# Problem Set 2 Solutions

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1 University of Cliforni, Berkeley Spring 2012 EE 42/100 Prof. A. Niknej Prolem Set 2 Solutions Plese note tht these re merely suggeste solutions. Mny of these prolems n e pprohe in ifferent wys. 1. In prolems like this, you my fin it helpful to rerw the iruit to ignore ll the irrelevnt terminls n suh tht everything is retngulr. Note tht sine we re fining eq from terminls n in eh of these, we must simplify the iruits towr these terminls. () Terminls n re onnete together, while e n f re irrelevnt. If you rerw the iruit s elow, you n see tht the two resistors in the first olumn re in prllel s, they re onnete together t oth pirs of terminls (ue to the short). This is lso true for the seon olumn of resistors. So eh of the pirs n e rerwn s one resistor with 1 2. e f /2 /2 It is esy to see tht ll the resistors re in series, so eq = 3. () Terminls e n f re shorte. Intuitively, eq shoul only inlue the two inner resistors, sine the other four re shorte out. Tht is, ny urrent flowing from to woul only go through 2 of resistne n proee etween e n f vi the short. e f e f 2 2 () The short etween n e gets ri of the one resistor etween the two noes. After tht, everything else simplifies s efore.

2 e f e f 2 8/3 2. The key to this prolem is tht the ler is infinite. Using this ft, we n onlue tht the equivlent resistne in shoul not hnge if we remove one rung of the ler. So we n reple the ler, sve for the first rung, with one resistor in. in in Now we n write n eqution in terms of n in n solve. in in = ( in ) = 2 in 2 in 2 in 2 2 = 0 Sine this is qurti eqution, there re two solutions. We know tht resistne nnot e negtive, so we isr the negtive solution. The positive solution is in = (1 3) 3. We hve two unknown noes, so we ll hve two nol equtions, one t noe 1 n the other t noe 2. V 1 V 2 V = 0 V 2 V 1 V = 0 Solving this system gives us V 1 = V n V 2 = 74.4 V. We n lso o this using superposition. Here we split the prolem into two suiruits, eh one with ifferent urrent soure zeroe out. ell tht the zeroe out urrent soure eomes n open iruit. Ω 1 2 8A 30Ω 20Ω 2

3 V 1 V 2 V 2 V 1 V = 0 V 2 20 = 0 This first iruit gives us V 1 = V n V 2 = 19.2 V. Ω Ω 20Ω -3A V 1 V 2 V 2 V 1 V 1 30 = 0 V = 0 The solutions re V 1 = 7.2 V n V 2 = 55.2 V. Clerly, the sum of these with the solutions for the first suiruit yiels the sme nswers s the originl iruit. 4. There re five noes in this iruit, lele elow. Pling the referene noe t v E reues the numer of unknowns to three, sine now v E = 0 V n v D = 5 V. Ω A B C Ω D 5V 18Ω 75Ω 1.2A E 30Ω The three nol equtions re the following: V A V C V A V B 18 V A 5 = 0 43V A 25V B 9V C = 45 V B V A 18 V C V A V B V C 75 V C V B = 0 25V A 31V B 6V C = 540 V C 30 = 0 3V A 2V B 10V C = 0 The solutions re V A = 32.8 V, V B = 47.6 V, n V C = 19.3 V. The rnh urrents n e foun using Ohm s lw n KCL: I AC = 0.27 A, I AB = 0.82 A, I BC = 0.38 A, I AD = 0.56 A, I CE = 0.64 A, I DE = 0.56 A. 3

