The MATLAB Notebook v1.6.… · Web viewThe 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;

The Analysis and Design of Linear Circuits, Sixth Edition Solutions Manual

Problem 2-1The current through a 33-kΩ resistor is 1.2 mA. Find the voltage across the resistor.

Solution:v = iR

clear allR = 33e3;ii = 1.2e-3;v = ii*R v = 39.6000e+000

Answer:v = 39.6 V

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

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Problem 2-2A 6.2-kΩ resistor dissipates 12 mW. Find the current through the resistor.

Solution:p = i2R

clear allformat short engR = 6.2e3;p = 12e-3;ii = sqrt(p/R) ii = 1.3912e-003

Answer:i = ±1.3912 mA




Problem 2-3The 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 allG = 1e-3;R = 1/G;v = 9;ii = v/R ii = 9.0000e-003

Answer:i = 9 mA




Problem 2-4In Figure P2-4 the resistor dissipates 25 mW. Find Rx.

Solution:

clear allp = 25e-3;v = 15;R = v^2/p R = 9.0000e+003

Answer:R = 9 kΩ




Problem 2-5In Figure P2-5 find Rx and the power delivered to the resistor.

Solution:clear allv = 100;ii = 10e-3;R = v/iip = v*ii R = 10.0000e+003p = 1.0000e+000

Answer:Rx = 10 kΩ, p = 1 W




Problem 2-6The 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 allformat short engii = [-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 i1v1 = 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 onplot(vv2,ii1,'g','LineWidth',1)grid onxlabel('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+003MaxError = 25.0000e-003MaxError2 = 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)

Cur

rent

(A)

NonlinearLinear

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 18.76250.5 37.525 18.76251 75.2 75.22 151.6 303.25 400 200010 950 9500

(b) ERRORMAX = 25 mV




Problem 2-7A 10-kΩ resistor has a power rating of ⅛W. Find the maximum voltage that can be applied to the resistor.

Solution:clear allR = 10e3;p = 1/8;v_max = sqrt(p*R) v_max = 35.3553e+000

Answer:vmax = 35.36 V




Problem 2-8A 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 allV = 500;p = 1/4;R = V^2/p R = 1.0000e+006

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 2-9Figure 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

(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 allv = [-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)xlabel('Voltage (V)')ylabel('Current (A)')grid onv = 5i5 = 2e-16*(exp(40*v)-1)v = -5iNeg5 = 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-003v = 5.0000e+000i5 = 144.5195e+069v = -5.0000e+000iNeg5 = -200.0000e-018




-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)

Cur

rent

(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 00.1 1.07E-14 1.07E-150.2 5.96E-13 1.19E-130.4 1.78E-09 7.11E-100.8 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 2-10In Figure P2-10 i2 = –2 A and i3 = 5 A. Find i1 and i4.

Solution:Apply KCL at Nodes B and C.

clear alli2 = -2;i3 = 5;i1 = -i2i4 = i2+i3 i1 = 2.0000e+000i4 = 3.0000e+000

Answer:i1 = 2 A and i4 = 3 A.




Problem 2-11For 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.

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: ;node B: ;node C:

KVL: loop 1-2: ;loop 2-3-4: ;loop 1-3-4:




Problem 2-12In Figure P2-11, i2 = –10 mA and i4 = 20 mA. Find i1 and i3.

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

clear alli2 = -10e-3;i4 = 20e-3;i3 = i4i1 = -i2-i3 i3 = 20.0000e-003i1 = -10.0000e-003

Answer:i1 = 10 mA and i3 = 20 mA.




Problem 2-13For 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.

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: ;node B: ;node C: ;node D:

KVL: loop 1-3-2: ;loop 2-4-5: ;loop 3-6-4:




Problem 2-14In Figure P2-13 v2 = 10 V, v3 = –10 V, and v4 = 3 V. Find v1, v5, and v6.

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

clear allv2 = 10;v3 = -10;v4 = 3;v1 = v2 - v3v5 = v2 - v4v6 = v4 - v3 v1 = 20.0000e+000v5 = 7.0000e+000v6 = 13.0000e+000

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




Problem 2-15The 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.

