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Full file at https://fratstock.euThe Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual

Copyright © 2008 John Wiley & Sons, Inc. All rights reserved.

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Problem 2-1

The current through a 33-kΩ resistor is 1.2 mA. Find the voltage across the resistor.

Solution:

v = iR

clear all

R = 33e3;

ii = 1.2e-3;

v = ii*R v =

39.6000e+000

Answer:

v = 39.6 V






Problem 2-2

A 6.2-kΩ resistor dissipates 12 mW. Find the current through the resistor.

Solution:

p = i2R

clear all

format short eng

R = 6.2e3;

p = 12e-3;

ii = sqrt(p/R) ii =

1.3912e-003

Answer:

i = ±1.3912 mA






Problem 2-3

The conductance of a particular resistor is 1 mS. Find the current through the resistor when

connected across a 9 V source.

Solution:

v = iR

clear all

G = 1e-3;

R = 1/G;

v = 9;

ii = v/R ii =

9.0000e-003

Answer:

i = 9 mA






Problem 2-4

In Figure P2-4 the resistor dissipates 25 mW. Find Rx.

RX15 V

P =25 mWX

Solution:

R

vp

2

clear all

p = 25e-3;

v = 15;

R = v^2/p R =

9.0000e+003

Answer:

R = 9 kΩ






Problem 2-5

In Figure P2-5 find Rx and the power delivered to the resistor.

R X

10 mA

100 V

Solution: clear all

v = 100;

ii = 10e-3;

R = v/ii

p = v*ii R =

10.0000e+003

p =

1.0000e+000

Answer:

Rx = 10 kΩ, p = 1 W






Problem 2-6

The i-v characteristic of a nonlinear resistor are v = 75i + 0.2i3.

(a) Calculate v and p for i = ±0.5, ±1, ±2, ±5, and ±10 A.

(b) Find the maximum error in v when the device is treated as a 75-Ω linear resistance on the

range |i| < 0.5 A.

Solution: clear all

format short eng

ii = [-10, -5, -2, -1, -0.5, 0.5, 1, 2, 5, 10];

v = 75*ii + 0.2*ii.^3;

p = v.*ii;

Results = [ii' v' p']

syms i1

v1 = 75*i1+0.2*i1^3;

v2 = 75*i1;

ii1 = -0.5:0.01:0.5;

vv1 = subs(v1,i1,ii1);

vv2 = subs(v2,i1,ii1);

plot(vv1,ii1,'b','LineWidth',3)

hold on

plot(vv2,ii1,'g','LineWidth',1)

grid on

xlabel('Voltage (V)')

ylabel('Current (A)')

legend('Nonlinear','Linear')

MaxError = max(vv1)-max(vv2)

MaxError2 = subs(v1-v2,i1,0.5) Results =

-10.0000e+000 -950.0000e+000 9.5000e+003

-5.0000e+000 -400.0000e+000 2.0000e+003

-2.0000e+000 -151.6000e+000 303.2000e+000

-1.0000e+000 -75.2000e+000 75.2000e+000

-500.0000e-003 -37.5250e+000 18.7625e+000

500.0000e-003 37.5250e+000 18.7625e+000

1.0000e+000 75.2000e+000 75.2000e+000

2.0000e+000 151.6000e+000 303.2000e+000

5.0000e+000 400.0000e+000 2.0000e+003

10.0000e+000 950.0000e+000 9.5000e+003

MaxError =

25.0000e-003

MaxError2 =

25.0000e-003






-40 -30 -20 -10 0 10 20 30 40-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

Voltage (V)

Curr

ent

(A)

Nonlinear

Linear

Answer:

(a)

i (A) v (V) p (W)

-10 -950 9500

-5 -400 2000

-2 -151.6 303.2

-1 -75.2 75.2

-0.5 -37.525 18.7625

0.5 37.525 18.7625

1 75.2 75.2

2 151.6 303.2

5 400 2000

10 950 9500

(b) ERRORMAX = 25 mV






Problem 2-7

A 10-kΩ resistor has a power rating of ⅛W. Find the maximum voltage that can be applied to

the resistor.

Solution: clear all

R = 10e3;

p = 1/8;

v_max = sqrt(p*R) v_max =

35.3553e+000

Answer:

vmax = 35.36 V






Problem 2-8

A certain type of film resistor is available with resistance values between 10 Ω and 100 MΩ.

The maximum ratings for all resistors of this type are 500 V and 1/4 W. Show that the voltage

rating is the controlling limit for R > 1 MΩ, and that the power rating is the controlling limit

when R < 1 MΩ.

Solution: clear all

V = 500;

p = 1/4;

R = V^2/p R =

1.0000e+006

R

vp

2

At R = 1 MΩ, both p and v can take their maximum values and there are no issues. For R >

1 MΩ, with a maximum voltage, the power must be less than 0.25 W, so the voltage rating on a

particular resistor will control the maximum allowable value for the power. For R < 1 MΩ, with

a maximum voltage, the power will be greater than 0.25 W, so the power rating on a particular

resistor will control the maximum allowable value for the voltage.

Answer:

Presented above.






Problem 2-9

Figure P2-9 shows the circuit symbol for a class of two-terminal devices called diodes. The i-v

relationship for a specific pn junction diode is Ae i v 1102 4016

(a) Use this equation to find i and p for v = 0, ±0.1, ±0.2, ±0.4, and ±0.8 V. Use these data to

plot the i-v characteristic of the element.

(b) Is the diode linear or nonlinear, bilateral or nonbilateral, and active or passive?

(c) Use the diode model to predict i and p for v = 5 V. Do you think the model applies to

voltages in this range? Explain.

(d) Repeat (c) for v = –5 V.

Solution: clear all

v = [-0.8, -0.4, -0.2, -0.1, 0 0.1, 0.2, 0.4, 0.8];

ii = 2e-16*(exp(40*v)-1);

p = v.*ii;

Results = [v' ii' p']

plot(v,ii,'b','LineWidth',3)



grid on

v = 5

i5 = 2e-16*(exp(40*v)-1)

v = -5

iNeg5 = 2e-16*(exp(40*v)-1) Results =

-800.0000e-003 -200.0000e-018 160.0000e-018

-400.0000e-003 -200.0000e-018 80.0000e-018

-200.0000e-003 -199.9329e-018 39.9866e-018

-100.0000e-003 -196.3369e-018 19.6337e-018

0.0000e-003 0.0000e-003 0.0000e-003

100.0000e-003 10.7196e-015 1.0720e-015

200.0000e-003 595.9916e-015 119.1983e-015

400.0000e-003 1.7772e-009 710.8888e-012

800.0000e-003 15.7926e-003 12.6341e-003

v =

5.0000e+000

i5 =

144.5195e+069

v =

-5.0000e+000

iNeg5 =

-200.0000e-018

+

-

v

i






-0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8-2

0

2

4

6

8

10

12

14

16x 10

-3

Voltage (V)

Curr

ent

(A)

Answer:

(a)

v (V) i (A) p (W)

-0.8 -2.00E-16 1.60E-16

-0.4 -2.00E-16 8.00E-17

-0.2 -2.00E-16 4.00E-17

-0.1 -1.96E-16 1.96E-17

0 0 0

0.1 1.07E-14 1.07E-15

0.2 5.96E-13 1.19E-13

0.4 1.78E-09 7.11E-10

0.8 1.58E-02 1.26E-02

(b) The plot in Part (a) shows that the device is nonlinear and nonbilateral. The power for the

device is always positive, so it is passive.

(c) For v = 5 V, i = 1.45 1071 A and p = 7.23 1071 W. The model is not valid because the

current and power are too large.

(d) For v = 5 V, i = −2.00 10−16 A and p = 1.00 10−15 W. The model is valid because the

current and power are both essentially zero.






Problem 2-10

In Figure P2-10 i2 = –2 A and i3 = 5 A. Find i1 and i4.

