Errata+Irwin Classical Electromagnetism Classical Electromagnetism PDF

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Classical Electromagnetism Classical Electromagnetism Classical Electromagnetism Classical Electromagnetism Classical Electromagnetism Classical Electromagnetism Classical Electromagnetism Classical Electromagnetism...

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Errata Irwin and Nelms, Engineering Circuit Analysis, 11e International Student Version

Pg 26 : In the Caption for Figure 2.4 it should be Circuits for Example 2.1 to 2.5. Pg33 : Learning Assessment E2.4 the question should be E2.4 Given the networks in Fig. E2.4, find (a) I1 in Fig. E2.4a and (b) IT in Fig. E2.4b. Pg36 : The equation before the solution should be −Vec – Vae + 16 − 12 = 0 Pg43: Example 2.19 solution the first equation to calculate voltage Vbd should be 10kI + Vbd + 3kI − 6 = 0 Pg 51: Example 2.26 1. Calculate the equivalent resistance Rabin the circuit below,

Sol: The 3 Ω and 6 Ω resistors are in parallel because they are connected to the same two nodes cand b. Their combined resistance is 3×6 =2Ω 3 Ω || 6 Ω = 3+6 Similarly, the 12 Ω and 4 Ω resistors are in parallel since they are connected to the same two nodes dand b. Hence, 12 × 4 =3Ω 12 Ω || 4 Ω = 12 + 4 Also the 1 Ω and 5 Ω resistors are in series; hence, their equivalent resistance is 1Ω+5Ω=6Ω With these three combinations, we can replace the given circuit with the circuit shown below:

Now in this circuit the 3 Ω resistance is parallel with 6 Ω which gives an equivalent resistance of 2 Ω (calculated above). This 2 Ω equivalent resistance is now in series with 1 Ω resistance which gives a combined resistance of 3Ω. Now 2 Ω and 3 Ω resistance are in parallel, => 2 Ω || 3Ω = (2 x 3) / (2 + 3) = 1.2 Ω The 1.2 Ω resistor is in series with 10 Ω, so that Rab = 1.2 Ω + 10 Ω = 11.2 Ω Pg66 : The first equation in the solution should be 4Io instead of 4mIo Pg68 : Design example 2.37a the equation should be P = 15 + 10 + 20 = 45W . Pg69 : Design example 2.38 the answers should be ⇒ Vr = 400 – 220V = Ir = 0.91 r ⇒ r = 197.8 Ω. Pg74 : The question number should be 2.22 not 20.22. Pg85: The question for 2.107 should be “Determine the value of V0 in the network in Fig. P2.107”. Pg87 : Question 2EF-3 c. should be 8Ω.

Pg116 The MATLAB solution is then >> R = [10 0 -6; 0 12 R= 10 0 0 12 -6 -3 >>

>>

-3;-6 -3 21] -6 -3 21

V = [-0.006; 0.006; 0] V= -0.0060 0.0060 0 I = inv(R)*V I = 1.0e-003* -0.6757 0.4685 -0.1261

Or I1 = −0.6757 mA I2 = 0.4685 mA I3 = −0.1261 mA Page 137: Problem 3.45: W should be replaced by Ω. Pg165: The caption for Figure 4.19 should be Circuits used in Example 4.10. Pg166: For the solution to Example 4.13 it should be If we chose R1 = 10 kΩ and R3 = 20 kΩ, then R2 = 50 kΩ and R4 = 20 kΩ. Pg168: Question 4.7 it should be “An op-amp based amplifier has ±24 V supplies and a gain of – 80 Over what input range is the amplifier linear?” Pg170: Question 4.24 the question should be “In the network in Fig. P4.24, determine the value of Vo. Page 211: Example 5.26: The ‘=’ sign should be replaced by ‘,’ Last line:

⇒R x =

R3 200 R= × 125=50 Ω R1 2 500

Example 6.3: 80J

Pg 236:

12F is correct, 10.12 should be corrected as 10-6.