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

(b) KCL: node A: ;node B: ;node C: ;node D:

clear alli1 = -20e-3;i2 = -12e-3;i3 = 50e-3;i4 = -i2-i3i5 = -i1-i2i6 = i3-i1 i4 = -38.0000e-003i5 = 32.0000e-003i6 = 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




(b) KCL: node A: ;node B: ;node C: ;node D:

(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 2-16In Figure P2-16, v2 = 10 V, v3 = 10 V, and v4 = 10 V. Find v1 and v5.

Solution:Apply KVL to the circuit.

clear allv2 = 10;v3 = 10;v4 = 10;v1 = v2+v3v5 = v3-v4 v1 = 20.0000e+000v5 = 0.0000e-003

Answer:v1 = 20 V and v5 = 0 V.




Problem 2-17In Figure P2-17 i2 = 10 mA, i3 = –15 mA, and i4 = 5 mA. Find i1 and i5.

Solution:Apply KCL to the circuit.

clear alli2 = 10e-3;i3 = -15e-3;i4 = 5e-3;i1 = i2-i3+i4i5 = i2-i1 i1 = 30.0000e-003i5 = -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?

Solution:(a) Voltage signs:

Elements 1 and 3: plus on bottom and minus on topElements 2 and 4: plus on top and minus on bottomElement 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: loop 3-4: loop 2-4-5:

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





Problem 2-19The 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 2-20Find vx and ix in Figure P2-20.

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

clear allformat short engis = 2e-3;ix = -isvx = 47e3*ix ix = -2.0000e-003vx = -94.0000e+000

Answer:vx = 94 V and ix = 2 mA.




Problem 2-21Find vx and ix in Figure P2-21.

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 alli10 = 1/2;v10 = 10*i10;v5 = v10;i5 = v5/5;ix = i5i4 = i10+i5;v4 = 4*i4;vx = v4+v10 ix = 1.0000e+000vx = 11.0000e+000

Answer:vx = 11 V and ix = 1 A




Problem 2-22In 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.

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:

clear allformat short engvs1 = 5;vs2 = 10;vs3 = 5;v1 = vs1-vs3v2 = vs1-vs2v3 = vs2-vs3 Copyright © 2008 John Wiley & Sons, Inc. All rights reserved.



v1 = 0.0000e-003v2 = -5.0000e+000v3 = 5.0000e+000

(c) v = iR

i1 = v1/50i2 = v2/100i3 = v3/100 i1 = 0.0000e-003i2 = -50.0000e-003i3 = 50.0000e-003

(d) KCL equations

is1 = -i1-i2is2 = i2-i3is3 = i1+i3 is1 = 50.0000e-003is2 = -100.0000e-003is3 = 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 2-23Find the power dissipated in the 1.5 kΩ resistor in Figure P2-23.

Solution:Label the elements.Element 1: 1-kΩ resistor with current flowing downElement 2: 500-Ω resistor with current flowing to the rightElement 3: 1.5-kΩ resistor with current flowing down

Write KCL, KVL, and Ohm's Law equations:i1 + i2 5 mA = 0i2 + i3 = 0v1 + v2 + v3 = 0v1 = 1000i1

v2 = 500i2

v3 = 1500i3

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

clear allformat short engEqn1 = '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




Problem 2-24Figure 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 rightElement 2: 2-kΩ resistor with positive voltage sign on the bottomUse Ohm's Law to compute i1.Use KCL at the center node to find ix.Use Ohm's Law to find vx.

clear allv1 = 20;R1 = 5e3;i1 = v1/R1;ix = i1+4e-3 - 6e-3Rx = 8e3;vx = ix*Rx ix = 2.0000e-003vx = 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 2-25In Figure P2-25 ix = –0.5 mA. Find the value of R.

Solution:Label the circuit elements.Element 1: Resistor R with current flowing down.Element 2: 10-kΩ resistor with current flowing from right to leftCompute 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 allRx = 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 2-26Figure 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.

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 allvx = -15-0ix = vx/100e3px = vx*ix vx = -15.0000e+000ix = -150.0000e-006px = 2.2500e-003

Answer:vx = 15 V, ix = 150 A, and px = 2.25 mW.Copyright © 2008 John Wiley & Sons, Inc. All rights reserved.






