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California State Polytechnic University, Pomona Electrical & Computer Eng. Dr. Z. Aliyazicioglu ECE 109 Introduction to Electrical Engineering Homework # Name :__________________________ Due date November 8, 2017 October 31, 2017
- Use mesh analysis to obtain I 0 in the following circuit.
Solution: KVL for I 1 : I 1 (^) = 2 mA
Super mesh − 6 + 1 kI (^) 2 + 2 kI 3 (^) + 2 ( k I 3 (^) − I 1 (^) ) + 1 ( k I (^) 2 − I 1 ) = 0 − 6 + 1 kI (^) 2 + 2 kI 3 (^) + 2 ( k I 3 (^) − 2 mA ) + 1 ( k I (^) 2 − 2 mA ) = 0 2000 I (^) 2 + 4000 I 3 = 6 + 4 + 2 2000 I (^) 2 + 4000 I 3 = 12
Also, over the super current source I (^) 3 − I (^) 2 = 4 mA or − 1000 I (^) 2 + 1000 I 3 = 4
2 3
I
I
2 3
I mA I mA
To find i
0 1 3 2 3.
i I I
mA
MATLAB code
A=[2000 4000;-1000 1000]; b=[12;4]; A\b ans = -0.
By inspection, write the mesh-current equation for the following circuit an find v 1 and v 2
600 Ω
50Ω 200 Ω
10V I^1
400 Ω
200 Ω
v 1 v 2
12V
6V (^600) Ω
I 2 I 3
I 4
Use PSpice to check your answer
Solution: Mesh-current by inspection
1 2 3 4
I
I
I
I
+^ −^ ^
^ =
− − + + ^ −
Simplify
1 2 3 4
I
I
I
I
−^
I 1 (^) = 40 mA , I (^) 2 = 0 A , I 3 (^) = − 10 mA , I (^) 4 = − 10 mA
To find v 1 and v 2
1 200 (^1 2 )
v I I mA V
2 600 (^2 3 )
v I I mA V
A=[250 - 200 0 0;- 200 1000 -600 -200;0 -600 1200 -600; -200 -600 1200]; b=[10;0;-6;-6]; A\b ans =
-0. -0. -0.
- Using mesh analysis, determine the power delivered by the independent source 6V in the following circuit.
20mA
6V
+- 6vx
50Ω 100Ω
50Ω
150Ω 300Ω (^) v+x
Solution Mesh Equations: 150 I 1 + 6 V (^) x = 0 (1) I (^) 2 = 20 mA (2) 50 I (^) 3 + 6 = 0 (3) (150 + 300) I 4 (^) − 150 I 2 − 6 = 0 (4)
Vx = 300 I 4
Using (3) 3 6 0. 50
I = − = − A
Using (4) 450 I (^) 4 − 150(0.02) − 6 = 0 450 I (^) 4 = 9
4
I = = A
The currents trough the independent 6V is I = I (^) 4 − I (^) 3 = 0.02 − −( 0.12) =0.14 A
The power delivered by the independent source 6V
P 6 (^) Ω = I * ( 6)− = 0.14 * ( 6)− =0.84 W
I
I 1 I^2
I (^3) I (^4)
- Using node analysis, find Vx in the following circuit.
1kΩ
2kΩ
2kΩ
4mA
V (^) B
V (^) x
2kΩ 6V
V (^) A
Solution At Node vb To write KCL i 1 (^) − i 2 (^) − i 3 = 0
v (^) A vB vB vB vx k k k
v (^) A − vB − 2 v (^) B − vB + vx = 0 v (^) A − 4 v (^) B + vx = 0 To write KCL for Super node 4 mA − i 1 (^) + i 3 (^) − i 4 = 0 4 0 0 2 2 2 mA v^ A^ vB^ vB^ v^ x^ vx k k k
− −^ + −^ − − − =
8 − v (^) A + vB + vB − vx − vx = 0 v (^) A − 2 v (^) B + 2 vx = 8
And KVL at super node v (^) A = 6 + vx v (^) A − vx = 6
In Matrix form
1 4 1 0 1 2 2 8 1 0 1 6
A B x
v v v
, 2 2 50 i =^ v , 3 (^3 ) i =^ v
Therefore
A B x
v V v V v V
v (^) x = 2.5 V
A=[1 - 4 1;1 - 2 2;1 0 - 1]; b=[0;8;6]; A\b ans =
i 1
i 2
i 3 i 4
