I 3 2 = I I 4 = 2A


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1 ECE 210 Eletril Ciruit Anlysis University of llinois t Chigo 2.13 We re ske to use KCL to fin urrents 1 4. The key point in pplying KCL in this prolem is to strt with noe where only one of the urrents is unknown. Let us lel the noes s,, n s shown. 2A 2 7A 4 1 3A 3 4A Now, let us pply KCL t noe, sine there is only one unknown ( 2 ). Writing n expression for the sum of the urrents out of ( out i = 0), we get = 0 2 = 10A Next, we pply KCL t noe where the unknown urrent is 4. Writing n expression for the sum of the urrents in equls the sum of the urrents out ( in i = out i), we get 2 = = 2A Next, we pply KCL t noe where in i = out i gives 1 2 = 2 1 = 2 2 = 12A Finlly, we pply KCL t noe where in i = out i gives 7 4 = 3 3 = 7 4 = 5A 2.14 We re ske to use KVL to fin voltges V 1 V 4. The key point in pplying KVL in this prolem is to strt with loop where only one of the voltges is unknown. Let us lel the noes s,,, n e s shown to help us efine the iretion of the pth of eh loop. 3V 4V V 3 V 1 V 2 e V4 2V Now, let us pply KVL lokwise roun the outer loop , sine there is only one unknown (V 2 ). Writing n expression for the sum of the voltge rises equls the sum of the voltge rops ( rise v = rop v), we get 3 V 2 = 4 5 V 2 = 6V 5V Dr. Vhe Cliskn 1 of 6 Poste: Ferury 7, 2009
2 ECE 210 Eletril Ciruit Anlysis University of llinois t Chigo Next, we pply KVL lokwise roun the right upper loop ee where the unknown voltge is V 1. Writing n expression for the sum of the voltge rises equls zero ( rise v = 0), we get V 1 V 2 2 = 0 V 1 = 2 V 2 = 8V Next, we pply KVL roun the lower right loop e where rise v = rop v gives V 4 = 2 5 V 4 = 7V Finlly, we pply KVL lokwise roun the lower left loop e where rop v = 0 gives 4 V 3 V 4 = 0 V 3 = V 4 4 = 11V 2.22 There re numer of things we n o efore we tully egin solving for the unknowns. First of ll, note tht the voltgeontrolle urrent soure epens on the voltge ross the 4Ω resistor tht hs een enote s V 0. The epenent soure hs urrent of vlue 2V 0 flowing upwrs; therefore, one we fin numeril vlue for V 0, we n etermine numeril vlue for the urrent of the epenent soure. Next, we introue new urrent 0 s enote in the figure elow. Aoring to Ohm s Lw, this urrent is relte to the voltge V 0 given y 0 = V 0 /4. Sine the 4Ω n the 6Ω resistors re in series, the sme urrent 0 flows through the 6Ω resistor s inite. This urrent flow proues voltge rop V 6 ross the 6Ω resistor given y V 6 = 6 0. This informtion will help us in writing KCL eqution t noe whih will eventully le to the etermintion of V 0. To etermine the power issipte in the epenent urrent soure, we nee to multiply the urrent (2V 0 ) n the voltge rop in the iretion of urrent flow. We introue new vrile lle V whih enotes this voltge rop s inite on the figure. One we etermine V, it will e simple tsk to etermine the issipte power in the epenent soure. 0 V 0 0 6Ω 0 V 6 4Ω 10A V 2V 0 Given the ove informtion, we n write KCL eqution t. Setting the sum of the urrents entering noe to zero yiels V 0 = 0 However, we know tht V 0 n 0 re relte from our isussion ove. Sustituting 0 = V 0 /4 in the ove expression n solving for V 0 gives V V 0 = 0 9V 0 = 40 V 0 = 40 9 V = 4.444V Using the numeril vlue for V 0, we n lso lulte urrent 0 0 = V = 10 9 A Sine we know 0, we n etermine the voltge ross the 6Ω resistor V 6 = 6 0 V 6 = 60 9 V Dr. Vhe Cliskn 2 of 6 Poste: Ferury 7, 2009
3 ECE 210 Eletril Ciruit Anlysis University of llinois t Chigo As we mentione in the eginning, we nee voltge V in orer to lulte the power issipte in the epenent soure. Applying KVL lokwise roun the outer loop ( rise v = rop v) gives V = V 6 V 0 V = = V Therefore the power issipte in the ontrolle urrent soure is ( p iss = vi = (V )(2V 0 ) = 100 ) ( [ 9 V 2 40 ] ) A 9 = W = 98.77W 2.36 The key point with prolems of this type is the rerwing of the the originl iruit multiple times using series n prllel resistor omintions. Along the wy, we mrk urrents n or voltges tht re useful in lter lultions. To solve for urrent, we nee to etermine the equivlent iruit to the right of terminls . One we know the equivlent resistne R eq, then urrent = /R eq. Looking t the Figure 1, we note tht the 50Ω n 30Ω resistor re in series n the 60Ω n resistors re in prllel. 