Electrical Engineering Chapter 4 Homework Returning Back The Original Circuit Note That

Techniques of Circuit Analysis

Assessment Problems

AP 4.1 [a] Redraw the circuit, labeling the reference node and the two node voltages:

The two node voltage equations are

Place these equations in standard form:

v1✓1

60 +1

15 +1

5◆+v2✓1

5◆= 15;

4

4–2 CHAPTER 4. Techniques of Circuit Analysis

AP 4.2 Redraw the circuit, choosing the node voltages and reference node as shown:

The two node voltage equations are:

4.5+v1

Place these equations in standard form:

v1✓1+1

8◆+v2✓1

8◆=4.5;

To find the voltage v, first find the current ithrough the series-connected 6 Ω

and 2 Ωresistors:

AP 4.3 [a] Redraw the circuit, choosing the node voltages and reference node as

shown:

The node voltage equations are:

Problems 4–3

The dependent source requires the following constraint equation:

Place these equations in standard form:

v1✓1

6+1

8+1

2◆+v2✓1

2◆+i1(3) = 50

6;

AP 4.4 Redraw the circuit and label the reference node and the node at which the

node voltage equation will be written:

The node voltage equation is

The constraint equation required by the dependent source is

4–4 CHAPTER 4. Techniques of Circuit Analysis

AP 4.5 Redraw the circuit identifying the three node voltages and the reference node:

Note that the dependent voltage source and the node voltages vand v2form a

supernode. The v1node voltage equation is

The constraint equation due to the dependent source is

The constraint equation due to the supernode is

Place this set of equations in standard form:

Problems 4–5

AP 4.6 Redraw the circuit identifying the reference node and the two unknown node

voltages. Note that the right-most node voltage is the sum of the 60 V source

and the dependent source voltage.

The node voltage equation at v1is

Place these two equations in standard form:

AP 4.7 [a] Redraw the circuit identifying the three mesh currents:

The mesh current equations are:

4–6 CHAPTER 4. Techniques of Circuit Analysis

Place these equations in standard form:

31i15i226i3= 80;

AP 4.8 [a] b= 8, n= 6, bn+ 1 = 3.

[b] Redraw the circuit identifying the three mesh currents:

The three mesh-current equations are

25 + 2(i1i2)+5(i1i3)+10 = 0;

Place these four equations in standard form:

7i12i25i3+0vφ= 15;

Problems 4–7

AP 4.9 Redraw the circuit identifying the three mesh currents:

The mesh current equations are:

25 + 6(iaib)+8(iaic) = 0;

The dependent source constraint equation is iφ=ia. We can substitute this

simple expression for iφinto the third mesh equation and place the equations

in standard form:

14ia6ib8ic= 25;

AP 4.10 Redraw the circuit identifying the mesh currents:

Since there is a current source on the perimeter of the i3mesh, we know that

i3=16 A. The remaining two mesh equations are

4–8 CHAPTER 4. Techniques of Circuit Analysis

11i12i2= 30;

AP 4.11 Redraw the circuit and identify the mesh currents:

There are current sources on the perimeters of both the ibmesh and the ic

mesh, so we know that

The remaining mesh current equation is

The dependent source requires the following constraint equation:

Place the mesh current equation and the dependent source equation is

standard form:

7ia2vφ= 55;

AP 4.12 Redraw the circuit and identify the mesh currents:

The 2 A current source is shared by the meshes iaand ib. Thus we combine

these meshes to form a supermesh and write the following equation:

The other mesh current equation is

The supermesh constraint equation is

Place these three equations in standard form:

2ia+4ib4ic= 10;

AP 4.13 Redraw the circuit and identify the reference node and the node voltage v1:

The node voltage equation is

4–10 CHAPTER 4. Techniques of Circuit Analysis

Rearranging and solving,

AP 4.14 Redraw the circuit and identify the mesh currents:

There is a current source on the perimeter of the i3mesh, so i3= 4 A. The

other two mesh current equations are

Substitute the constraint equation into the second mesh equation and place

the resulting two mesh equations in standard form:

Solving,

i1= 9 A; i2=6 A; i3= 4 A; ix=94 = 5 A;

AP 4.15 [a] Redraw the circuit with a helpful voltage and current labeled:

