Chapter 02

CHAPTER Exercises E2.1 (a) R2, R3, and R4 are in parallel Furthermore R1 is in series with the combination of the other resistors Thus we have: Req = R1 + = 3Ω / R2 + / R3 + / R4 (b) R3 and R4 are in parallel Furthermore, R2 is in series with the combination of R3, and R4 Finally R1 is in parallel with the combination of the other resistors Thus we have: =5Ω Req = / R1 + /[R2 + /(1 / R3 + / R4 )] (c) R1 and R2 are in parallel Furthermore, R3, and R4 are in parallel Finally, the two parallel combinations are in series 1 + = 52.1 Ω Req = / R1 + / R2 / R3 + / R4 (d) R1 and R2 are in series Furthermore, R3 is in parallel with the series combination of R1 and R2 = 1.5 kΩ Req = / R3 + /(R1 + R2 ) E2.2 (a) First we combine R2, R3, and R4 in parallel Then R1 is in series with the parallel combination Req = = 9.231 Ω / R2 + / R3 + / R4 i1 = 20 V 20 = = 1.04 A R1 + Req 10 + 9.231 v eq = Req i1 = 9.600 V i2 = v eq / R2 = 0.480 A i4 = v eq / R4 = 0.240 A i3 = v eq / R3 = 0.320 A (b) R1 and R2 are in series Furthermore, R3, and R4 are in series Finally, the two series combinations are in parallel Req = R1 + R2 = 20 Ω v eq = × Req = 20 V Req = R3 + R4 = 20 Ω Req = i1 = v eq / Req = A / Req 1 = 10 Ω + / Req i2 = v eq / Req = A (c) R3, and R4 are in series The combination of R3 and R4 is in parallel with R2 Finally the combination of R2, R3, and R4 is in series with R1 Req = R3 + R4 = 40 Ω Req = v = i1Req = 20 V E2.3 (a) v = v s = 20 Ω / Req + / R2 i1 = vs =1A R1 + Req i2 = v / R2 = 0.5 A i3 = v / Req = 0.5 A R1 = 10 V v = v s R2 R1 + R2 + R3 + R4 R1 + R2 + R3 + R4 Similarly, we find v = 30 V and v = 60 V = 20 V (b) First combine R2 and R3 in parallel: Req = (1 / R2 + R3 ) = 2.917 Ω R1 = 6.05 V Similarly, we find R1 + Req + R4 Then we have v = v s v2 = vs E2.4 Req = 5.88 V and v = 8.07 V R1 + Req + R4 (a) First combine R1 and R2 in series: Req = R1 + R2 = 30 Ω Then we have Req R3 15 30 i1 = is = = A and i3 = is = = A R3 + Req 15 + 30 R3 + Req 15 + 30 (b) The current division principle applies to two resistances in parallel Therefore, to determine i1, first combine R2 and R3 in parallel: Req = Req 1/(1/R2 + 1/R3) = Ω Then we have i1 = is = = 1A R1 + Req 10 + Similarly, i2 = A and i3 = A E2.5 E2.6 E2.7 Write KVL for the loop consisting of v1, vy , and v2 The result is -v1 - vy + v2 = from which we obtain vy = v2 - v1 Similarly we obtain vz = v3 - v1 v1 − v3 v1 − v2 v − v1 v2 v2 − v3 + = ia Node 2: + + =0 R1 R2 R3 R4 R2 v v − v2 v3 − v1 Node 3: + + + ib = R5 R4 R1 Node 1: Following the step-by-step method in the book, we obtain 1  +  R1 R2  −  R2    E2.8 − R2 + R2 R3 − R4    v   − is     v =  − R4     1  v   is  +  R4 R5  + R4 Instructions for various calculators vary The MATLAB solution is given in the book following this exercise E2.9 (a) Writing the node equations we obtain: v − v3 v1 v1 − v2 Node 1: + + =0 20 10 v −v3 v − v1 Node 2: + 10 + =0 10 v − v1 v3 v3 − v2 Node 3: + + =0 20 10 (b) Simplifying the equations we obtain: 0.35v − 0.10v − 0.05v = − 0.10v + 0.30v − 0.20v = −10 − 0.05v − 0.20v + 0.35v = (c) and (d) Solving using Matlab: >>clear >>G = [0.35 -0.1 -0.05; -0.10 0.30 -0.20; -0.05 -0.20 0.35]; >>I = [0; -10; 0]; >>V = G\I V= -27.2727 -72.7273 -45.4545 >>Ix = (V(1) - V(3))/20 Ix = 0.9091 E2.10 Using determinants we can solve for the unknown voltages as follows: − + v1 = = = 10.32 V 0.7 − 0.2 0.35 − 0.04 − v2 = − + = = 6.129 V 0.7 − 0.2 0.35 − 0.04 − Many other methods exist for solving linear equations E2.11 First write KCL equations at nodes and 2: Node 1: v − 10 v v1 − v2 =0 10 v − 10 v v − v Node 2: + + =0 10 10 + + Then, simplify the