Problem 2-27Find the equivalent resistance REQ in Figure P2-27.

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

clear allR1 = 75;R2 = 300;R3 = 100;R4 = 200;% Combine R3 and R4 in seriesR34 = R3+R4;% Combine R2 in parallel with the series combination of R3 and R4R234 = 1/(1/R2 + 1/R34);% Combine R1 in series with the other combinationR1234 = R1 + R234;Req = R1234 Req = 225.0000e+000

Answer:REQ = 225 Ω




Problem 2-28Find the equivalent resistance REQ in Figure P2-28.

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

clear allR1 = 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 2-29Find REQ in Figure P2-29 when the switch is open. Repeat when the switch is closed.

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 allR1 = 100;R2 = 200;R3 = 50;R4 = 50;disp('Switch Open')Req_open = R1 + 1/(1/R2 + 1/R3) + R4disp('Switch Closed')Req_closed = R1 + 0 + R4 Switch OpenReq_open = 190.0000e+000Switch ClosedReq_closed = 150.0000e+000

Answer:Switch open: REQ = 190 Ω.Switch closed: REQ = 150 Ω.




Problem 2-30Show 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.

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 2100 A D200 A B150 A C50 C D25 A+B+C D

33.3 B+C D




133.3 A B+C




Problem 2-31In Figure P2-31 find the equivalent resistance between terminals A-B, A-C, A-D, B-C, B-D, and C-D.

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 allRab = 1/(1/40 + 1/(40+80)) + 60Rac = 1/(1/40 + 1/(40+80)) + 30Rad = 1/(1/40 + 1/(40+80)) + 10Rbc = 60 + 1/(1/(40+40) + 1/80) + 30Rbd = 60 + 1/(1/(40+40) + 1/80) + 10Rcd = 30 + 0 + 10 Rab = 90.0000e+000Rac = 60.0000e+000Rad = 40.0000e+000Rbc = 130.0000e+000Rbd = 110.0000e+000Rcd = 40.0000e+000

Answer:RAB = 90 ΩRAC = 60 ΩRAD = 40 ΩRBC = 130 ΩRBD = 110 ΩRCD = 40 Ω







Problem 2-32Select a value of RL in Figure P2-32 so that REQ = 25 kΩ. Repeat for REQ = 20 kΩ.

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

clear allsyms RL Req positiveEqn = '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 2-33Using 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 seriesR = 500 Ω: two resistors in parallelR = 1.5 kΩ: one resistor in series with a parallel combination of two resistorsR = 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 allR = 1000;R2000 = R + RR500 = 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+003R500 = 500.0000e+000R1500 = 1.5000e+003R333 = 333.3333e+000R250 = 250.0000e+000R400 = 400.0000e+000





Problem 2-34Find the equivalent practical voltage source at terminals A and B in Figure P2-34

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 2-35In Figure P2-35 the i-v characteristic of network N is v + 50i = 5 V. Find the equivalent practical current source for the network.

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 2-36Select the value of Rx in Figure P2-36 so that REQ = 60 kΩ.

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

clear allsyms Rx Req positiveEqn = '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+003Check_Req = 60.0000e+003

Answer:Rx = 21.78 kΩ.




Problem 2-37Two 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?

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 allR = 10e3;Rmax = R + RRmin = 1/(1/R + 1/R) Rmax = 20.0000e+003Rmin = 5.0000e+003

Answer:5 kΩ REQ 20 kΩ




Problem 2-38Select the value of R in Figure P2-38 so that RA-B = RL.

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 allsyms R Rab RL positiveEqn = '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*RLCheck_Rab =RL

Answer:R = RL/3.




Problem 2-39What is the range of REQ in Figure P2-39?

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

clear allRp = [0 4e3];Req = 1e3 + 1./(1./Rp + 1/4e3) Req = 1.0000e+003 3.0000e+003

Answer:1 kΩ REQ 3 kΩ




Problem 2-40Find the equivalent resistance between terminals A and B in Figure P2-40

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 2-41Use voltage division in Figure P2-41 to obtain an expression for vL in terms of R, RL, and vS.