BC

i3

i4

i2i1

A 1 2

3

4

Solution:

Apply KCL at Nodes B and C.

clear all

i2 = -2;

i3 = 5;

i1 = -i2

i4 = i2+i3 i1 =

2.0000e+000

i4 =

3.0000e+000

Answer:

i1 = 2 A and i4 = 3 A.






Problem 2-11

For the circuit in Figure P2-11:

(a) Identify the nodes and at least two loops.

(b) Identify any elements connected in series or in parallel.

(c) Write KCL and KVL connection equations for the circuit.

1

3i

4

31i 4i

BA

C

2

2i

Solution:

There are three nodes and three loops.

Answer:

(a) nodes: A, B, C; loops: 1-2; 2-3-4; 1-3-4

(b) series: 3 and 4; parallel: 1 and 2

(c) KCL: node A: 0321 iii ;

node B: 043 ii ;

node C: 0421 iii

KVL: loop 1-2: 021 vv ;

loop 2-3-4: 0432 vvv ;

loop 1-3-4: 0431 vvv






Problem 2-12

In Figure P2-11, i2 = –10 mA and i4 = 20 mA. Find i1 and i3.

1

3i

4

31i 4i

BA

C

2

2i

Solution:

Use the KCL equations developed in the solution to Problem 2-11.

clear all

i2 = -10e-3;

i4 = 20e-3;

i3 = i4

i1 = -i2-i3 i3 =

20.0000e-003

i1 =

-10.0000e-003

Answer:

i1 = 10 mA and i3 = 20 mA.






Problem 2-13

For the circuit in Figure P2-13:

(a) Identify the nodes and at least three loops in the circuit.

(b) Identify any elements connected in series or in parallel.

(c) Write KCL and KVL connection equations for the circuit.

2

3i

6

32i 6i

BA

CD5

5i5

v

6v

3v

2v

1i1v

4v4

1

4i

Solution:

There are four nodes and at least five loops. There are only three independent KVL equations.

Answer:

(a) nodes: A, B, C, D;

loops: 1-3-2; 2-4-5; 3-6-4; 1-6-5; 2-3-6-5; 1-6-4-2; 1-3-4-5

(b) series: none; parallel: none

(c) KCL: node A: 0432 iii ;

node B: 0631 iii ;

node C: 0521 iii ;

node D: 0654 iii

KVL: loop 1-3-2: 0231 vvv ;

loop 2-4-5: 0542 vvv ;

loop 3-6-4: 0463 vvv






Problem 2-14

In Figure P2-13 v2 = 10 V, v3 = –10 V, and v4 = 3 V. Find v1, v5, and v6.

2

3i

6

32i 6i

BA

CD5

5i5

v

6v

3v

2v

1i1v

4v4

1

4i

Solution:

Use the KVL equations developed in the solution to Problem 2-13.

clear all

v2 = 10;

v3 = -10;

v4 = 3;

v1 = v2 - v3

v5 = v2 - v4

v6 = v4 - v3 v1 =

20.0000e+000

v5 =

7.0000e+000

v6 =

13.0000e+000

Answer:

v1 = 20 V, v5 = 7 V, and v6 = 13 V.






Problem 2-15

The circuit in Figure P2-15 is organized around the three signal lines A, B, and C.

(a) Identify the nodes and at least three loops in the circuit.

(b) Write KCL connection equations for the circuit.

(c) If i1 = –20 mA, i2 = –12 mA, and i3 = 50 mA, find i4, i5, and i6

(d) Show that the circuit in Figure P2-15 is identical to that in Figure P2-13.

3i

6i

B

A

C

D

4i 5i

2i1i

3 61 2 4 5

Solution:

(a) There are four nodes and at least five loops.

(b) KCL: node A: 0432 iii ;

node B: 0631 iii ;

node C: 0521 iii ;

node D: 0654 iii

clear all

i1 = -20e-3;

i2 = -12e-3;

i3 = 50e-3;

i4 = -i2-i3

i5 = -i1-i2

i6 = i3-i1 i4 =

-38.0000e-003

i5 =

32.0000e-003

i6 =

70.0000e-003

Answer:

(a) nodes: A, B, C, D;

loops: 1-3-2; 2-4-5; 3-6-4; 1-6-5; 2-3-6-5; 1-6-4-2; 1-3-4-5

2

3i

6

32i 6i

BA

CD5

5i5

v

6v

3v

2v

1i1v

4v

4

1

4i






(b) KCL: node A: 0432 iii ;

node B: 0631 iii ;

node C: 0521 iii ;

node D: 0654 iii

(c) i4 = −38 mA; i5 = 32 mA; i6 = 70 mA

(d) The circuits have the same nodes, connections, and current directions, so they must be

equivalent.






Problem 2-16

In Figure P2-16, v2 = 10 V, v3 = 10 V, and v4 = 10 V. Find v1 and v5.

v2 v4

v3 v5v1 1

2

3

4

5

Solution:

Apply KVL to the circuit.

clear all

v2 = 10;

v3 = 10;

v4 = 10;

v1 = v2+v3

v5 = v3-v4 v1 =

20.0000e+000

v5 =

0.0000e-003

Answer:

v1 = 20 V and v5 = 0 V.






Problem 2-17

In Figure P2-17 i2 = 10 mA, i3 = –15 mA, and i4 = 5 mA. Find i1 and i5.

A

B C

i

1

i3

i2

i4

i5

1

2

3

4

5

Solution:

Apply KCL to the circuit.

clear all

i2 = 10e-3;

i3 = -15e-3;

i4 = 5e-3;

i1 = i2-i3+i4

i5 = i2-i1 i1 =

30.0000e-003

i5 =

-20.0000e-003

Answer:

i1 = 30 mA and i5 = −20 mA






Problem 2-18

(a) Use the passive sign convention to assign voltage variables consistent with the currents in

Figure P2-17. Write three KVL connection equations using these voltage variables.

(b) If v3 = 0 V, what can be said about the voltages across all the other elements?

A

B C

i

1

i3

i2

i4

i5

1

2

3

4

5

Solution:

(a) Voltage signs:

Elements 1 and 3: plus on bottom and minus on top

Elements 2 and 4: plus on top and minus on bottom

Element 5: plus on left and minus on right

Write the KVL equations for the loops formed by 1-2, 3-4, and 2-4-5

loop 1-2: 021 vv

loop 3-4: 043 vv

loop 2-4-5: 0542 vvv

(b) If v3 = 0 V, then v4 = 0 V. In addition, v2 = v5 and v1 = v5.

Answer:

Presented above.






Problem 2-19

The KCL equations for a three-node circuit are:

Node A – i1 + i2 – i4 = 0

Node B – i2 – i3 + i5 = 0

Node C i1 + i3 + i4 – i5 = 0

Draw the circuit diagram and indicate the reference directions for the element currents.

Answer:

R1 R2

R3

R4

R5

B

A

C

i1 i2 i4

i3

i5






Problem 2-20

Find vx and ix in Figure P2-20.

vX2 mA

47 kΩ

iX

33 kΩ

Solution:

Use KCL to find the current and Ohm's Law to find the voltage.

clear all

format short eng

is = 2e-3;

ix = -is

vx = 47e3*ix ix =

-2.0000e-003

vx =

-94.0000e+000

Answer:

vx = 94 V and ix = 2 mA.






Problem 2-21

Find vx and ix in Figure P2-21.

Rest of

the

circuit

v10 Ω

i

X

X

5 Ω

4 Ω

½ A

Solution:

Find the voltage across the 10-Ω resistor using Ohm's Law. The 10-Ω and 5-Ω resistors are in

parallel, so they have the same voltage. Find the current through the 5-Ω resistor. The current

through the 4-Ω resistor is the sum of the currents through the other two resistors. Find the

voltage across the 4-Ω resistor. Then vx is the sum of the voltages across the 4-Ω and 10-Ω

resistors.

clear all

i10 = 1/2;

v10 = 10*i10;

v5 = v10;

i5 = v5/5;

ix = i5

i4 = i10+i5;

v4 = 4*i4;

vx = v4+v10 ix =

1.0000e+000

vx =

11.0000e+000

Answer:

vx = 11 V and ix = 1 A






Problem 2-22

In Figure P 2-22:

(a) Assign a voltage and current variable to every element.

