电阻电路例题
Question 1：
Please use source transformation to find the current I in the following circuit. [8 Marks]
Ω
I
Solution:
Use source transformation to redraw the circuit, we will get:
Ω
8V
[5 Marks]
So:
158
0.5347
I A -=
=++ [3 Marks]
Question 2:
Please use source transformation to find the current 2i in the following circuit. [8 Marks]
R 3
4
I s
Solution:
Redraw the original circuit as the following:
R 3
I s
[5 Marks]
Then the current 2i is:
3223
s
R I i R R =
+ [3 Marks]
Question 3: 15 Marks
Use the node-voltage method to find the branch current o I and the power dissipated in the 18 Ω resistor in the circuit in Figure P3.
Ω
12I 0
Figure P3
Solution:
Ω
12I 0
Select the lower node as the reference node and define the node voltages of node 1 and node 2 as 1V and 2V . Apply KCL for node 1 and node 2, and we get
112
00.560126
--+
++=V V V I [3 Marks]
20
21212606618
---++=V I V V V [3 Marks] The controlling current can be expressed as:
206
=V
I [3 Marks]
Solving for 1V and 2V gives
218V =V [2 Marks]
Then, the branch current 2018
3A 66
=
==V I [2 Marks] The power dissipated in the 18 Ω resistor is:
22
20181218123181818W 1818Ω
--?????
=?=?= ? ?
????
V I P [2 Marks]
Question 4: 12 Marks
Use mesh current method to find the current I in the circuit in Fig.1.
2
Fig. 1
Solution:
Assume the reference directions of three mesh currents I 1, I 2 and I 3 are clockwise, then, we will get:
2
From the circuit, we know I 3=2A, only I 1, I 2 are unknown. [2 Marks] Build KVL equations for the appointed super mesh, we will get:
2
1123(2)6230I I I -+++= [4 Marks]
As to the current source 4A, we can get another equation:
124I I -= [4 Marks]
So we can get:
1213I A
I A
=??
=-? [1 Marks] So:
12211I I A =-=-= [1 Marks]
Question 5: 15 Marks
Use the mesh-current method to find the power dissipated in the 6 Ω resistor in the circuit in Figure P3.
60 V
D
Figure P3
Solution:
60 V
D
Define mesh currents for the two meshes as 1I and 2I , and build two mesh-current equations as:
()112861200.50D +-+-=I I I V [4 Marks] ()2221246012060++-+-=I I I I [4 Marks] The controlling voltage can be expressed as:
24D =V I [3 Marks] Solving for 1I and 2I gives
18A =-I , and 21A =I [2 Marks] The power dissipated in the 6 Ω resistor is:
()()2
2
6126816486W Ω=-?=--?=P I I [2 Marks]
Question 6: 13 Marks
Find
the current I in the circuit in Fig. 3 by using the Thevenin equivalent of the remaining circuit except for the resistor of 14Ω.
Fig. 3
Solution:
The circuit without 14Ω branch is as following:
In order to find I, we will find the Thevenin equivalent circuit between terminal a and b. [2 Marks]
Step 1: Find Thevenin equivalent voltage. Obviously, the Thevenin equivalent voltage is:
2.520
(
)12.57.52.510205
Th ab V V V ==-?=-++ [4 Marks]
Step 2: Find equivalent resistor.
Turn off all independent sources, we will get:
So the equivalent resistor is: 2.5//1020//56Th R =+=Ω
[4 Marks] So we can get the equivalent circuit:
[1 Mark]
So the current I is: 7.5
0.37514614
Th Th V I A R -=
==-++ [2 Marks]
Question 7: 12 Marks
The operational amplifier in the circuit shown in Figure 6 is ideal. Find the current i o
and the voltage v o in the circuit.
12V
i 0-0
Figure 6
Solution:
12V
i 0-0
Define node voltages v a for node a as shown in the circuit above.
The voltages at the input nodes of an ideal operational amplifier are equal due to virtual short, so:
2V a v =-
[3 Marks]
Apply KCL at node a:
()()
2122084
o v ----+=
[3 Marks]
So, we get:
030V v =-
[3 Marks]
Apply Ohm ’s Law to the 8k Ω resistor, we can get the current i 0:
()0
o 2302mA mA =3.5mA 88
v i -----=
=
[3 Marks]
Question 8: 12 Marks
The operational amplifier in the circuit shown in Figure 6 is ideal. Find the current i 0.
Figure 6
Solution:
0+-v 0
Define node voltages v 0, v 1 and v 2 as shown in the circuit above. Due to virtual short for the operational amplifier, we get:
16V v =
[3 Marks]
Due to virtual short for the operational amplifier, we get:
216=36V v v =
[3 Marks]
Then, the voltage v 0 can be obtained:
o 1263642V v v v =+=+=
[3 Marks]
Apply Ohm ’s Law to the 10k Ω resistor, we can get the current i 0:
0o 42mA =4.2mA 1010
v i =
=
[3 Marks]
Question 9:
Please find the equivalent capacitance between terminal a and b. [7 Marks]
2F
Solution:
()()3220153220 2.53220153220ab C +???
+??
?++??==+???
++?
?++??
[7 Marks]