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 allsyms vs R RL Req vLReq = 1/(1/R + 1/RL);vL = simple(Req*vs/(Req+R)) vL =vs*RL/(2*RL+R)

Answer:




Problem 2-42Use current division in Figure P2-42 to find vL if R =10 Ω, RL = 20 Ω, and iS = 2.5 mA.

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 allR = 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 2-43Find ix and iy in Figure P2-43.

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 allReq1 = 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-003iy = 1.2000e+000

Answer:ix = 500 mA and iy = 1.2 A.




Problem 2-44The 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?

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

Rx1k

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 allsyms Rx RL Req vLEqn1 = '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+003V_Check = 5.0000e+000

Answer:




RX = 2.3383 kΩ.




Problem 2-45Find the range of values of vO in Figure P2-45.

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 allRp = [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 2-46Figure 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.

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.

and

clear allsyms vs Ra Rb Rc RxEqn = 'Rb*vs/(Ra+Rb) - Rx*vs/(Rc+Rx)';Soln = solve(Eqn,'Rx') Soln =Rb*Rc/Ra

Answer:




Problem 2-47Ideally, 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.

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 allR1 = 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 2-48Select values for R1, R2, and R3 in Figure P 2-48 so the voltage divider produces the two output voltages shown.

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 allvs = 15;R1 = 5e3;R2 = 7e3;R3 = 3e3;v23 = (R2+R3)*vs/(R1+R2+R3)v3 = R3*vs/(R1+R2+R3) v23 = 10.0000e+000v3 = 3.0000e+000

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




Problem 2-49Select a value of Rx in Figure P2-49 so that vL = 3 V.

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 allR = 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 2-50Select 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.

Solution:

clear allvL = [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 2-51Use circuit reduction to find vx and ix in Figure P2-51.

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

clear allsyms R is ix vx Req1 Req2Req1 = 2*R + R;Req2 = 1/(1/3/R+1/Req1);vx = is*Req2ix = vx/3/R vx =3/2*is*Rix =1/2*is

Answer:vx = 1.5 R iS, ix = 0.5 iS





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 allsyms R vs vx ix Req1 Req2Req1 = 1/(1/R + 1/R/2);Req2 = 2*R + 2*Req1;is = vs/Req2;vx = Req1*isix = (1/R/2)*(-is)/(1/R + 1/R/2) vx =1/5*vsix =-1/10*vs/R

Answer:




Problem 2-53Use circuit reduction to find vx and the power provided by the source in Figure P2-53.

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 allReq1 = 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 transformationvs = is*Rs;vx = Req3*vs/(Rs + 20 + Req3) ps = 6.4773e+000vx = 2.2727e+000

Answer:vx = 2.2727 V and pS = 6.4773 W.





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 allvs = 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-006vx = 2.0000e+000

Answer:vx = 2 V and ix = 500 A.




Problem 2-55Use source transformation to find ix in Figure P2-55.

Solution:Perform a source transformation on the current source and the 30-Ω resistor and then use Ohm's Law to find ix.

clear allvs1 = 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 2-56Select a value for RX so that ix = 0 A in Figure P2-56.

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 allvs1 = 24;vs2 = 12;Rs2 = 30;is2 = vs2/Rs2Rx = vs1/is2 is2 = 400.0000e-003Rx = 60.0000e+000

Answer:RX = 60 Ω.




Problem 2-57Use source transformations in Figure P2-57 to relate vO to v1, v2, and v3.

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

clear allsyms v1 v2 v3 R i1 i2 i3 Req ieq voi1 = 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:



v1/RR R R

v2/R v3/R

(v1+v2+v3)/RR/3


Problem 2-58The current through RL in Figure P2-58 is 40 mA. Use source transformations to find RL.

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 alliL = 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 2-59Select RX so that 25 V are across it in Figure P2-59.

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 allvx = 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 2-60The 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.

