo (t ) = − vC (t ) − 5000 &times; 10 −3 = −5 + 4 e −2 t − 5 = 4 e −2 t − 10 V .
5
NCTU 2008 Course: Electric Circuit(I)
Midterm-2 (Cap7 to Chap10)
6. Consider the following third-order circuit with input voltage v(t).
T
(a) Write the state equation x&amp; (t ) = Ax (t ) + b v(t ) , where x (t ) = [x1 (t ) x 2 (t ) x3 (t )] contains
three state variables x1(t), x2(t) and x3(t) shown in the circuit.
(10%)
(b) Let the inductor current be the output, i.e., y(t)=i(t)=x3(t). Then, the circuit can be described
by a third-order differential equation given as
(6%)
&amp;y&amp;&amp;(t ) + a 2 &amp;y&amp;(t ) + a1 y&amp; (t ) + a 0 y (t ) = b2 v&amp;&amp;(t ) + b1v&amp;(t ) + b0 v(t )
Find the coefficients ai and bi, i=0,1,2.
1Ω
1F
1Ω
v(t)
+
x1(t)
i(t)
−
+
−
+
x2(t)
1F
1Ω
1H
x3(t)
−
Sol:
1 v1
○
2 v2
○
1Ω
1F
1Ω
v(t)
+
−
3 v3
○
1Ω
+
x1(t)
iC1
1F
iC2
i(t)
−
+
vL
+
x2(t)
−
1H
x3(t)
−
(a)
From nodal analysis and by treating state variables as sources, we have node equations
v1 = v(t )
v 2 = x1 (t ) + x 2 (t )
v3 = x 2 (t )
Based on the component equations, we have
6
NCTU 2008 Course: Electric Circuit(I)
Midterm-2 (Cap7 to Chap10)
v1 − v 2
− i (t ) = v(t ) − ( x1 (t ) + x 2 (t )) − x3 (t )
1
= − x1 (t ) − x 2 (t ) − x3 (t ) + v(t )
x&amp;1 (t ) = iC 1 =
v1 − v3 v3
−
1
1
= − x1 (t ) − x 2 (t ) − x3 (t ) + v(t ) + v(t ) − x 2 (t ) − x 2 (t )
x&amp; 2 (t ) = iC 2 = i1 +
= − x1 (t ) − 3 x 2 (t ) − x3 (t ) + 2 v(t )
x&amp; 3 (t ) = v L = v 2 = x1 (t ) + x 2 (t )
Rearraging the above equations yields
 x&amp;1 (t ) − 1 − 1 − 1  x1 (t ) 1 
 x&amp; (t ) = − 1 − 3 − 1 ⋅  x (t ) + 2 v(t )
 2  
  2   
 x&amp; 3 (t )  1
1
0   x3 (t ) 0
(b)
Since y(t)=i(t)=x3(t), we have
y (t ) = x3 (t )
y&amp; (t ) = x&amp; 3 (t ) = x1 (t ) + x 2 (t )
&amp;y&amp;(t ) = x&amp;1 (t ) + x&amp; 2 (t )
= − x1 (t ) − x 2 (t ) − x3 (t ) + v(t ) − x1 (t ) − 3 x 2 (t ) − x3 (t ) + 2 v(t )
= −2 x1 (t ) − 4 x 2 (t ) − 2 x3 (t ) + 3 v(t )
&amp;y&amp;&amp;(t ) = −2 x&amp;1 (t ) − 4 x&amp; 2 (t ) − 2 x&amp; 3 (t ) + 3 v&amp;(t )
= 2 x1 (t ) + 2 x 2 (t ) + 2 x3 (t ) − 2 v(t )
+ 4 x1 (t ) + 12 x 2 (t ) + 4 x3 (t ) − 8 v(t ) − 2 x1 (t ) − 2 x 2 (t ) + 3 v&amp;(t )
= 4 x1 (t ) + 12 x 2 (t ) + 6 x3 (t ) − 10 v(t ) + 3 v&amp;(t )
According to the characteristic equation
λ +1
λI − A =
1
−1
1
1
λ + 3 1 = λ 3 + 4λ 2 + 4 λ + 2
−1 λ
the third-order differential equation can be derived as
&amp;y&amp;&amp;(t ) + 4 &amp;y&amp;(t ) + 4 y&amp; (t ) + 2 y (t ) = 3 v&amp;(t ) + 2 v(t )
Hence, we obtain a2=4, a1=4, a0=2, b2=0, b1=3, b0=2.
7
NCTU 2008 Course: Electric Circuit(I)
Midterm-2 (Cap7 to Chap10)
7. Complete the design of R in the following
10kΩ
oscillator and calculate its oscillation
frequency.
8kΩ
−
+
(15%)
2&micro;F
R
2&micro;F
5kΩ
Sol:
10kΩ
It is true that v1 and vo must be in phase.
Therefore, their phasors must satisfy
the following condition:
8kΩ
1
−1
V1
8
R + jω C
=
=
1
1
Vo 10 + 8
+ r+
−1
jω C
R + jω C
−
+
Hence,
R
R
1 + j ω RC
=
R
1

1 
+ r+
(1 + j ω RC )
R +  r +
jω C
1 + j ω RC
j
C
ω


R
4
⇒ =
1 
9

2 R + r + j  ω rRC −

ωC 

R
 4
 9 = 2 R+ r
⇒
1
 ω rRC −
=0
ωC

4
=
9
⇒ R = 4 r = 20kΩ
⇒ R = 20kΩ
and
and
f =
ω=
r
v1
2&micro;F
C
1
1
=
= 50
2
2 rC
rRC
50
Hz
2π
8
vo
R
5kΩ
2&micro;F
C
```