(b) Use KVL to find the voltage across each resistor.

(c) Use Ohm's law to find the current through each resistor.

(d) Use KCL to find the current through each voltage source.

100 Ω

5V 10V

50 Ω

5V

100 Ω

v4 v5

6v

A CB

6i

5i4i

2i1i 3i

Solution:

(a) For each of the three resistors, the voltage positive sign is on the left and the negative sign is

on the right. The current flows from left to right through each element.

Element 1: 50-Ω resistor.

Element 2: left 100-Ω resistor.

Element 3: right 100-Ω resistor.

The left voltage source is vS1, with iS1 flowing down.

The center voltage source is vS2, with iS2 flowing down.

The right voltage source is vS3, with iS3 flowing down.

(b) KVL equations:

0S311S vvv

02S21S vvv

03S32S vvv

clear all

format short eng

vs1 = 5;

vs2 = 10;

vs3 = 5;

v1 = vs1-vs3

v2 = vs1-vs2

v3 = vs2-vs3






v1 =

0.0000e-003

v2 =

-5.0000e+000

v3 =

5.0000e+000

(c) v = iR

i1 = v1/50

i2 = v2/100

i3 = v3/100 i1 =

0.0000e-003

i2 =

-50.0000e-003

i3 =

50.0000e-003

(d) KCL equations

01S21 iii

0S232 iii

0S331 iii

is1 = -i1-i2

is2 = i2-i3

is3 = i1+i3 is1 =

50.0000e-003

is2 =

-100.0000e-003

is3 =

50.0000e-003

Answer:

(a) Presented above.

(b) v1 = 0 V, v2 = 5 V, and v3 = 5 V

(c) il = 0 mA, i2 = 50 mA, and i3 = 50 mA

(d) iS1 = 50 mA, iS2 = 100 mA, and iS3 = 50 mA






Problem 2-23

Find the power dissipated in the 1.5 kΩ resistor in Figure P2-23.

5 mA

PL

500 Ω

1 kΩ 1.5 kΩ

Solution:

Label the elements.

Element 1: 1-kΩ resistor with current flowing down

Element 2: 500-Ω resistor with current flowing to the right

Element 3: 1.5-kΩ resistor with current flowing down

Write KCL, KVL, and Ohm's Law equations:

i1 + i2 5 mA = 0

i2 + i3 = 0

v1 + v2 + v3 = 0

v1 = 1000i1

v2 = 500i2

v3 = 1500i3

Solve the equations for v3 and i3 and then compute the power p3 = v3 i3.

clear all

format short eng

Eqn1 = 'i1+i2-5e-3';

Eqn2 = '-i2+i3';

Eqn3 = '-v1+v2+v3';

Eqn4 = 'v1-1000*i1';

Eqn5 = 'v2-500*i2';

Eqn6 = 'v3-1500*i3';

Soln = solve(Eqn1,Eqn2,Eqn3,Eqn4,Eqn5,Eqn6,'v1','v2','v3','i1','i2','i3');

v3 = Soln.v3;

i3 = Soln.i3;

p3 = double(v3*i3) p3 =

4.1667e-003

Answer:

PL = 4.167 mW






vx

5 k4 mA

2 k

6 mA

20 V

12 V

ix

Rest of the Circuit

8 k

vA

Problem 2-24

Figure P2-24 shows a subcircuit connected to the rest of the circuit at

four points.

(a) Use element and connection constraints to find vx and ix.

(b) Show that the sum of the currents into the rest of the circuit

is zero.

(c) Find the voltage vA with respect to the ground in the

circuit.

Solution:

(a) Label the elements.

Element 1: 5-kΩ resistor with current flowing from left to right

Element 2: 2-kΩ resistor with positive voltage sign on the bottom

Use Ohm's Law to compute i1.

Use KCL at the center node to find ix.

Use Ohm's Law to find vx.

clear all

v1 = 20;

R1 = 5e3;

i1 = v1/R1;

ix = i1+4e-3 - 6e-3

Rx = 8e3;

vx = ix*Rx ix =

2.0000e-003

vx =

16.0000e+000

(b) The sum of the currents into the rest of the circuit is i1 + i2 4 mA + ix.

i2 = 6e-3;

Current_Out = -i1+i2-4e-3+ix Current_Out =

0.0000e-003

(c) vA = 12 + vx v2

R2 = 2e3;

v2 = i2*R2;

vA = 12+vx-v2 vA =

16.0000e+000

Answer:

(a) vx = 16 V and ix = 2 mA.






(b) iOUT = 0 mA.

(c) vA = 16 V.






Problem 2-25

In Figure P2-25 ix = –0.5 mA. Find the value of R.

R

10 k

4 V 15 V

i x

Rest of the Circuit

10 k

Solution:

Label the circuit elements.

Element 1: Resistor R with current flowing down.

Element 2: 10-kΩ resistor with current flowing from right to left

Compute the voltage vx using Ohm's Law.

Compute the voltage v1 using KVL.

Compute the voltage v2 using KVL.

Compute the current i2 using Ohm's Law.

Compute the current i1 using KCL.

Compute the resistance R1 = R using Ohm's Law.

clear all

Rx = 10e3;

R2 = 10e3;

ix = -0.5e-3;

vx = ix*Rx;

v1 = 4-vx;

v2 = 15-v1;

i2 = v2/R2;

i1 = ix+i2;

R1 = v1/i1;

R = R1 R =

90.0000e+003

Answer:

R = 90 kΩ






Problem 2-26

Figure P2-26 shows a resistor with one terminal connected to ground and the other connected to

an arrow. The arrow symbol is used to indicate a connection to one terminal of a voltage source

whose other terminal is connected to ground. The label next to the arrow indicates the source

voltage at the ungrounded terminal. Find the voltage across, current through, and power

dissipated in the resistor.

i100 kΩ

v

-15 V

Solution:

The voltage across the resistor is the voltage on the right side (15 V) minus the voltage on the

left side (0 V), so vx = 15 0 = 15 V. Using Ohm's Law, the current is ix = vx / Rx = 150 A.

The power px = vx ix = 2.25 mW.

clear all

vx = -15-0

ix = vx/100e3

px = vx*ix vx =

-15.0000e+000

ix =

-150.0000e-006

px =

2.2500e-003

Answer:

vx = 15 V, ix = 150 A, and px = 2.25 mW.











Problem 2-27

Find the equivalent resistance REQ in Figure P2-27.

REQ

75

300

100

200

Solution:

Combine the resistors working from right to left in the circuit.

clear all

R1 = 75;

R2 = 300;

R3 = 100;

R4 = 200;

% Combine R3 and R4 in series

R34 = R3+R4;

% Combine R2 in parallel with the series combination of R3 and R4

R234 = 1/(1/R2 + 1/R34);

% Combine R1 in series with the other combination

R1234 = R1 + R234;

Req = R1234 Req =

225.0000e+000

Answer:

REQ = 225 Ω






Problem 2-28

Find the equivalent resistance REQ in Figure P2-28.

REQ3.3 k

1.5 k2.2 k4.7 k

Solution:

Combine the resistors working from right to left in the circuit.

clear all

R1 = 4.7e3;

R2 = 3.3e3;

R3 = 1.5e3;

R4 = 2.2e3;

R34 = 1/(1/R3 + 1/R4);

R234 = R2 + R34;

R1234 = 1/(1/R1 + 1/R234);

Req = R1234 Req =

2.2157e+003

Answer:

REQ = 2.2157 kΩ






Problem 2-29

Find REQ in Figure P2-29 when the switch is open. Repeat when the switch is closed.