equations to obtain: 8v − v = 50 and − v + 4v = 10 Solving manually or with a calculator, we find v1 = 6.77 V and v2 = 4.19 V The MATLAB session using the symbolic approach is: >> clear [V1,V2] = solve('(V1-10)/2+(V1)/5 +(V1 - V2)/10 = 0' , '(V2-10)/10 +V2/5 +(V2-V1)/10 = 0') V1 = 210/31 V2 = 130/31 Next, we solve using the numerical approach >> clear G = [8 -1; -1 4]; I = [50; 10]; V = G\I V= 6.7742 4.1935 E2.12 The equation for the supernode enclosing the 15-V source is: v3 − v2 v3 − v1 v1 v2 + = + R3 R1 R2 R4 This equation can be readily shown to be equivalent to Equation 2.37 in the book (Keep in mind that v3 = -15 V.) E2.13 Write KVL from the reference to node then through the 10-V source to node then back to the reference node: − v + 10 + v = Then write KCL equations First for a supernode enclosing the 10-V source, we have: v1 v1 − v3 v2 − v3 + + =1 R1 R2 R3 Node 3: v3 v3 − v1 v3 − v2 + + =0 R4 R2 R3 Reference node: v1 v3 + =1 R1 R4 An independent set consists of the KVL equation and any two of the KCL equations E2.14 (a) Select the reference node at the left-hand end of the voltage source as shown at right Then write a KCL equation at node v v − 10 + +1 = R1 R2 Substituting values for the resistances and solving, we find v1 = 3.33 V 10 − v Then we have ia = = 1.333 A R2 (b) Select the reference node and assign node voltages as shown Then write KCL equations at nodes and v − 25 v v − v + + =0 R2 R4 R3 v − 25 v − v v + + =0 R1 R3 R5 Substituting values for the resistances and solving, we find v1 = 13.79 V v −v2 = -0.259 A and v = 18.97 V Then we have ib = R3 E2.15 (a) Select the reference node and node voltage as shown Then write a KCL equation at node 1, resulting in v v − 10 + − 2ix = 5 Then use ix = (10 − v ) / to substitute and solve We find v1 = 7.5 V 10 − v Then we have ix = = 0.5 A (b) Choose the reference node and node voltages shown: Then write KCL equations at nodes and 2: v1 + v − 2i y +3= v2 + v − 2i y 10 =3 Finally use i y = v / to substitute and solve This yields v = 11.54 V and i y = 2.31 A E2.16 >> clear >> [V1 V2 V3] = solve('V3/R4 + (V3 - V2)/R3 + (V3 - V1)/R1 = 0', 'V1/R2 + V3/R4 = Is', 'V1 = (1/2)*(V3 - V1) + V2' ,'V1','V2','V3'); >> pretty(V1), pretty(V2), pretty(V3) R2 Is (2 R3 R1 + R4 R1 + R4 R3) R3 R1 + R4 R1 + R1 R2 + R4 R3 + R3 R2 R2 Is (3 R3 R1 + R4 R1 + R4 R3) R3 R1 + R4 R1 + R1 R2 + R4 R3 + R3 R2 Is R2 R4 (3 R1 + R3) R3 R1 + R4 R1 + R1 R2 + R4 R3 + R3 R2 E2.17 Refer to Figure 2.33b in the book (a) Two mesh currents flow through R2: i1 flows downward and i4 flows upward Thus the current flowing in R2 referenced upward is i4 - i1 (b) Similarly, mesh current i1 flows to the left through R4 and mesh current i2 flows to the right, so the total current referenced to the right is i2 - i1 (c) Mesh current i3 flows downward through R8 and mesh current i4 flows upward, so the total current referenced downward is i3 - i4 (d) Finally, the total current referenced upward through R8 is i4 - i3 E2.18 Refer to Figure 2.33b in the book Following each mesh current in turn, we have R1i1 + R2 (i1 − i4 ) + R4 (i1 − i2 ) − v A = R5i2 + R4 (i2 − i1 ) + R6 (i2 − i3 ) = R7i3 + R6 (i3 − i2 ) + R8 (i3 − i4 ) = R3i4 + R2 (i4 − i1 ) + R8 (i4 − i3 ) = In matrix form, these equations become − R4 − R2 (R1 + R2 + R4 )  i1  v A    i    − R4 − R6 (R4 + R5 + R6 )    =     i3    − R6 − R8 (R6 + R7 + R8 )      − R2 − R8 (R2 + R3 + R8 ) i     E2.19 We choose the mesh currents as shown: Then, the mesh equations are: 5i + 10(i1 − i2 ) = 100 and 10(i2 − i1 ) + 7i2 + 3i2 = Simplifying and solving these equations, we find that i1 = 10 A and i2 = A The net