EE2111 Homework #2 A03
Assigned 04 September 03
Due 10 September 03
Problem 1. Hambly P2.37
Solution:
First, we can write
ix 
Then since v1  v2  5ix , we have
v1  v2 
or
v1  2v2 .
v1
10
5v1 v1

10 2
Now we will do nodal analysis at nodes I and 2. For node 1 we
have
v
 5  1  i12  0
10
Applying KCL at node 2 we have
 i12 
v2
3 0
20
Adding these 2 equations we get
v
v
5 1  2 3 0
10 20
or
2v1  v2  160
Thus we solve the 2 equations
v1  2v2  0
2v1  v2  160
to get v1  64 V and v2  32 V.
Finally
ix 
v1
 6.4 A.
10
Problem 2. Hambly P2.39
Solution: Label the left node 1 and the right node 2 and insert a 1
A current source between a and b. There is a dependent current
source in this problem. We have
v
v v
is  x  1 2
5
5
Now apply KCL at nodes 1 and 2
v1 v1  v2

1
20
10
v2  v1 v2 v1  v2
 
0
10
5
5
or
3v1  2v2  20
v2  v1  0
Solve these 2 equations to find v1  4 V The equivalent resistance
is equal in value to v1 hence Req  4  .
Problem 3. Hambly P2.40
Solution: This is just like problem 2.39. We insert a fictitious 1A
Current source between points a and b and apply nodal analysis.
The dependent source equation is ix  v1  v2 and the KCL
equations at nodes 1 and 2 are
v1 v1  v2

1
5
1
v2  v1 v2
  2v1  v2   0
1
2
or
6v1  5v2  5
 6v1  7v2  0
Solving, we find v1  2.917 V and thus the equivalent resistance is
Req  2.917  .
Problem 4. Hambly P2.45
Solution: The mesh currents are i1 and i2 . We apply KVL around
the 2 loops to get
14i1  8i2  10
 8i1  16i2  0
Solving we find i1  1 A and i2  0.5 A.
Problem 5. Hambly P2.46
Solution: Label the left hand loop current i1 and the right hand
loop current i2 . Now apply KVL around the 2 loops:
40i1  20i2  10
 20i1  40i2  0
Solving we find i1  0.333 A and i2  0.1667 A.
Thus v  20i1  i2   3.333 V.
Problem 6. Hambly P2.47
Solution: Label the left hand loop current i1 and the right hand
loop current i2 . The mesh equations are
i1  10
 15i1  20i2  0
Solving we find i2  7.5 A. The current i3 shown in Figure P2.30
is i3  i3  7.5 A.
Problem 7. Hambly P2.52
Solution: We can get the Thevenin Equivalent resistance by
eliminating the voltage source and combining resistances in series
and parallel to find Re  5.198  . We can solve for the open
circuit voltage by either nodal or mesh analysis. I will use nodal
analysis. Label the nodes 1, 2, and 3 from left to right. Since node
1 is connected to a voltage source whose other end is attached to
the reference node, we have v1  25 V. Applying KCL at the other
2 nodes, we find
v2  25 v2 v2  v3


0
10
20
5
v3  25 v3  v2 v3

 0
25
5
15
or
7v2  4v3  50
 15v2  23v3  75
I solved these equations using MATLAB and found v2  14.3564
V and v3  12.6238 V. The Thevenin voltage is vt  v3  12.6238
V.
The Norton equivalent is a current source with value
v 12.6238
in  t 
 2.429 A in parallel with Rt .
Rt
5.198
Problem 8. Hambly P2.55
Solution: The Thevenin voltage is equal to the open circuit voltage
which is 20 V. When we apply a 1 kΩ load, the load voltage is 5 V
and the current through the load is
5
il 
 5 mA. The voltage drop across the Thevenin resistance
1000
15
 15 kΩ.
is therefore 15 volts, thus Rt il  15 and so Rt 
5  103
Problem 9. Hambly P2.56
Solution: The load resistance closes a loop with the Thevenin
equivalent circuit. For Rl  100 Ω we have a 10 V load voltage,
thus
v
10
il  l 
 0.1
Rl 100
Appling KVL around the loop we find
vt  0.1Rt  10
Similarly for the 200 Ω load, we find
v
12
il  l 
 0.06
Rl 200
and
vt  0.06Rt  12
Solving the system
vt  0.1Rt  10
vt  0.06 Rt  12
we find Rt  50  and vt  15 V.
Problem 10. Hambly P2.57
Solution: Under open circuit conditions we have
20  voc
ix 
5
Apply KCL at node a
voc
 ix  0.5ix  0
10
Solving these 2 equations for voc we find voc  10 V.
20
= 4 A.
5
Applying KCL at node a we have isc  ix  0.5ix  2 A. Therefore
the Thevenin resistance is:
v
Rt  oc  5 
isc
Under short circuit conditions, the current ix 