REQ

200

100

50 50

Solution:

With the switch open, the 200-Ω and 50-Ω resistors are in parallel. With the switch closed, the

200-Ω and 50-Ω resistors are in parallel with a short circuit, so their equivalent resistance is zero.

clear all

R1 = 100;

R2 = 200;

R3 = 50;

R4 = 50;

disp('Switch Open')

Req_open = R1 + 1/(1/R2 + 1/R3) + R4

disp('Switch Closed')

Req_closed = R1 + 0 + R4 Switch Open

Req_open =

190.0000e+000

Switch Closed

Req_closed =

150.0000e+000

Answer:

Switch open: REQ = 190 Ω.

Switch closed: REQ = 150 Ω.






Problem 2-30

Show how the circuit in Figure P2-30 could be connected to achieve a resistance of 100 Ω, 200

Ω, 150 Ω, 50 Ω, 25 Ω, 33.3 Ω, and 133.3 Ω.

A

D

100

B

C

50

100

Solution:

The required resistor combinations are described below.

100 Ω: A single 100-Ω resistor.

200 Ω: Two 100-Ω resistors in series.

150 Ω: A 100-Ω resistor in series with a 50-Ω resistor.

50 Ω: A single 50-Ω resistor.

25 Ω: A parallel combination of two 100-Ω resistors and a 50-Ω resistor

33.3 Ω: A parallel combination of a 100-Ω resistor and a 50-Ω resistor

133.3 Ω: A 100-Ω resistor in series with a parallel combination of a 100-Ω resistor and a 50-Ω

resistor.

Answer:

The following table summarizes the required connections. A plus sign indicates the nodes are

connected together at one of the terminals.

Resistance (Ω) Terminal 1 Terminal 2

100 A D

200 A B

150 A C

50 C D

25 A+B+C D

33.3 B+C D






133.3 A B+C






Problem 2-31

In Figure P2-31 find the equivalent resistance between terminals A-B, A-C, A-D, B-C, B-D, and

C-D.

B C

D

40 Ω

40 Ω 30 Ω

10 Ω80 Ω60 Ω

RC-D is shown.

Solution:

For each pair of end terminals, combine the appropriate resistors in series and parallel to get the

equivalent resistance.

clear all

Rab = 1/(1/40 + 1/(40+80)) + 60

Rac = 1/(1/40 + 1/(40+80)) + 30

Rad = 1/(1/40 + 1/(40+80)) + 10

Rbc = 60 + 1/(1/(40+40) + 1/80) + 30

Rbd = 60 + 1/(1/(40+40) + 1/80) + 10

Rcd = 30 + 0 + 10 Rab =

90.0000e+000

Rac =

60.0000e+000

Rad =

40.0000e+000

Rbc =

130.0000e+000

Rbd =

110.0000e+000

Rcd =

40.0000e+000

Answer:

RAB = 90 Ω

RAC = 60 Ω

RAD = 40 Ω

RBC = 130 Ω

RBD = 110 Ω

RCD = 40 Ω











Problem 2-32

Select a value of RL in Figure P2-32 so that REQ = 25 kΩ. Repeat for REQ = 20 kΩ.

REQ

10 k

10 k

10 k

RL

Solution:

Find an expression for RL in terms of REQ and then solve for RL for both values of REQ.

clear all

syms RL Req positive

Eqn = 'Req - (10e3 + 1/(1/10e3 + 1/RL) + 10e3)';

Soln = solve(Eqn,'RL')

Req_values = [25e3 20e3];

RL_values = subs(Soln,Req,Req_values) Soln =

-10000.*(Req-20000.)/(Req-30000.)

RL_values =

10.0000e+003 0.0000e-003

Answer:

For REQ = 25 kΩ, RL = 10 kΩ.

For REQ = 20 kΩ, RL = 0 kΩ.






Problem 2-33

Using no more than four 1-kΩ resistors, show how the following equivalent resistors can be

constructed: 2 kΩ, 500 Ω, 1.5 kΩ, 333 Ω, 250 Ω, and 400 Ω.

Solution:

R = 2 kΩ: two resistors in series

R = 500 Ω: two resistors in parallel

R = 1.5 kΩ: one resistor in series with a parallel combination of two resistors

R = 333 Ω: three resistors in parallel;

R = 250 Ω: four resistors in parallel;

R = 400 Ω: two resistors in series in parallel with two resistors in parallel

clear all

R = 1000;

R2000 = R + R

R500 = 1/(1/R + 1/R)

R1500 = R + 1/(1/R + 1/R)

R333 = 1/(1/R + 1/R + 1/R)

R250 = 1/(1/R + 1/R + 1/R + 1/R)

R400 = 1/(1/(R+R) + 1/R + 1/R) R2000 =

2.0000e+003

R500 =

500.0000e+000

R1500 =

1.5000e+003

R333 =

333.3333e+000

R250 =

250.0000e+000

R400 =

400.0000e+000

Answer:

Presented above.






Problem 2-34

Find the equivalent practical voltage source at terminals A and B in Figure P2-34

A

5 A

5 Ω

10 Ω

B

Solution:

A current source in series with a resistor is equivalent to just the current source, so we can

remove the 5-Ω resistor. That leaves a 5-A current source in parallel with a 10-Ω resistor. The

current source and parallel resistor can be converted into a voltage source in series with the same

resistor. The value for the voltage source follows Ohm's Law, so vS = (5 A)(10 Ω) = 50 V.

5

105Adc

105Adc

50Vdc

10

Answer:

vS = 50 V and RS = 10 Ω.






Problem 2-35

In Figure P2-35 the i-v characteristic of network N is v + 50i = 5 V. Find the equivalent practical

current source for the network.

N

i

v

A

B

Solution:

When the circuit is open between nodes A and B, there is no current, i = 0 A, and the voltage

must be v = 5 V. When a short is placed between nodes A and B, the voltage is zero, v = 0 V,

and the current is i = 100 mA. The corresponding practical current source will have a current iS

= 100 mA and a parallel resistance R = (5 V)/(100 mA) = 50 Ω.

50100mA

Answer:

iS = 100 mA and RS = 50 Ω.






Problem 2-36

Select the value of Rx in Figure P2-36 so that REQ = 60 kΩ.

REQ

47 k

22 k

RX

10 k

Solution:

Find an expression for REQ in terms of Rx and then solve for Rx in terms of REQ.

clear all

syms Rx Req positive

Eqn = 'Req - (47e3 + 1/(1/22e3 + 1/(Rx+10e3)))';

Soln = solve(Eqn,'Rx');

Rx_value = subs(Soln,Req,60e3)

Check_Req = 47e3 + 1/(1/22e3 + 1/(Rx_value+10e3)) Rx_value =

21.7778e+003

Check_Req =

60.0000e+003

Answer:

Rx = 21.78 kΩ.






Problem 2-37

Two 10-kΩ potentiometers (a variable resistor whose value between the two ends is 10 kΩ and

between one end and the wiper – the third terminal – can range from 0 Ω to 10 kΩ) are

connected as shown in Figure P2-37. What is the range of REQ?

REQ

10 k

10 k

Solution:

At the limits of their settings, the two potentiometers are either in series or parallel. These

represent the maximum and minimum equivalent resistances that the combination can take.

clear all

R = 10e3;

Rmax = R + R

Rmin = 1/(1/R + 1/R) Rmax =

20.0000e+003

Rmin =

5.0000e+003

Answer:

5 kΩ REQ 20 kΩ






Problem 2-38

Select the value of R in Figure P2-38 so that RA-B = RL.

A

B

R R

4R R L

RA-B

Solution:

Find an expression for RA-B in terms of R and RL. Set RA-B equal to RL. Solve for R in terms of

RL and choose the positive solution for the resistance.

clear all

syms R Rab RL positive

Eqn = 'RL - (R + 1/(1/4/R + 1/(R+RL)))';

Soln = solve(Eqn,'R')

R = RL/3;

Check_Rab = R + 1/(1/4/R + 1/(R+RL)) Soln =

-1/3*RL

1/3*RL

Check_Rab =

RL

Answer:

R = RL/3.






Problem 2-39

What is the range of REQ in Figure P2-39?

