# ECE35_SP17_HW03_Solution.pdf

## Full text

(1)

th

th

### P 4.11-2

(2)

Figure P4.3-3

P4.3-3. Determine the values of the power supplied by each of the sources in the circuit shown in Figure P4.3-3.

Solution: First, label the node voltages, and express the resistor currents in terms of the node voltages.

Identify the super node corresponding to the 24 V source

Apply KCL to the super node to get

a

a a

a a

a

## 

a

(3)

Figure P 4.3-7

### Solution:

Apply KCL at nodes 1 and 2 to get

1 1 1 2

1 2

10

23 3 150 1000 3000 5000

v v v v

v v

 

    

2 1 3 3

1 3

10

-4 19 50 4000 5000 2000

v v v v

v v

 

    

Solving, e.g. using MATLAB, gives

1

1 1

2

### 

Then b 1 2

7.06 4.12

0.588 mA

5000 5000

v v

i     

Apply KCL at the top node to get

1 2

a

10 10 7.06 10 4.12 10

4.41 mA

1000 4000 1000 4000

v v

(4)

### Determine the values of the node voltages of the circuit shown in Figure P 4.3-9.

Figure P 4.3-9

Solution:

Express the voltage source voltages as functions of the node voltages to get

2 1 5 and 4 15

vvv

Apply KCL to the supernode corresponding to the 5 V source to get

1 3 2

1 2 3

15

1.25 0 80 5 2 5

8 20

v v v

v v v

 

      

Apply KCL at node 3 to get

1 3 3 3

1 3

15

15 28 150

8 40 12

v v v v

v v

 

     

Solving, e.g. using MATLAB, gives

1 1

2 2

3 3

1 1 0 5 22.4

5 2 5 80 27.4

15 0 28 150 17.4

v v

v v

v v

   

     

   

   

   

     

   

     

        

So the node voltages are:

1 22.4 V, 2 27.4 V, 3 17.4 V, and 4 15

(5)

Figure P 4.3-12

### Solution:

Express the voltage source voltages in terms of the node voltages:

2 1 8 and 3 1 12

vvvv

Apply KVL to the supernode to get

2 1 3

2 1 3

0 2 5 4 0

10 4 5

v v v

v v v

      

so

1

1

1

## 

1

64

2 8 5 4 12 0 V

11

v v v v

       

The node voltages are

1

2

3

5.818 V

2.182 V

6.182 V

v v v

(6)

x = 2.4 A

Figure P 4.4-5

### Solution:

First, express the controlling current of the CCVS in terms of the node voltages: 2

2

x

v i

Next, express the controlled voltage in terms of the node voltages:

2

2 2

24

12 3 3 V

2 5

x

v

v i v

    

(7)

Figure P 4.4-11

### Solution:

This circuit contains two ungrounded voltage sources, both incident to node x. In such a circuit it is necessary to merge the supernodes corresponding to the two ungrounded voltage sources into a single supernode. That single supernode separates the two voltage sources and their nodes from the rest of the circuit. It consists of the two resistors and the current source. Apply KCL to this supernode to get

x x

x

20

4 0 10 V

2 10

v v

v

    

### .

The power supplied by the dependent source is

0.1vx

## 

30

 30 W

(8)

### Solution:

Express the controlling voltage and current of the dependent sources in terms of the node voltages:

a 4

vv and b 3 4 2

### 

Express the voltage source voltages in terms of the node voltages:

1 s

vV and v2v3 Ava Av4

Apply KCL to the supernode corresponding to the dependent voltage source

2 1 3 4

s 2 1 2 2 1 3 1 4 1 2 s

1 2

### 

Apply KCL at node 4:

### 

3 4 3 4 4 2

3 4

2 2 3 3

1 1 0

v v v v v R

B B v B v

R R R R

 

 

       

 

Organizing these equations into matrix form:

1 s

2

2 2 1 1

3 1 2 s

2 4 3

1 0 0 0

0 1 1

0

0 0 1 1 0

v V

A

v

R R R R

v R R I

R v B B R                                                

With the given values:

1 1

2 2

3 3

4 4

(9)

Figure P 4.5-2

### Solution:

The mesh equations are:

Top mesh: 4 (2 3) R(2) 10 (2 4)0 so R = 12 .