Solution:The solution is presented in the commented MATLAB script below.

clear allR = 10e3;% Source transformation to a current sourcevs = 50;Rs = 10e3;is = vs/Rs% Combine parallel resistorsReq = 1/(1/Rs + 1/R)% Source transformation back to a voltage sourcevs2 = is*Req% Unknown resistor rangeRp = [8e3 80e3];% Voltage division to calculate vxvx = R*vs2./(R+Rp+Req) is = 5.0000e-003Req = 5.0000e+003vs2 = 25.0000e+000vx = 10.8696e+000 2.6316e+000

Answer:2.6316 V vx 10.8696 V




Problem 2-61A circuit is found to have the following element and connection equations:v1 = 24 Vv2 = 8k i2

v3 = 5k i3

v4 = 4k i4

v5 = 16k i5

–v1 + v2 + v3 = 0–v3 + v4 + v5 = 0i1 + 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 allA = [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 2-62Consider the circuit of Figure P2-62. Use MATLAB to find all of the voltages and currents in the circuit.

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 allformat short eng% Set the matrix to be all zeros and then assign individual values directlyA = 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%AB = zeros(14,1);B(1) = 120;% Display the B matrix%B% Solve for the output vectorx = A\B;Results = x';% Create vectors of the voltages and currentsCopyright © 2008 John Wiley & Sons, Inc. All rights reserved.



Voltages = x(1:7)Currents = x(8:14)% Compute the associated powersPowers = Voltages.*CurrentsPTotal = sum(Powers) Voltages = 120.0000e+000 77.5537e+000 42.4463e+000 30.5455e+000 11.9008e+000 8.9917e+000 2.9091e+000Currents = -5.1702e-003 5.1702e-003 4.2446e-003 925.6198e-006 793.3884e-006 132.2314e-006 132.2314e-006Powers = -620.4298e-003 400.9720e-003 180.1687e-003 28.2735e-003 9.4420e-003 1.1890e-003 384.6732e-006PTotal = -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 400.97202 42.4463 4.2446 180.16873 30.5455 0.9256 28.27354 11.9008 0.7934 9.44205 8.9917 0.1322 1.18906 2.9091 0.1322 0.3847




Problem 2-63Consider 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.





Problem 2-64Use 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 2-65Nonlinear 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 alldisp('Part (a)')% Set the range of voltages to plotv = -200:1:200;% Compute the current through the resistoriR = v/50;% Compute the current through the varistoriV = 2.6e-5*v.^3;% Sum the two path currents to get the total currentiTotal = iR + iV;% Plot the i-v characteristicplot(v,iTotal,'b','LineWidth',3)xlabel('Voltage (V)')ylabel('Current (A)')grid on Part (a)




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

-200

-150

-100

-50

0

50

100

150

200

250

Voltage (V)

Cur

rent

(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 allsyms vEqn = '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 allsyms vEqn = '2.6e-5*v^3 - v/5'; Soln = double(solve(Eqn,'v')) Soln = 0 87.7058 -87.7058

|v| > 87.7058 V

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



(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 2-66Center 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.

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.

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

Answer:




Problem 2-67Active Transducer (A)Figure P2-67 shows an active transducer whose resistance R(VT) varies with the transducer voltage VT as R(VT) = 0.5 VT

2 + 1. The transducer supplies a current to a 10-Ω load. At what voltage will the load current equal 100 mA?

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 allformat short engsyms iL VtrEqn = '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+000iL_Check = 100.0000e-003 100.0000e-003

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




Problem 2-68Programmable 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.

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 2-69Analog 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.

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 allIFS = 400e-6;RM = 25;Req10 = 10/IFS;R2 = Req10 - RMReq100 = 100/IFS;R1 = Req100-Req10 R2 = 24.9750e+003R1 = 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 allReq = 1/(1/20e3 + 1/250e3);vM = Req*50/(Req + 30e3)ErrorPercent = 100*(20-vM)/20 vM = 19.0840e+000ErrorPercent = 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 allReq = 1/(1/20e3 + 1/1e6);vM = Req*50/(Req + 30e3)ErrorPercent = 100*(20-vM)/20 vM = 19.7628e+000ErrorPercent = 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 2-70MATLAB 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 = 500R2 = 4000R3 = 2000

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



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The MATLAB Notebook v1.6.… · Web viewThe 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;