4 kΩ

REQ

1 kΩ

4 kΩ

Solution:

The potentiometer varies between 0 and 4 kΩ. Find REQ with the potentiometer to its maximum

and minimum values.

clear all

Rp = [0 4e3];

Req = 1e3 + 1./(1./Rp + 1/4e3) Req =

1.0000e+003 3.0000e+003

Answer:

1 kΩ REQ 3 kΩ






Problem 2-40

Find the equivalent resistance between terminals A and B in Figure P2-40

A BR R R

Solution:

Place a voltage source between Nodes A and B and redraw the circuit as the equivalent circuit

shown below

Vs

R1 R2

R3

0

Use KVL to show that each resistor experiences a voltage drop equal to the voltage at the source:

for R1, the drop is from left to right, for R2, the drop is from right to left and for R3, and the drop

is from top to bottom. Therefore, each resistor has current flowing in the direction of the voltage

drop equal to vS/R. Applying KCL at the node above the voltage source, the current flowing out

of the voltage source is iS = vS/R + vS/R + vS/R = 3 vS/R. The equivalent resistance is the ratio of

vS to iS, REQ = vS/iS = R/3.

Alternatively, rearrangement of the circuit shows that the three resistors are connected in

parallel, which yields the same final result.

Answer:

REQ = vS/iS = R/3.






Problem 2-41

Use voltage division in Figure P2-41 to obtain an expression for vL in terms of R, RL, and vS.

vLR

R

RLv

S

Solution:

Find an equivalent resistance for the two resistors in parallel. The voltage vL appears across this

parallel combination, so use voltage division with the equivalent resistance and the resistor in

series with the source.

clear all

syms vs R RL Req vL

Req = 1/(1/R + 1/RL);

vL = simple(Req*vs/(Req+R)) vL =

vs*RL/(2*RL+R)

Answer:

L

SLL

2RR

vRv






Problem 2-42

Use current division in Figure P2-42 to find vL if R =10 Ω, RL = 20 Ω, and iS = 2.5 mA.

vLR

R

RLiS

Solution:

Use current division to find the current through the series combination of R and RL and then use

Ohm's Law to compute the voltage vL.

clear all

R = 10;

RL = 20;

is = 2.5e-3;

Req = R + RL;

iL = (1/Req)*is/(1/Req + 1/R);

vL = iL*RL vL =

12.5000e-003

Answer:

vL = 12.5 mV.






Problem 2-43

Find ix and iy in Figure P2-43.

20

5

6 30 i y2 A

ix

Solution:

Combine the 20-Ω and 5-Ω resistors in parallel and then combine that result in series with the 6-

Ω resistor. Use current division to find ix. Use KCL to find the current entering the parallel

combination of the 20-Ω and 5-Ω resistors and then use current division again to find the current

through the 5-Ω resistor, which is iy.

clear all

Req1 = 1/(1/20 + 1/5);

Req2 = Req1 + 6;

is = 2;

ix = (1/30)*is/(1/30 + 1/Req2)

iReq1 = is - ix;

iy = (1/5)*iReq1/(1/5 + 1/20) ix =

500.0000e-003

iy =

1.2000e+000

Answer:

ix = 500 mA and iy = 1.2 A.






Problem 2-44

The 1-kΩ load in Figure P2-44 needs 5 V across it to operate correctly. Where should the wiper

on the potentiometer be set (RX) to obtain the desired output voltage?

5 V

5 kΩ24 V

1 kΩ}RX

Solution:

Redraw the circuit as shown below to clearly identify the potentiometer settings and the resistor

values on each side of the wiper.

5k - Rx

Rx

1k

24Vdc

+5 V

-

Find an equivalent resistance for the parallel combination. Use voltage division to find an

expression for the voltage across the parallel combination in terms of RX. Solve for RX when the

voltage across the parallel combination is 5 V. Choose the positive solution.

clear all

syms Rx RL Req vL

Eqn1 = 'Req - 1/(1/Rx + 1/1e3)';

Eqn2 = '5 - (Req*24/(Req + 5e3 - Rx))';

Soln = solve(Eqn1,Eqn2,'Req','Rx');

Rx = double(Soln.Rx)

Rx = max(Rx);

Req_Check = 1/(1/Rx + 1/1e3);

V_Check = Req_Check*24/(Req_Check + 5e3 - Rx) Rx =

2.3383e+003

-2.1383e+003

V_Check =

5.0000e+000

Answer:

RX = 2.3383 kΩ.











Problem 2-45

Find the range of values of vO in Figure P2-45.

100 Ω

vO

100 Ω

100 Ω24 V

Solution:

Find an expression for vO in terms of the value of the potentiometer and then evaluate the

expression for the maximum and minimum values of the potentiometer.

clear all

Rp = [0,100];

Req = 1./(1./Rp + 1/100);

vs = 24;

vo = Req*vs./(Req + 100) vo =

0.0000e-003 8.0000e+000

Answer:

0 V vO 8 V






Problem 2-46

Figure P2-46 shows a voltage bridge circuit, that is, two voltage dividers in parallel with a source

vS. One resistor RX is variable. The goal is often to “balance” the bridge by making vX = 0 V.

Derive an expression for RX in terms of the other resistors for when the bridge is balanced.

VS

RB

vX

RA RC

RX

Solution:

Let the node between resistors RA and RB have a voltage v1 and let the node between resistors RC

and RX have a voltage v2. The goal is to make v1 equal v2. Use voltage division to derive

expressions for the two voltages. Set the expressions equal and solve for RX.

BA

SB1

RR

vRv

and

XC

SX2

RR

vRv

clear all

syms vs Ra Rb Rc Rx

Eqn = 'Rb*vs/(Ra+Rb) - Rx*vs/(Rc+Rx)';

Soln = solve(Eqn,'Rx') Soln =

Rb*Rc/Ra

Answer:

A

CBX

R

RRR






Problem 2-47

Ideally, a voltmeter has infinite internal resistance and can be placed across any device to read

the voltage without affecting the result. A particular digital multimeter (DMM), a common

laboratory tool, is connected across the circuit shown in Figure P2-47. The expected voltage was

5.73 V. However, the DMM reads 3.81 V. The large, but finite, internal resistance of the DMM

was “loading” the circuit and causing a wrong measurement to be made. Find the value of the

internal resistance of this DMM.

DMM4.7 MΩ

6.3 MΩ10 V RM

Solution:

The DMM is in parallel with the 6.3-MΩ resistor, so they share the same voltage. Calculate the

current through the 6.3-MΩ resistor. The remaining voltage must drop across the 4.7-MΩ

resistor, so use that voltage and Ohm's Law to calculate the current through the 4.7-MΩ resistor.

Use KCL to find the current through the DMM and then Ohm's Law to find the internal

resistance of the DMM.

clear all

R1 = 4.7e6;

R2 = 6.3e6;

vs = 10;

v2 = 3.81;

i2 = v2/R2;

v1 = vs - v2;

i1 = v1/R1;

iDMM = i1 - i2;

RDMM = v2/iDMM RDMM =

5.3492e+006

Answer:

RDMM = 5.3492 MΩ.






Problem 2-48

Select values for R1, R2, and R3 in Figure P 2-48 so the voltage divider produces the two output

voltages shown.

10 V

15 V

R1

3 VR3

R2

Solution:

If we select the resistors such that the total equivalent resistance seen by the source is 15 kΩ,

then 1 mA will flow through each resistor. We can then see that R1 = 5kΩ, R2 = 7 kΩ, and R3 = 3

kΩ will solve the problem.

clear all

vs = 15;

R1 = 5e3;

R2 = 7e3;

R3 = 3e3;

v23 = (R2+R3)*vs/(R1+R2+R3)

v3 = R3*vs/(R1+R2+R3) v23 =

10.0000e+000

v3 =

3.0000e+000

Answer:

R1 = 5kΩ, R2 = 7 kΩ, and R3 = 3 kΩ. Other solutions are possible.






Problem 2-49

Select a value of Rx in Figure P2-49 so that vL = 3 V.