Transform the 120 V source in series with the 20 Ωresistor into a 6 A

resistor. The result is the following circuit:

Combine the three current sources into a single current source, using

KCL, and combine the 20 Ω, 5 Ω, and 6 Ωresistors in parallel. The

Use voltage division in the circuit on the right to calculate vas follows:

[b] Calculate iin the circuit on the right using Ohm’s law:

4–12 CHAPTER 4. Techniques of Circuit Analysis

resistor. Use this fact to calculate the current in the 120 V source, ia:

AP 4.16 To find RTh, replace the 72 V source with a short circuit:

Note that the 5 Ωand 20 Ωresistors are in parallel, with an equivalent

Use node voltage analysis to find vTh. Begin by redrawing the circuit and

labeling the node voltages:

The node voltage equations are

Place these equations in standard form:

v1✓1

5+1

20 +1

8◆+vTh ✓1

8◆=72

5;

Problems 4–13

AP 4.17 We begin by performing a source transformation, turning the parallel

combination of the 15 A source and 8 Ωresistor into a series combination of a

120 V source and an 8 Ωresistor, as shown in the figure on the left. Next,

AP 4.18 Find the Th´evenin equivalent with respect to A, B using source

transformations. To begin, convert the series combination of the 36 V source

and 12 kΩresistor into a parallel combination of a 3 mA source and 12 kΩ

resistor. The resulting circuit is shown below:

Now combine the two parallel current sources and the two parallel resistors to

give a 3 + 18 = 15 mA source in parallel with a 12 kk60 k= 10 kΩresistor.

4–14 CHAPTER 4. Techniques of Circuit Analysis

equivalent and use voltage division to calculate the meter reading vAB:

AP 4.19 Begin by calculating the open circuit voltage, which is also vTh, from the

circuit below:

Summing the currents away from the node labeled vTh We have

Substituting the second equation into the first and solving for vTh yields

Write a KCL equation at the middle node:

Problems 4–15

Substitute the second equation into the first equation:

AP 4.20 Begin by calculating the open circuit voltage, which is also vTh, using the

node voltage method in the circuit below:

The node voltage equations are

The dependent source constraint equation is

Substitute the constraint equation into the node voltage equations and put the

two equations in standard form:

Now use the test source method to calculate the test current and thus RTh.

4–16 CHAPTER 4. Techniques of Circuit Analysis

Write a KCL equation at the rightmost node:

The dependent source constraint equation is

AP 4.21 First find the Th´evenin equivalent circuit. To find vTh, create an open circuit

between nodes a and b and use the node voltage method with the circuit

below:

The node voltage equations are:

The dependent source constraint equation is

Problems 4–17

Now create a short circuit between nodes a and b and use the mesh current

method with the circuit below:

The mesh current equations are

The dependent source constraint equation is

Place these four equations in standard form:

4i14i2+0isc +vφ= 80;

[b] The Th´evenin voltage, vTh = 120 V, splits equally between the Th´evenin

resistance and the load resistance, so

AP 4.22 Sustituting the value R= 3 Ωinto the circuit and identifying three mesh

currents we have the circuit below:

The mesh current equations are:

100 + 4(i1i2)+vφ+ 20 = 0;

The dependent source constraint equation is

Place these four equations in standard form:

4i14i2+0i3+vφ= 80;

Solving, i1= 30 A, i2= 20 A, i3= 20 A, and vφ= 40 V.

Problems 4–19

Problems

P 4.1

[a] 11 branches, 8 branches with resistors, 2 branches with independent

[b] The current is unknown in every branch except the one containing the 8 A

P 4.2 [a] From Problem 4.1(d) there are 8 essential branches where the current is

[b] From Problem 4.1(f), there are 4 essential nodes, so we can apply KCL at

4–20 CHAPTER 4. Techniques of Circuit Analysis

[d] We must avoid using the topmost mesh and the leftmost mesh. Each of

P 4.3 [a] There are eight circuit components, seven resistors and the voltage source.

[b]

There are three essential nodes in this circuit, identified by the boxes. At

[c] Sum the currents at any two of the three essential nodes a, b, and c. Using

nodes a and c we get

[d] There are three meshes in this circuit: one on the left with the

components vs,R1,R2and R3; one on the top right with components R2,

[e] vs+R1i1+R2i2+R3i3= 0;