NCTU 2008 Course Electric Circuit(I)
Midterm-1 (Cap1 to Chap6)
1. Under dc condition, determine the load current iL and the energy stored in the 5mF capacitor.
1kΩ
4V
5kΩ
+
vx
1kΩ
+
−
−
3kΩ
4mF
2vx
(12%)
4mA
iL
2kΩ
5mF
Solution:
1kΩ
4V
5kΩ
1kΩ
+
−
+
vx
3kΩ
4mF
i=0
−
2vx
4mA
iL
2kΩ
5mF
Under dc condition, we replace each capacitor with an open circuit. Then, no current through
the 5kΩ resistor and the voltage across the 4kΩ resistor is vx. According to the principle of
voltage division, we obtain
vx = 4 ×
1
= 2V
1+1
Therefore, the load current is
i L = 2 v x − 4 × 10 −3 = 2 × 2 − 4 × 10 −3 = 3.996A
The energy stored in the 5mF capacitor is
1
1
E = Cv x2 = × 5 × 10 −3 × 2 2 = 10mJ
2
2
2. Find the Thevenin equivalent circuit to the left of terminals a and b.
(15%)
25Ω
5V
+
−
25Ω
a
20Ω
6Ω
10Ω
R
15Ω
b
Solution:
According to the transformation formula, we have
R1 =
10 × 6 + 6 × 15 + 15 × 10 300
=
= 50 Ω
6
6
1
NCTU 2008 Course Electric Circuit(I)
Midterm-1 (Cap1 to Chap6)
R2 =
10 × 6 + 6 × 15 + 15 × 10 300
=
= 20 Ω
15
15
R3 =
10 × 6 + 6 × 15 + 15 × 10 300
=
= 30 Ω
10
10
R2
6Ω
10Ω
R1
R3
15Ω
The circuit can be changed into
25Ω
5V
+
−
20Ω
25Ω
a
10Ω
20Ω
R
30Ω
50Ω
b
With source transformation, we have
10Ω
0.2A
a
R
25Ω
25Ω
30Ω
50Ω
10Ω
b
i.e.
10Ω
0.2A
10Ω
a
R
30Ω
b
2
NCTU 2008 Course Electric Circuit(I)
Midterm-1 (Cap1 to Chap6)
Further, we obtain
10Ω
10Ω
2V
+
−
a
+
vTh
30Ω
R
−
b
With R→∞, the Thevenin voltage is determined from the voltage across the terminals a and b,
which is
vTh =
30
30
×2 =
× 2 = 1.2 V
50
30 + 10 + 10
By deleting the 2V source, the equivalent resistance to the left of terminals a and b is
1
60
RTh =
=
= 12 Ω
1
1
2+3
+
30 10 + 10
a
12Ω
1.2V
Finally, the Thevenin circuit is shown
+
R
−
On the right.
b
3. Based on the mesh-current method, write a set of mesh current equations in the following matrix
T
form : E ⋅ i = Bu , where E is a 4×4 square matrix, i = [i1 i2 i3 i4 ] is a 4×1 column
vector containing all the mesh currents, B is a 4×2 matrix, and u = [v s
vector composed of the independent sources.
is ]
T
is a 2×1 column
(Note that you have to assign the mesh currents and number them from i1 to i4 in your own way.)
(15%)
Ix
4Ω
5Ω
vs
2Ω
+
−
is
1Ω
3Ix
Solution:
3
NCTU 2008 Course Electric Circuit(I)
Midterm-1 (Cap1 to Chap6)
ix
5Ω
vs
4Ω
4
□
is
1
□
+
−
i3
3
□
i4
2Ω
i1
2
□
i2
3ix
1Ω
First, number the mesh currents i1, i2, i3, and i4.
1 ,□
2 ) and (□
3 ,□
4 ). The mesh current equations are
There are two supermesh, (□
1 □
2 :
Supermesh□
KCL: i1 − i2 = 3 i x = 3 i4 ⇒ i1 − i2 − 3 i4 = 0
KVL: − v s + 5(i1 − i3 ) + 2(i2 − i3 ) + i2 = 0
⇒ 5 i1 + 3 i2 − 7 i3 = v s
(1)
(2)
3 □
4 :
Supermesh□
KCL: i3 − i4 = i s
KVL: 5(i3 − i1 ) + 4 i4 + 2(i3 − i2 ) = 0
⇒ − 5 i1 − 2 i2 + 7 i3 + 4 i4 = 0
(3)
(4)
From (1)-(4), we have
 1 − 1 0 − 3  i1  0 0
5
3 − 7 0  i2  1 0 v s 

⋅
=
⋅
0
0
1 − 1 i3  0 1  i s 
 {
   

i
5
2
7
4
0
0
−
−
 23 u
144424443 {
4
1
i
E
B
4. (A) When the load R is not connected, the current through 6Ω resistor is i0=1A.
Determine the source voltage vs.
(4%)
(B) If the load R is connected, what is the maximum power can be transferred to R?
8Ω
vs
+
−
2Ω
(8%)
2Ω
8Ω
6Ω
R
i0
4
NCTU 2008 Course Electric Circuit(I)
Midterm-1 (Cap1 to Chap6)
Solution:
(A) According to the principle of current division, the currents in the circuit
are obtained as below:
2Ω
8Ω
vs
2A
+
−
8Ω
6Ω
1A
1A
(B)
2Ω
Hence the voltage source is v s = 2 × 8 + 1 × 8 = 24 V .