12 V vL

1 k R X

1 k 1 k

Solution:

The voltage vL appears across a parallel combination of two 1-kΩ resistors, which have an

equivalent resistance of 500 Ω. The current through the equivalent resistor must be 6 mA. That

same current must flow through the series combination of the 1-kΩ resistor and Rx and the pair

must drop a total of 9 V. The 1-kΩ resistor with 6 mA of current drops 6 V, so Rx must drop 3

V. To carry 6 mA of current and drop 3 V, Rx = 500 Ω.

clear all

R = 1e3;

vs = 12;

Req = 1/(1/R+1/R);

vL = 3;

iReq = vL/Req;

is = iReq;

ix = is;

vx = vs-is*R-vL;

Rx = vx/ix Rx =

500.0000e+000

Answer:

Rx = 500 Ω.






Problem 2-50

Select a value of Rx in Figure P2-50 so that vL = 2 V. Repeat for 4 V and 6 V. Caution: Rx must

be positive.

12 V vL50 Ω

100

R X

Solution:

clear all

vL = [2 4 6];

vs = 12;

R1 = 100;

RL = 50;

v1 = vs-vL;

i1 = v1/R1;

iL = vL/RL;

ix = i1-iL;

vx = vL;

Rx = vx./ix;

Rx = Rx' Rx =

33.3333e+000

Inf

-100.0000e+000

Answer:

For vL = 2 V, select Rx = 33.33 Ω.

For vL = 4 V, make Rx an open circuit.

For vL = 6 V, there is no positive value for Rx that will meet this specification.






Problem 2-51

Use circuit reduction to find vx and ix in Figure P2-51.

vx3R

2R

RiS

ix

Solution:

Find the equivalent resistance of the entire circuit. Use Ohm's Law first to compute vx and then

again to compute ix.

clear all

syms R is ix vx Req1 Req2

Req1 = 2*R + R;

Req2 = 1/(1/3/R+1/Req1);

vx = is*Req2

ix = vx/3/R vx =

3/2*is*R

ix =

1/2*is

Answer:

vx = 1.5 R iS, ix = 0.5 iS






Problem 2-52


vx 2R R

ix

2R

2R

vS

R

Solution:

Find the equivalent resistance of the entire circuit and compute the current through the source.

Use the source current and the equivalent resistance seen by vx to compute vx. Use the source

current and current division to calculate ix.

clear all

syms R vs vx ix Req1 Req2

Req1 = 1/(1/R + 1/R/2);

Req2 = 2*R + 2*Req1;

is = vs/Req2;

vx = Req1*is

ix = (1/R/2)*(-is)/(1/R + 1/R/2) vx =

1/5*vs

ix =

-1/10*vs/R

Answer:

5

Sx

vv

R

vi

10

Sx






Problem 2-53

Use circuit reduction to find vx and the power provided by the source in Figure P2-53.

10 Ωv

20 Ω 20 Ω

X

¾ A

20 Ω 10 Ω10 Ω

Solution:

First, find the power provided by the source by combining all of the resistors to determine an

overall equivalent resistance. We can then calculate p = i2R. To find vx, combine the resistors to

the right of vx to determine their equivalent resistance and then use a source transformation to

convert the problem into a voltage division problem. Note: You cannot find the power provide

by the original source by finding the power provided by an equivalent transformed source.

clear all

Req1 = 1/(1/10 + 1/10);

Req2 = 20+Req1;

Req3 = 1/(1/10 + 1/Req2);

Req4 = 20+Req3;

Req = 1/(1/20 + 1/Req4);

is = 3/4;

Rs = 20;

ps = is^2*Req

% Source transformation

vs = is*Rs;

vx = Req3*vs/(Rs + 20 + Req3) ps =

6.4773e+000

vx =

2.2727e+000

Answer:

vx = 2.2727 V and pS = 6.4773 W.






Problem 2-54


8 kΩ

iX

12 kΩ12 V

6 kΩ

4 kΩ vX

Solution:

Perform a source transformation and then use current division to find ix and the current

associated with vx. Use Ohm's Law to find vx.

clear all

vs = 12;

Rs = 6e3;

is = vs/Rs;

ix = (1/12e3)*is/(1/12e3 + 1/6e3 + 1/12e3)

ivx = (1/12e3)*is/(1/12e3 + 1/6e3 + 1/12e3);

vx = ivx*4e3 ix =

500.0000e-006

vx =

2.0000e+000

Answer:

vx = 2 V and ix = 500 A.






Problem 2-55

Use source transformation to find ix in Figure P2-55.

50 Ω

30 Ω 200 mA10 V

ix

Solution:

Perform a source transformation on the current source and the 30-Ω resistor and then use Ohm's

Law to find ix.

clear all

vs1 = 10;

Rs1 = 50;

is2 = 200e-3;

Rs2 = 30;

vs2 = is2*Rs2;

ix = (vs1-vs2)/(Rs1+Rs2) ix =

50.0000e-003

Answer:

ix = 50 mA.






Problem 2-56

Select a value for RX so that ix = 0 A in Figure P2-56.

30 ΩRX

20 Ω-24 ViX=0

12 V

Solution:

Perform source transformations on both voltage sources to get the following equivalent circuit:

24/Rx400mAdc

Rx 20 30

All three resistors share the same voltage because they are in parallel. If ix = 0 A, then there is

no voltage drop across any of the resistors and no current flowing through them. All of the

current from the right source must flow through the left source, so we can compute RX.

clear all

vs1 = 24;

vs2 = 12;

Rs2 = 30;

is2 = vs2/Rs2

Rx = vs1/is2 is2 =

400.0000e-003

Rx =

60.0000e+000

Answer:

RX = 60 Ω.






Problem 2-57

Use source transformations in Figure P2-57 to relate vO to v1, v2, and v3.

R

v

v

1

R

v2

R

v3

O

Solution:

Convert each voltage source and resistor pair into a current source in parallel with a resistor to

get the following circuit:

We can then add the three current

sources, since they are in parallel, and

combine the three parallel resistors to get

the following circuit:

The output voltage is then the product of the equivalent current source

and the equivalent resistance

33

321321O

vvvR

R

vvvv

clear all

syms v1 v2 v3 R i1 i2 i3 Req ieq vo

i1 = v1/R;

i2 = v2/R;

i3 = v3/R;

Req = 1/(1/R + 1/R + 1/R);

ieq = i1 + i2 + i3;

vo = simple(ieq*Req) vo =

1/3*v1+1/3*v2+1/3*v3

Answer:

3

321O

vvvv

v1/RR R R

v2/R v3/R

(v1+v2+v3) /RR/3






Problem 2-58

The current through RL in Figure P2-58 is 40 mA. Use source transformations to find RL.

RL

50 Ω100 Ω

100 Ω10 V RL

50 Ω

100 Ω100 m A 100 Ω

50 Ω

RL

50 Ω50 Ω

5 V

Solution:

Perform a source transformation and then combine the two 100-Ω resistors in parallel to get a

50-Ω equivalent resistor. The resulting current source has a value of 100 mA. Since 40 mA

flows through RL, the remaining 60 mA must flow through the equivalent 50-Ω resistor and it

causes a voltage drop of 3 V. The series path with the original 50-Ω resistor and RL must also

drop a total of 3 V, so the total equivalent resistance of the path must be (3 V)/(40 mA) = 75 Ω.

Therefore RL = 25 Ω.

clear all

iL = 40e-3;

vs = 10;

Rs = 100;

is = vs/Rs;

Req = 1/(1/100 + 1/100);

iReq = is - iL;

vReq = iReq*Req;

v50L = vReq;

R50L = v50L/iL;

RL = R50L - 50 RL =

25.0000e+000

Answer:

RL = 25 Ω.






Problem 2-59

Select RX so that 25 V are across it in Figure P2-59.