4 5. The iruit hs four unknown noes, lele elow. 100Ω 0Ω 1V V A V B V C 5V 10 V x 80Ω V x 5Ω V x 90Ω In ition, we must lso write supernoe eqution t the 1 oure. The epenent urrent soure oes not present prolem, sine we tret V x s one of the unknowns. The nol equtions t V A n V x re V A V A 80 V A V B 0 10V x = 0 V x V C 5 The two equtions t the supernoe re V x 90 = 0 V B V A 0 V C V x 5 V B V C = 1 The solution to these equtions is V A = V, V B = V, V C = V, V x = V. 6. Note tht the left prt of the iruit is inepenent of the right prt. Sine there is no omplete loop onneting the two prts, no urrent n flow etween them; the wire onneting them only serves s ommon groun. The equivlent resistne on the left is = 9 Ω. So the totl urrent from is just Vs. Also, we hve tht I 9 x is just hlf of tht, sine the two prllel rnhes hve equl resistne. So I x = Vs. 18 On the right hlf, the two resistors re in prllel, so the equivlent resistne is just 2 Ω. Sine the urrent going through them is given y AI x = AVs, the voltge ross 18 them is V out = I eq = AVs Vout. Solving for A n plugging in 9 V in = 9, we hve tht A = The steps to fining the equivlent iruits re to fin V o, I s, n/or eq. Sine there is epenent soure in this iruit, it is esier to fin the first two. V o is simply the voltge ifferene etween n. By inspetion, this is just = 0 V o = (100 kω)βi x = (100 kω)β 10 kω = 10β I s is foun y shorting the terminls n solving for I. Note tht shorting out the 100 kω resistor will leve us with I = βi x = β Vs (note the sign!). 10 kω 4

5 Thus, our esire vlues re V T h = 10β n I N = βvs. Also, the Thévenin n 10 kω Norton resistne is just given y T h = N = V T h I N = 100 kω. 8. () The mximum power trnsfer theorem tells us tht the lo resistne shoul e equl to the Thévenin resistne of the iruit. In this se, it is trivil to see tht tht L = T h = s. If you wnt to review the erivtion of this result, plese onsult the leture notes or textook. () We wnt the power ross the lo, P L, to e equl to 80% of the power elivere y the voltge soure, P s. The former n e foun using P L = I 2 L, wheres the ltter is P s = I. In this iruit, I = So putting everything together, we hve Solving for L gives us L = 4 s. s L Vs 2 L ( s L ) = 0.8 Vs 2 2 s L () Power is epenent on oth voltge n urrent. While the nswer to prt () gives us higher voltge ross L n greter frtion of the totl power, the totl power is tully muh smller in this se. This is euse the urrent is 5 times smller ue to the greter resistne, while the voltge ross L is only 1.6 times lrger. 9. Intuitively we know tht losing the swith puts 2 n 3 in prllel, lowering the effetive resistne. So it must e in the open se tht v o tkes the greter voltge vlue, 26 V. In this senrio, the iruit looks like the following: Ω 39V 2 v o = 26V Solving for this iruit, using voltge ivier, nol nlysis, or otherwise, gives us 2 = 100 Ω. Now if we lose the swith, v o = 24 V. The vlue of 2 remins unhnge, so ll tht s remining is to solve for 3. Ω 39V 100 Ω v o 3 5

6 We n write nol eqution t v o = 24 V: The solution is 3 = 400 Ω. v o 39 v o 100 v o 3 = () When the rige is lne, no urrent flows ross the glvnometer. By KCL, the urrent ross 1 n 2 re the sme, s well s the urrent ross 3 n 4. In ition, the voltges v n v must e equl ue to the 0 urrent. 1 G 3 I 1 I 2 2 x By KVL, the voltge rops ross 1 n 3 must then e the sme. The sme resoning goes for 2 n x. I 1 1 = I 2 3 I 1 2 = I 2 x Sustituting for the urrents n solving for x, we hve x = () Given our vlues for 1 n 3, we n mesure x = 2 2. Sine the highest vlue tht 2 n tke is 1000 Ω, the mximum resistne tht n e mesure for x is just 0 Ω. As 2 omes in inrements of 10 Ω, the mesurement n e me with n ury to within 20 Ω. () The Thévenin equivlent is foun y fining the voltge n equivlent resistne ross the open terminls n. Note tht here we nnot ssume the rige is lne, s we re looking t the generl se. v o is simply the ifferene in potentil etween noes n, oth of whih n e foun vi voltge ivier: 2 v = 1 2 v = 3 x ( 2 V T h = v v = ) x x The equivlent resistne n e foun y zeroing out the voltge soure, mking it short, n fining eq. 6 x

7 1 3 2 x We hve 1 n 2 in prllel with eh other, n 3 n x in prllel with eh other. These two pirs re in series, so the equivlent resistne is T h = ( 1 2 ) ( 3 x ). 7

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