Example 6.4:

Page 241: Example 6.6 t

E ( t ) =∫ p (t ) dt + E ( 0 ) 0 t

3 ¿∫ 3906.25 t dt+ 0 0

¿ 976.56 t 4 J

0  t  4 ms

Now energy at t = 4 ms is given by E(4 ms) =

−3 4

976.56(4 10 )

= 249.9 nJ

Page 255: Example 6.20 Design Example 6.20: Design an op amp integrator as shown in the figure, in which the input is an impulse of 1V height and 1ms width. The input and output of this op amp is related as t

V o=−10∫ V i (t )dt . 0

Solution: This is an integrator, whose input output relation is: t −1 V 0= V (t)dt ∫ CR 0 i Comparing the above equation with the given relation, we get Choosing R = 10K, the value of C comes out to be C = 10F.

Pg276: The solution for Example 7.3 should be :

CR=0.1

vc(t) = 80e −106t/100 = 80 e −104t V, t > 0

Thus t = 34.66 μs

Pg281 The equation i(t) = 9/2 + 5/6e-t/0.15mA Pg295: For equation 7.20, t in the equation should be no italics. Page 310: Problem 7.1: ‘V’ should be corrected as ‘A’ Pg333: For example 8.5. The question should be “Once again, let us determine the current in the RL circuit examined in Example 8.4. However, rather than apply VMcos ωt, we will apply VMe jωt. Pg335: Example 8.7. The question should be “Again, we consider the RL circuit in Example 8.4. Let us use phasors to determine the expression for the current.” Page 342: Example 8.16 ‘L’ should be removed ‘capacitance’ should be replaced by ‘impedance’.

Page 370: Problem 8.9, 8.12: Zin has to be corrected as Z. Problem 8.14: Z has to be corrected as ZT.

Pg371: Question 8.25 the equation should be : v2(t ) = 0.7571 cos (2t + 66.7°) v3(t ) = 0.6064 cos (2t − 69.8°)

Page 388: Example 9.3:

I 10 =

20 ×5.423 ∠ 49.40 =4.851 ∠−14.04 10+ j20

Example 9.4:

a. Peak current = 28√ 2=39.6 A −1 b. θload =co s ( 0.812 ) =35.71 (since lagging PF). Assume ang(v) = 0. 2 2300 √ ¿ ¿ p ( t )=¿ At t = 2.5ms, then p(t) = 71.89kW. c. P = V eff I eff cosθ=(2300 )( 28 ) cos ( 35.71 )=52.29 kW . d. S=V eff I eff ∠θ=64.4 ∠ 35.71 kVA e. Apparent power = |S| = 64.4 kVA Q=V eff I eff sinθ =37.59 kVAR f. Pg 396 : Example 9.9 V th =

j10 ×100 =20+ j 40 20+ j10

Pg414: The integrals of IaA and IbB can be determined graphically from Fig. 9.21b.

Pg417: The sentences just before example 9.21 should be “This secondary path is called a fault. For example, the fault path in Fig. 9.24 is through Joe and the concrete floor. The GFI detects this fault and opens the circuit in response. Its principle of operation is illustrated by the following example.” Pg419: Example 9.23 2nd Paragraph, 1st line. “The action in Fig. 9.27a is modeled as shown in Fig. 9.27b.”

Pg422: Design example 9.27. The 400 should be 4000.

Pg 423: The solution for Design Example 9.28. The 540 below should be 2402

Page 430: Problem 9.66 A composite load consists of three loads connected in parallel. One draws 100W at a PF of 0.92 lagging, another takes 250W at a PF of 0.8 lagging, and the third requires 150W at a unity PF. Find the effective current drawn from 115 V (rms) supply by this load. Also find the overall PF. Pg433: The question 9.90 should be : What value of capacitance must be placed in parallel with the 18-kW load in Figure 9.84 to raise the power factor of this load to 0.9 lagging? Page 434, Problem 9.100 A 5 kW load operates at 60 Hz, 240 V rms and has a power factor of 0.866 lagging. We wish to create a power factor of at least 0.975 lagging using a single capacitor. Find the value of the capacitor to meet this requirement.

Pg460: 7th line from the top. Therefore, one of the resulting equivalent circuits for the network in Fig. 10.22a is as shown Pg462: 2nd Paragraph, 1st line. The circuit in Fig. 10.24a thus reduces to that shown in Fig. 10.24c. Forming an equivalent… Pg463: The figure number mention here should be Fig.10.26. Pg465: Example 10.18 it should be ZTh - ZL / n2

Pg475: Question 10.53 In the network in Fig. P10.53, if

A find Vs.