500 Ω RX

200 Ω100 V

800 Ω

1 kΩ 1 kΩ

25 V

Solution:

Reduce the circuit on both sides of RX. To the right of RX, the equivalent resistance is 500 Ω. To

the left of RX, after a source transformations, a parallel resistor combination, and another source

transformation, we have a voltage source vS =

66.7 V in series with a 333-Ω resistor. The

resulting equivalent circuit is shown below:

The two known resistors can be combined in series to reduce the circuit further:

If RX drops 25 V, then the 833-Ω resistor must drop (66.7 25) V

= 41.7 V. The current through the circuit must be 41.7/833 = 50

mA. We can then calculate RX = (25 V)/(50 mA) = 500 Ω.

clear all

vx = 25;

Req_right = 1/(1/1000 + 1/(800+200));

vs = 100;

Rs = 500;

is = vs/Rs;

Req_left = 1/(1/500 + 1/1000);

vs2 = is*Req_left;

Req = Req_left + Req_right;

vReq = vs2 - vx;

iReq = vReq/Req;

ix = iReq;

Rx = vx/ix Rx =

500.0000e+000

Answer:

RX = 500 Ω.

333 Rx

50066.7Vdc

Rx

83366.7Vdc











Problem 2-60

The box in the circuit in Figure P2-60 is a resistor whose value can be anywhere between 8 kΩ

and 80 kΩ. Use circuit reduction to find the range of values of vx.

10 kΩ

50 V 10 kΩ 10 kΩ vx

Solution:

The solution is presented in the commented MATLAB script below.

clear all

R = 10e3;

% Source transformation to a current source

vs = 50;

Rs = 10e3;

is = vs/Rs

% Combine parallel resistors

Req = 1/(1/Rs + 1/R)

% Source transformation back to a voltage source

vs2 = is*Req

% Unknown resistor range

Rp = [8e3 80e3];

% Voltage division to calculate vx

vx = R*vs2./(R+Rp+Req) is =

5.0000e-003

Req =

5.0000e+003

vs2 =

25.0000e+000

vx =

10.8696e+000 2.6316e+000

Answer:

2.6316 V vx 10.8696 V






Problem 2-61

A circuit is found to have the following element and connection equations:

v1 = 24 V

v2 = 8k i2

v3 = 5k i3

v4 = 4k i4

v5 = 16k i5

–v1 + v2 + v3 = 0

–v3 + v4 + v5 = 0

i1 + i2 = 0

–i2 + i3 +i4 = 0

–i4 + i5 = 0

Use MATLAB to solve for all of the unknown voltages and currents associated with this circuit.

Sketch one possible schematic that matches the given equations.

Solution:

Write the equations in matrix form using a vector of unknowns as:

x = [v1, v2, v3, v4, v5, i1, i2, i3, i4, i5].

clear all

A = [1 0 0 0 0 0 0 0 0 0;

0 1 0 0 0 0 -8000 0 0 0;

0 0 1 0 0 0 0 -5000 0 0;

0 0 0 1 0 0 0 0 -4000 0;

0 0 0 0 1 0 0 0 0 -16000;

-1 1 1 0 0 0 0 0 0 0;

0 0 -1 1 1 0 0 0 0 0;

0 0 0 0 0 1 1 0 0 0;

0 0 0 0 0 0 -1 1 1 0;

0 0 0 0 0 0 0 0 -1 1];

B = [24 0 0 0 0 0 0 0 0 0]';

x = A\B x =

24.0000

16.0000

8.0000

1.6000

6.4000

-0.0020

0.0020

0.0016

0.0004

0.0004






One possible circuit design that matches the given equations is shown below.

24Vdc

8k

5k

4k

16k

Answer:

[v1, v2, v3, v4, v5, i1, i2, i3, i4, i5]

= [24 V, 16 V, 8 V, 1.6 V, 6.4 V, 2 mA, 2 mA, 1.6 mA, 0.4 mA, 0.4 mA]

The schematic is shown above.






Problem 2-62

Consider the circuit of Figure P2-62. Use MATLAB to find all of the voltages and currents in

the circuit.

33 kΩ

22 kΩ120 V

68 kΩ

10 kΩ 15 kΩ

15 kΩ

Solution:

Use KVL, KCL, and Ohm's Law to create a set of 14 equations to solve for all of the voltages

and currents. Also compute the powers for each element. The unknown vector is:

x = [vS, v1, v2, v3, v4, v5, v6, iS, i1, i2, i3, i4, i5, i6]

The equations and solution are presented in the following MATLAB code.

clear all

format short eng

% Set the matrix to be all zeros and then assign individual values directly

A = zeros(14,14);

A(1,1) = 1;

A(2,2) = 1; A(2,9) = -15e3;

A(3,3) = 1; A(3,10) = -10e3;

A(4,4) = 1; A(4,11) = -33e3;

A(5,5) = 1; A(5,12) = -15e3;

A(6,6) = 1; A(6,13) = -68e3;

A(7,7) = 1; A(7,14) = -22e3;

A(8,8) = 1; A(8,9) = 1;

A(9,9) = 1; A(9,10) = -1; A(9,11) = -1;

A(10,11) = 1; A(10,12) = -1; A(10,13) = -1;

A(11,13) = 1; A(11,14) = -1;

A(12,1) = -1; A(12,2) = 1; A(12,3) = 1;

A(13,3) = -1; A(13,4) = 1; A(13,5) = 1;

A(14,5) = -1; A(14,6) = 1; A(14,7) = 1;

% Display the A matrix

%A

B = zeros(14,1);

B(1) = 120;

% Display the B matrix

%B

% Solve for the output vector

x = A\B;

Results = x';

% Create vectors of the voltages and currents






Voltages = x(1:7)

Currents = x(8:14)

% Compute the associated powers

Powers = Voltages.*Currents

PTotal = sum(Powers) Voltages =

120.0000e+000

77.5537e+000

42.4463e+000

30.5455e+000

11.9008e+000

8.9917e+000

2.9091e+000

Currents =

-5.1702e-003

5.1702e-003

4.2446e-003

925.6198e-006

793.3884e-006

132.2314e-006

132.2314e-006

Powers =

-620.4298e-003

400.9720e-003

180.1687e-003

28.2735e-003

9.4420e-003

1.1890e-003

384.6732e-006

PTotal =

-101.2103e-018

Answer:

The results are summarized in the table below:

Element Voltage (V) Current (mA) Power (mW)

Source 120.0000 -5.1702 -620.4298

1 77.5537 5.1702 400.9720

2 42.4463 4.2446 180.1687

3 30.5455 0.9256 28.2735

4 11.9008 0.7934 9.4420

5 8.9917 0.1322 1.1890

6 2.9091 0.1322 0.3847






Problem 2-63

Consider the circuit of Figure P2-62 again. Use OrCAD to find all of the voltages and currents

in the circuit.

Solution:

The results of the OrCAD simulation are shown below.

Answer:

Presented above.






Problem 2-64

Use OrCAD to find all of the power dissipated or provided in the circuit of Figure P2-62. Verify

that the sum of all power in the circuit is zero.

Solution:

The results of the OrCAD simulation are shown below.

pTOTAL = 620.4 + 401.0 + 180.2 + 28.27 + 9.442 + 1.189 + 0.3847 = 0 mW

Answer:

The schematic is shown above. The sum of the powers is p = 0 mW.






Problem 2-65

Nonlinear Device Characteristics (A)

The circuit in Figure P2-65 is a parallel combination of a 50-Ω linear resistor and a varistor

whose i-v characteristic is iV = 2.6 10-5v3. For a small voltage, the

varistor current is quite small compared to the resistor current. For

large voltages, the varistor dominates because its current increases

more rapidly with voltage.

(a) Plot the i-v characteristic of the parallel combination.

(b) State whether the parallel combination is linear or nonlinear,

active or passive, and bilateral or nonbilateral.

(c) Find the range of voltages over which the resistor current is at

least 10 times as large as the varistor current.

(d) Find the range of voltages over which the varistor current is at least 10 times as large as the

resistor current.

Solution:

The solution is presented in the following commented MATLAB code.