By setting R→∞, the Thevenin voltage is the voltage across the 6Ω resistor, i.e.,
vTh = 1 × 6 = 6 V . The equivalent resistance RTh to the left of R is obtained by the
following circuit:
8Ω
4Ω 2Ω
3Ω 2Ω
8Ω
6Ω
RTh=5Ω
Since the equivalent resistance RTh=5Ω, the maximum power transferred to the load is
under the condition R=5Ω. Therefore, the maximum power is
 6 6
1
  v
Pmax = v R ⋅ i R =  vTh  ⋅  Th  = ⋅
= 1 .8 J .
 2   R + RTh  2 5 + 5
5. Use superposition to find v.
(12%)
1Ω
60Ω
8V
2A
40Ω
40Ω
4V
+−
+−
40Ω
40Ω
Solution:
5
60Ω
+
v
−
NCTU 2008 Course Electric Circuit(I)
Midterm-1 (Cap1 to Chap6)
The voltage v can be solved by the following three cases:
1Ω
60Ω
6V
2A
40Ω
40Ω
4V
+−
+−
20Ω
40Ω
The voltage v1 = −4 ×
1Ω
60
= −3 V .
60 + 20
60Ω
40Ω
1Ω
4V
+−
+−
2A
24Ω
60Ω
40Ω
+
v2
−
24
= −3 V .
40 + 24
60Ω
4V
+−
+−
40Ω
15Ω
40Ω
40Ω
8V
2A
40Ω
40Ω
8V
The voltage v 2 = −8 ×
60Ω
40Ω
+
v1
−
40Ω
60Ω
+
v3
−
15Ω
The voltage v3 = 2 × 15 = 30 V .
Hence, v = v1 + v 2 + v3 = −3 − 3 + 30 = 24 V .
6. Let v(t ) = 12 + 5 e −2 t V.
As t→∞, determine the total energy stored in the three capacitors and
the total energy stored in the three inductors.
(12%)
6
NCTU 2008 Course Electric Circuit(I)
Midterm-1 (Cap1 to Chap6)
12Ω
2mF
4Ω
v(t)
45mH
+
−
4mF
1mF
100mH
300mH
Solution:
First, rearrange the circuit as below:
12Ω
2mF
4Ω
v(t)
45mH
+
−
75mH
5mF
100mH
300mH
As t→∞, v(t ) = 12 , i.e., the circuit is under dc condition. Therefore, all the capacitors are open
and all the inductors are short. The circuit is changed into
v(t)
+
−
4Ω
+
0V
0A
−
12Ω
1A
2mF
120mH
+
12V
5mF
−
Hence, the total energy stored in the capacitors is EC =
total energy stored in the inductors is E L =
1
× 5 × 10 −3 × 12 2 = 0.36 J and the
2
1
× 120 × 10 −3 × 12 = 0.06 J .
2
7
NCTU 2008 Course Electric Circuit(I)
Midterm-1 (Cap1 to Chap6)
7. Find the Norton equivalent circuits with respect to terminals a and b.
(12%)
15V
30Ω
+
vx
−
+
−
a
0.2vx
RL
50Ω
60Ω
b
Solution:
First, find the Norton current iN from the following circuit:
15V
30Ω
+
−
a
0.2vx
+
vx
−
iN
50Ω
60Ω
b
The voltage vx can be calculated by
i N = 0.2 v x = 0.4 A .
v x − 15 v x
+
+ 0.2 v x = 0 ⇒ v x = 2 . The Norton current is
30
60
The equivalent resistance is determined by adding an extra voltage source V as below:
1
○
30Ω
+
vx
−
I
0.2vx
50Ω
60Ω
+ V
−
vx vx
+
+ 0.2 v x = 0 ⇒ v x = 0 . Hence, V=50I. That means the
30 60
V
equivalent resistance Req = = 50 Ω. The Norton circuit is shown as below:
I
1 , we have
From node ○
a
0.4A
R
50Ω
b
8
NCTU 2008 Course Electric Circuit(I)
Midterm-1 (Cap1 to Chap6)
8. The following circuit contains ideal op amps, find the output voltage v0.
1kΩ
(10%)
3kΩ
3kΩ
5kΩ
−
4V
+
−
−
+
10V
12kΩ
+
−
4kΩ
+
+
−
+
2kΩ
10kΩ
1kΩ
vo
−
10kΩ
Solution:
1kΩ
3kΩ
3kΩ
5kΩ
−
4V
+
−
−
3
○
+
10V
+
−
12kΩ
+
4kΩ
1
○
2
○
+
−
+
2kΩ
10kΩ
10kΩ
1kΩ
vo
−
1 , ○
2 and ○
3 and their node voltages v1, v2, and v3. Then, we have
Assign nodes ○
1
4
v3 = − × 4 = −
3
3
3
3
3  4 5
17
v 2 = − v3 − × 10 = − ×  −  − = −
5
12
5  3 2
10
v1 =
2
1  17 
17
v2 = ×  −  = −
4+2
3  10 
30
17
 17 
v0 = 2 v1 = 2 ×  −  = −
V
15
 30 
9