Pg481: 1st and 2nd line Finally, let us examine the instantaneous power generated by a threephase system. Assume that the voltages in Fig. 11.4 are…. Pg482: Line 13th from the bottom. Fig. 11.5 are also balanced, there are two possible equivalent configurations for the load. Pg485: 6th line from the top. The phase relationships between the line and phase voltages are shown in Fig. 11.10. From…

Pg492: The answers for example 11.6 should be

= 39.11A.

Page495: Example11.9: There should be an ‘=’ sign at the marked point. = Pg497: The equation should be :

Pg499: The last line should be “The network in Fig. 11.21b indicates that” Pg502: Example 11.17, 2nd paragraph 1st line. It should be The network in Fig. 11.19b indicates that… Pg504: The correct equation should be Scap = Snew − Sold = j14.736 MVA Page 507: Problem 11.31: ‘ranch’ should be corrected as ‘branch’. Problem 11.39: 215 should be corrected as 16. Page 506: Problem 11.15 the question should be A balanced three phase generator has an abc phase sequence with phase voltage V an =255 ∠ 0o V . The generator feeds an induction motor which may be represented by a balanced wye connected load with an impedance of 12+ j 5 Ω per phase. Find out the line currents and load voltages.

Pg508: Question 11.54 the equation should be

V rms

Pg 515: Example 12.2 Magnitude of the transfer function is V2 1 = 2 V 1 √ 1 +(ω 510 3 10 10−9 )2

||

ω=20000 f=

rad =2 πf sec

ω =3.1831 kHz 2π

Pg 519: Example 12.4 For zeros s = -2, -1.5 For poles s = -2  j

Page 526: Example12.6 So magnitude is

| jω2 |−20 log|jω|

|G( jω)|db =20 log 1+

Pg537: Line 12.16 it should be Since I-YV1 and the voltage across the resistor is VR = IR Pg541: Example 12.15

is change to 1/LC = 106

Pg 554: Example 12.22 The question should be If the values of the circuit parameters in Fig 12.37 are R = 2 Ω, L = 1 H, and C = 12 F, let us determine the values of the elements if the circuit is magnitude scaled by a factor KM = 102 and frequency scaled by a factor KF = 106. Pg555: 2nd last line. shown in Fig. 12.39b, except that the output voltage is taken across the resistor. The voltage Pg559: The caption for Figure 13.43 Bode plots for the netowrk in Fig. 12.42. Pg560: The caption for Figure 12.45 Transient analysis of the network in Fig. 12.44.

Pg 567: Example 12.30 ‘k’ should be corrected as ‘K’

k ω2 Rf = =9.2308 Ri ω1 +ω 2 If we select Ri = 10 k then Rf = 9.23 Ri = 92.3 k

Pg574: Question 12.55. The automatic gain control circuit in Fig. P12.55 is used to limit the transconductance, i0/υin. Pg575: Question 12.66. The question should be Design a series RLC resonant circuit with v0 = 40 rad/s and BW = 10 rad/s. Pg595: Example 13.13. f (∞) = lim sF(s) s→O

Pg611: 2nd Equation the I1 should be I2. 4th Equation 6/2 is wrong it should be 6/S

Pg643: Equation 14.31 y(t) = XM ∣ H( jω0) ∣ ejω0tejϕ (jω0) + ∙ ∙ ∙ = XM ∣ H( jω0) ∣ e j(ω0t + ϕ ( jω0) ) + ∙ ∙ ∙ Pg 646: Example 14.20 Question: There should be a minus sign in front of the equation. Solution: (Vs – 0)Y3 = (0 – V0)(Y1 + Y2)  Y3Vs = -(Y1+Y2)Vo

Pg662: See marked change

Pg680: Example 15.12 Pg701: Question 15.32 the Figure number P15.32 should be P15.33 Pg701: Question 15.32 question to be replaced with

The J should be –J.

The voltage and current through an element are, respectively, 0 0 0 v (t )=30 cos (t +25 ) +10 cos ( 2 t +35 ) +4 cos (3 t−10 ) V i( t )=2 cos t +cos ( 2 t+10 0) A

(a) Find the average power delivered to the element. (b) Plot the power spectrum.

Pg705:

Should now be Ɵ = tan−1y/x...