(a) clear all

disp('Part (a)')

% Set the range of voltages to plot

v = -200:1:200;

% Compute the current through the resistor

iR = v/50;

% Compute the current through the varistor

iV = 2.6e-5*v.^3;

% Sum the two path currents to get the total current

iTotal = iR + iV;

% Plot the i-v characteristic

plot(v,iTotal,'b','LineWidth',3)



grid on Part (a)

50 Ωv

i i v






-200 -150 -100 -50 0 50 100 150 200-250

-200

-150

-100

-50

0

50

100

150

200

250

Voltage (V)

Curr

ent

(A)

(b) The plot created in Part (a) shows that the parallel combination is nonlinear, passive, and

bilateral.

(c) Solve the equations for positive voltages and realize that the full range will include negative

voltages.

clear all

syms v

Eqn = 'v/50 - 10*(2.6e-5*v^3)';

Soln = double(solve(Eqn,'v')) Soln =

0

8.7706

-8.7706

8.7706 V < v < 8.7706 V

(d) Solve the equations for positive voltages and realize that the full range will include negative

voltages.

clear all

syms v

Eqn = '2.6e-5*v^3 - v/5';

Soln = double(solve(Eqn,'v')) Soln =

0

87.7058

-87.7058

|v| > 87.7058 V

Answer:






(a) The plot is shown above.

(b) The parallel combination is nonlinear, passive, and bilateral.

(c) 8.7706 V < v < 8.7706 V

(d) |v| > 87.7058 V






Problem 2-66

Center Tapped Voltage Divider (A)

Figure P2-66 shows a voltage divider with the center tap connected to ground. Derive equations

relating vA and vB to vS, R1, and R2.

v

R

R

1

2

S

vA

vB

A

B

i

Solution:

Using the passive sign convention, iS = iA = iB. We can calculate the magnitude of the current

by combining the resistors in series and using Ohm's Law.

21

SA

RR

vi

Now apply Ohm's Law to each resistor to find vA and vB.

21

S11AA

RR

vRRiv

21

S22BB

RR

vRRiv

Answer:

21

S1A

RR

vRv

21

S2B

RR

vRv






Problem 2-67

Active Transducer (A)

Figure P2-67 shows an active transducer whose resistance R(VT) varies with the transducer

voltage VT as R(VT) = 0.5 VT2 + 1. The transducer supplies a current to a 10-Ω load. At what

voltage will the load current equal 100 mA?

iLR(VT)

10 ΩVT

Transducer

Solution:

Develop an expression for the current in terms of the transducer voltage and solve for the voltage

that will make the current equal 100 mA.

clear all

format short eng

syms iL Vtr

Eqn = 'iL - Vtr/(0.5*Vtr^2 + 1 + 10)';

Soln = solve(Eqn,'Vtr');

Vtr_Soln = double(subs(Soln,iL,100e-3))

iL_Check = Vtr_Soln./(0.5*Vtr_Soln.^2+1+10) Vtr_Soln =

18.8318e+000

1.1682e+000

iL_Check =

100.0000e-003

100.0000e-003

Answer:

There are two answers: VT = 1.1682 V or VT = 18.8318 V.






Problem 2-68

Programmable Voltage Divider (A)

Figure P2-68 shows a programmable voltage divider in which digital inputs b0 and b1 control

complementary analog switches connecting a multitap voltage divider to the analog output vO.

The switch positions in the figure apply when digital inputs are low. When inputs go high the

switch positions reverse. Find the analog output voltage for (b1,b0) = (0,0), (0,1), (1,0), and (1,1)

when VREF = 12 V.

O

v

b

O

vREF

1b

R

R

R

R

Solution:

There are four equal resistors in series with a voltage source, so each drops one quarter of the

total voltage, or 3 V in this case. As we cycle through the four combinations of the digital

inputs, the switches connect the output voltage to be across zero, one, two, or three resistors, in

that order. The output voltages are therefore 0 V, 3 V, 6 V, and 9 V.

Answer:

For (b1,b0) = (0,0), vO = 0 V.

For (b1,b0) = (0,1), vO = 3 V.

For (b1,b0) = (1,0), vO = 6 V.

For (b1,b0) = (1,1), vO = 9 V.






Problem 2-69

Analog Voltmeter Design (A, D, E)

Figure P2-69(a) shows a voltmeter circuit consisting of a D'Arsonval meter, two series resistors,

and a two-position selector switch. A current of IFS = 400 μA produces full-scale deflection of

the D'Arsonval meter, whose internal resistance is RM = 25 Ω.

(a) (D) Select the series resistance R1 and R2 so a voltage vx = 100 V produces full-scale

deflection when the switch is in position A, and voltage vx = 10 V produces full-scale deflection

when the switch is in position B.

(b) (A) What is the voltage across the 20-kΩ resistor in Figure P2-69(b)? What is the voltage

when the voltmeter in part (a) is set to position A and connected across the 20-kΩ resistor?

What is the percentage error introduced connecting the voltmeter?

(c) (E) A different D'Arsonval meter is available with an internal resistance of 100 Ω and a full-

scale deflection current of 100 μA. If the voltmeter in part (a) is redesigned using this

D'Arsonval meter, would the error found in part (b) be smaller or larger? Explain.

1

B

R

A

R

R2

M

vx

30 k

50 V

20 k VM

(b)(a)

Solution:

(a) Solve for R2 first, such that a 10-V input causes 400 A to flow through the two resistors.

Then solve for R1, such that a 100-V input causes 400 A to flow through all three resistors.

clear all

IFS = 400e-6;

RM = 25;

Req10 = 10/IFS;

R2 = Req10 - RM

Req100 = 100/IFS;

R1 = Req100-Req10 R2 =

24.9750e+003

R1 =

225.0000e+003






(b) Using voltage division, the voltage across the 20-kΩ resistor is 20 V when the voltmeter is

not connected. When the voltmeter is set in position A and connected in parallel to the 20-kΩ

resistor, it is equivalent to placing a 250-kΩ resistor in parallel with the 20-kΩ resistor. We can

then find the voltage using voltage division and compute the error.

clear all

Req = 1/(1/20e3 + 1/250e3);

vM = Req*50/(Req + 30e3)

ErrorPercent = 100*(20-vM)/20 vM =

19.0840e+000

ErrorPercent =

4.5802e+000

(c) With a full-scale deflection current of 100 A for an applied voltage of 100 V, (switch in

position A,) the total resistance of the meter must be 1 MΩ. The increased meter resistance will

draw less current when it is connected to the 20-kΩ resistor and have a smaller impact on the

voltage. The error will decrease. The following calculations verify the results with numerical

values.

clear all

Req = 1/(1/20e3 + 1/1e6);

vM = Req*50/(Req + 30e3)

ErrorPercent = 100*(20-vM)/20 vM =

19.7628e+000

ErrorPercent =

1.1858e+000

Answer:

(a) R1 = 225 kΩ and R2 = 24.975 kΩ.

(b) vACTUAL = 20 V, vMEAS = 19.084 V, Error = 4.58%

(c) The error will be smaller as explained and verified above.






Problem 2-70

MATLAB Function for Parallel Equivalent Resistors (A)

Create a MATLAB function to compute the equivalent resistance of a set of resistors connected

in parallel. The function has a single input, which is a vector containing the values of all of the

resistors in parallel, and it has a single output, which is the equivalent resistance. Name the

function “EQparallel” and test it with at least three different resistor combinations. At least one

test should have three or more resistor values.

Solution:

Create a MATLAB script (m-file) named EQparallel.m that contains the following code:

function Zp = EQparallel(Z)

% Compute the equivalent parallel impedance of a list of impedances

Zinv = 1./Z; Zp = 1/sum(Zinv);

Save the function file and change the MATLAB path to include the location of the function file.

Run the following commands to check the function.

R1 = EQparallel([1000 1000]) R2 = EQparallel([5e3 20e3]) R3 = EQparallel([4e3 5e3 20e3])

The results are:

R1 = 500

R2 = 4000

R3 = 2000

Answer:

The MATLAB function and examples are presented above. Run the code outside of this Word

document.

Full file at ://fratstock.eu/sample/Solutions-Manual-The... · Full file at The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual Copyright © 2008 John Wiley

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