Download Problem 1

機二丙 電工學 第一次 期中考
Problem 1:
7:30 – 9:00 PM
2012/ 3/28
Total: 104
Points
Basic Circuit Concepts (12 Pts)
(a). C.2.32 The unknown devices shown in the figure below absorb the indicated powers.
unknown current ix. (6 Pts)
P3=6W
P3=1W
P2=-3W
P4=12W
(b). C. 2.27 Determining voltages v1 an v2 in the following circuit. (6 Pts)
ix
Determining the
機二丙 電工學 第一次 期中考
Problem 2:
2012/ 3/28
7:30 – 9:00 PM
Total: 104
Points
Resistive Circuits ( 18 Pts)
(a). In the following figure, please find the powers delivered by the current source and the voltage source,
respectively. (6 Pts)
SOL:
i2 
10 V
 2A
5
i1  i2  3  1 A
Pcurrent  source  3 A  10 V  30 W
Pvoltage  source  10i1  10 W
電流源產能功率 30W
電壓源產能功率-10W
(b). H 2.40 Suppose we need to design a voltage-divider circuit to provide an output voltage vo=5V from a 15-V
source shown below.
The current taken from the 15-V source is to be 200mA. Please find the values of R1 and
R2. Now, suppose that a load resistance of 200 is connected across the output terminals. Please find the value of
vo.
(6 Pts)
機二丙 電工學 第一次 期中考
7:30 – 9:00 PM
2012/ 3/28
Total: 104
Points
SOL:
(a) R1  R2 
R2
15 V
 75 
0.2 A
R1  R2
 15  5
Solving, we find R2  25  and R1  50  .
(b)
The equivalent resistance for the parallel combination of R2 and the load is
1
Req 
 22.22 
1 25  1 200
Then, using the voltage division principle, we have
vo 
Problem 3:
Req
 15 V  4.615 V
R1  Req
Kirchhoff’s Laws and Circuit Analysis (node + mesh) (36 Pts)
(a) 2.58 Solve for the power delivered to the 8- resistance and for the node voltage of the system shown below.
(12 Pts)
SOL: First, we can write
ix  
v1
10
.
KCL at the reference node, we have
ix and simplifying, we have
Then writing KVL, we have v1  5ix  v 2  0 .
v2
20
 ix  8 .
Writing
Using the first equation to substitute for
機二丙 電工學 第一次 期中考
2012/ 3/28
7:30 – 9:00 PM
Total: 104
Points
1.5v 1  v 2  0
2v1  v 2  160
Solving, we find v 1  45.71 V, v 2  68.57 V, and ix  
power delivered to the 8- resistance is P 
(b) Consider the circuit shown below.
(i):
Please find
v1
10
 4.571 A . Finally, the
(v 1  v 2 )2
 65.31 W.
8
(12 Pts)
current i1, i2, and i3 (4 Pts)
(ii): voltage across the 9.5, 20.5, and 10 load resistance.
(4 Pts)
(iii): if the 1  neutral wire is broken, determine the voltage across the 10  load resistance. (4 Pts)
機二丙 電工學 第一次 期中考
2012/ 3/28
7:30 – 9:00 PM
Total: 104
Points
(c.) For the following circuit, please write a set of mesh-current equations by using the symbol accordingly.
va = -50V, vb = 150V. (12 Pts)
Where
機二丙 電工學 第一次 期中考
Problem 4:
2012/ 3/28
7:30 – 9:00 PM
Total: 104
Equivalent Circuits and Power Transfer (26 Pts)
(a) H. 2.88 Find the Thevenin equivalent circuit for the circuit shown below. (12 Pts)
Points
機二丙 電工學 第一次 期中考
2012/ 3/28
7:30 – 9:00 PM
Total: 104
Points
SOL:
Open-circuit conditions:
ix 
vx
15  v x
5
10  10
have Vt  v oc  v x
 ix  0.5ix  0
Solving, we find v x  10 V and then we
10
 5 V.
10  10
Short-circuit conditions:
ix 
vx
15  v x
5
have isc 
vx
10
10
 ix  0.5ix  0
Solving, we find v x  7.5 V and then we
 0.75 A. Then, we have Rt  voc isc  6.67  .
Thus the equivalents are:
(b) Determine both the Thevenin and Norton equivalent circuit with respect to terminals a and b in the following
circuit (10 Points).
In addition, please find the maximum power transfer on a load Rab connected to the
機二丙 電工學 第一次 期中考
7:30 – 9:00 PM
2012/ 3/28
Total: 104
Points
terminals a and b. (4 Points)
266 Ohms
410V
Problem 5:
(a) H 3.22a
205/133 A
Capacitor and Resistor (12 Pts)
Please find the equivalent capacitance between terminal x and y for the circuit shown below. (6 Pts)
機二丙 電工學 第一次 期中考
2012/ 3/28
SOL: Ceq  5 
1.
H 3.18
7:30 – 9:00 PM
Total: 104
Points
1
1

 6.667  F
1 2  1 1 1 2  1 1  1
Please use any methods to show that the equivalent capacitance of N capacitors in series is
1
1
1
1
(6 Pts)
 
 ..... 
CT C1 C2
CN