Bottom, right mesh: 8 (4 3) 10 (4 2)   v2 0

so v2 = 28 V.

Bottom left mesh  v1 4 (3 2) 8 (3 4)   0

(10)

Figure P 4.5-6

60  300  50 

### 

so the simplified circuit is

The mesh equations are

## 

1 1 2

200i 50 ii 120

## 

2 1 2

100i  8 50 ii 0 or

1 1

2 2

### 

The power supplied by the 12 V source is

1

### 0.48 W

. The power supplied by the 8 V

source is

2

### 0.32 W

. The power absorbed by the 30  resistor is

###   

2 2

1 30 0.04 30 0.048 W

(11)

Figure P 4.6-5

### Label the mesh currents:

Express the current source current in terms of the mesh currents:

3 1 2 1 3 2

i  ii  i

Supermesh: 6i13i35

i2i3

## 

 8 0  6i15i28i38

Lower, left mesh:   12 8 5

i2i3

## 

0  5i2  4 5i3

Eliminating i1 and i2 from the supermesh equation:

3

3

## 

3 3

6 i   2 4 5i 8i 8  9i 24

The voltage measured by the meter is: 3

24

3 3 8 V

9

i  

(12)

Figure P 4.6-8

### . Express the currents to the supermesh to get

1 3 2

i  i

Apply KVL to the supermesh to get

3 3

3

1 2

## 

1 2 3

4 ii  1 i  3 1 ii 0  i 5i 5i 3

Apply KVL to mesh 2 to get

## 

###  

2 2 3 2 1 1 2 3

2i 4 ii  1 ii 0  1 i 7i 4i 0

Solving, e.g. using MATLAB, gives

1 1

2 2

3 3

1 0 1 2 3

1 5 5 3 1

1 7 4 0 1

i i

i i

i i

   

     

   

   

   

     

   

      

(13)

Figure P 4.7-7

### Solution:

Express va and ib, the controlling voltage and current of the dependent sources, in terms of the mesh

currents

## 

a 5 2 3 and b 2

vii i  i

Next express 20 ib and 3 va, the controlled voltages of the dependent sources, in terms of the mesh

currents

b 2

20 i  20 i

3 va 15

i2i3

## 

Apply KVL to the meshes

2 3

2

## 

1

15 i i 20 i 10 i 0

     

20 i2

5 i2 i3

20 i2 0

     

2 3

2 3

## 

10 5 ii 15 ii 0

These equations can be written in matrix form

1

2

3

10 35 15 0

0 45 5 0

0 10 10 10

i i i

  

   

 

 

 

   

 

    

    

Solving, e.g. using MATLAB, gives

1 1.25 A, 2 0.125 A, and 3 1.125 A

(14)

Figure P 4.7-12

### Solution:

Express the controlling voltage and current of the dependent sources in terms of the mesh currents:

## 

a 3 1 2

vR ii and ib  i3 i2

Express the current source currents in terms of the mesh currents:

2 s

i  I and i1 i3 B ib B i

3i2

## 

Consequently

1

3 s

### B I

Apply KVL to the supermesh corresponding to the dependent current source

R i1 3A R i3

1i2

R2

i3i2

R i3

1i2

Vs 0

A1

R i3 1

R2

A1

R i3

2R1R2

## 

i3Vs

Organizing these equations into matrix form:

## 

1 s

2 s

3 s

3 2 3 1 2

0 1 0

1 0 1

1 1

i I

B i B I

i V

A R R A R R R

                               

With the given values:

1 1

2 2

3 3

0 1 0 2 0.8276

1 0 4 6 2 A

60 80 50 25 1.7069

(15)

### Solution:

Apply KCL at node a:

2 0

4 2

20 4 4

2 4 0

4 2

b a a

vv v

 

  

 

 

    

 

## References

Related subjects : two-to-one merge