24Ω Ω 25Ω 30Ω V 0 60Ω Figure 1: Comining the 50Ω n 30Ω resistor in series results in single 80Ω resistor. We mrk the urrent tht is flowing through oth these resistors s 80. The urrent flowing through the 24Ω resistor is mrke s 40. Notie tht when we omine two resistors in series, the noe etween the resistors is eliminte; therefore, the voltge mrke V 0 ross the 30Ω resistor in the originl iruit is not visile in the simplifie iruit. The omintion of the 60Ω n resistors is given y = (60)(20)/(6020) = 15Ω. The resulting iruit is shown in Figure 2. 24Ω Ω 80Ω 15Ω Figure 2: Now the 25Ω n 15Ω resistors re in series resulting in 40Ω resistor. The n 80Ω resistors re in prllel; therefore, the resulting equivlent resistne is = (20)(80)/(20 80) = 16Ω. The simplifie Dr. Vhe Cliskn 3 of 6 Poste: Ferury 7, 2009
4 ECE 210 Eletril Ciruit Anlysis University of llinois t Chigo iruit is shown in Figure 3. Note tht the urrent 80 oes not pper in this simplifie iruit sine its rnh hs een omine y the prllel omintion. 24Ω 40 40Ω 16Ω Figure 3: Now the 24Ω n 16Ω resistors re in series resulting in 40Ω resistor. With this simplifition, the iruit reues to the one shown in Figure 4(). t is now simple mtter to etermine the equivlent resistor t terminls . First, we see tht the two 40Ω resistors re in prllel with the omintion equl to =. The iruit reues to the one shown in Figure 4(). Note tht the urrent mrke 40 oes not pper here s its rnh hs een eliminte in the prllel omintion Ω 40Ω 30Ω () () () Figure 4: Now we only hve two resistors left n they re in series; therefore, the equivlent resistne t terminls  is simply R eq = = 30Ω. The finl simplifie iruit is shown in Figure 4(). Using Ohm s Lw, the urrent tht is flowing out of the voltge soure is given y = R eq = 30Ω = 1 2 A = 0.5A To etermine V 0, we hve to work our wy kwrs through the equivlent iruits we hve rwn using the ft tht the urrent is known n equl to 0.5A. We will strt with the equivlent iruit shown in Figure 4() where the urrent is flowing through the resistor n will split n ontinue flowing through the two 40Ω resistors. The urrent mrke 40 n e etermine y using urrent ivision 40 = = 1 (0.5A) = 0.25A 2 Rell tht 40 through the 40Ω resistor is tully the urrent through the series omintion of the 24Ω n the 16Ω resistors s shown in Figure 3. We lso see tht the 16Ω resistor shown in Figure 3 is tully the prllel omintion of 80Ω n resistors s shown in Figure 2. From looking t Figure 2, it is ler Dr. Vhe Cliskn 4 of 6 Poste: Ferury 7, 2009
5 ECE 210 Eletril Ciruit Anlysis University of llinois t Chigo tht urrent 40 will ivie n flow into the 80Ω n resistors. The urrent flowing through the 80Ω resistor n e written using urrent ivision 80 = = 1 (0.25A) = 0.05A = 50mA 5 One gin, we relize tht the 80Ω resistor represents the series omintion of the 50Ω n 30Ω resistors s shown in Figure 1 where 80 flows through eh of these two resistors. With this informtion, the voltge V 0 ross the 30Ω resistor n e etermine using Ohm s Lw V 0 = (30Ω) 80 = (30Ω)(50mA) = 1.5V 2.39 For prolems tht require the lultion of n equivlent resistne etween terminls, one shoul first ientify ll noes in the iruit. One the noes hve een estlishe, the iruit shoul e rerwn to revel the simplifie topologil struture. With the simplifie struture, it is usully quite simple to etermine the equivlent resistne. () entify noes, n ; ssign ritrry lotions to the noes; rerw eh resistor etween their respetive noes in the new shemti. R 1 R 1 R 2 2kΩ 2kΩ 1kΩ 2kΩ R 2 1kΩ 1kΩ 1kΩ 2kΩ The equivlent resistne t terminls  is given y R eq = [ (R 1 R 2 )] = 1kΩ [2kΩ (2kΩ 1kΩ)] = 1kΩ [2kΩ 23 ] kω = 1kΩ 8 3 kω R eq = 8 kω = Ω 11 Dr. Vhe Cliskn 5 of 6 Poste: Ferury 7, 2009
6 ECE 210 Eletril Ciruit Anlysis University of llinois t Chigo () entify noes, n ; ssign ritrry lotions to the noes; rerw eh resistor etween their respetive noes in the new shemti. R 1 R 2 4kΩ 6kΩ R 2 4kΩ R 1 6kΩ The equivlent resistne t terminls  is given y R eq = R 2 [R 1 ( )] = 4kΩ [6kΩ ( )] = 4kΩ [6kΩ 6kΩ] = 4kΩ R eq = 3kΩ 2.47 We follow the sme proeure s in Prolem 2.39 n note tht noes n f re tully the sme noe. 5Ω 5Ω 6Ω 8Ω e, f 3Ω 8Ω e 3Ω f 6Ω The equivlent resistne t terminls  is given y R eq = (5Ω ) (3Ω 6Ω) 8Ω = 4Ω 2Ω 8Ω R eq = 24Ω Dr. Vhe Cliskn 6 of 6 Poste: Ferury 7, 2009
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