current flowing downward through the 10-Ω resistance is i1 − i2 = A To solve by node voltages, we select the reference node and node voltage shown (We not need to assign a node voltage to the connection between the 7-Ω resistance and the 3-Ω resistance because we can treat the series combination as a single 10-Ω resistance.) The node equation is (v − 10) / + v / 10 + v / 10 = Solving we find that v1 = 50 V Thus we again find that the current through the 10-Ω resistance is i = v / 10 = A Combining resistances in series and parallel, we find that the resistance “seen” by the voltage source is 10 Ω Thus the current through the source and 5-Ω resistance is (100 V)/(10 Ω) = 10 A This current splits equally between the 10-Ω resistance and the series combination of Ω and Ω E2.20 First, we assign the mesh currents as shown Then we write KVL equations following each mesh current: 2(i1 − i3 ) + 5(i1 − i2 ) = 10 5i2 + 5(i2 − i1 ) + 10(i2 − i3 ) = 10i3 + 10(i3 − i2 ) + 2(i3 − i1 ) = Simplifying and solving, we find that i1 = 2.194 A, i2 = 0.839 A, and i3 = 0.581 A Thus the current in the 2-Ω resistance referenced to the right is i1 - i3 = 2.194 - 0.581 = 1.613 A E2.21 Following the step-by-step process, we obtain − R3 − R2  i1   v A  (R2 + R3 )  −R  i  =  − v  (R3 + R4 )     B   − R2 (R1 + R2 ) i3   v B  10 E2.22 Refer to Figure 2.39 in the book In terms of the mesh currents, the current directed to the right in the 5-A current source is i1, however by the definition of the current source, the current is A directed to the left Thus, we conclude that i1 = -5 A Then we write a KVL equation following i2, which results in 10(i2 − i1 ) + 5i2 = 100 E2.23 Refer to Figure 2.40 in the book First, for the current source, we have i2 − i1 = Then, we write a KVL equation going around the perimeter of the entire circuit: 5i1 + 10i2 + 20 − 10 = Simplifying and solving these equations we obtain i1 = -4/3 A and i2 = -1/3 A E2.24 (a) As usual, we select the mesh currents flowing clockwise around the meshes as shown Then for the current source, we have i2 = -1 A This is because we defined the mesh current i2 as the current referenced downward through the current source However, we know that the current through this source is A flowing upward Next we write a KVL equation around mesh 1: 10i1 − 10 + 5(i1 − i2 ) = Solving, we find that i1 = 1/3 A Referring to Figure 2.30a in the book we see that the value of the current ia referenced downward through the Ω resistance is to be found In terms of the mesh currents, we have ia = i1 − i2 = / A 11 (b) As usual, we select the mesh currents flowing clockwise around the meshes as shown Then we write a KVL equation for each mesh − 25 + 10(i1 − i3 ) + 10(i1 − i2 ) = 10(i2 − i1 ) + 20(i2 − i3 ) + 20i2 = 10(i3 − i1 ) + 5i3 + 20(i3 − i2 ) = Simplifying and solving, we find i1 = 2.3276 A, i2 = 0.9483 A, and i3 = 1.2069 A Finally, we have ib = i2 - i3 = -0.2586 A E2.25 (a) KVL mesh 1: − 10 + 5i1 + 5(i1 − i2 ) = For the current source: i2 = −2ix However, ix and i1 are the same current, so we also have i1 = ix Simplifying and solving, we find ix = i1 = 0.5 A (b) First for the current source, we have: i1 = A Writing KVL around meshes and 3, we have: 2(i2 − i1 ) + 2i y + 5i2 = 10(i3 − i1 ) + 5i3 − 2i y = However i3 and iy are the same current: i y = i3 Simplifying and solving, we find that i3 = i y = 2.31 A 12 E2.26 Under open-circuit conditions, A circulates clockwise through the current source and the 10-Ω resistance The voltage across the 10-Ω resistance is 50 V No current flows through the 40-Ω resistance so the open circuit voltage is Vt = 50 V With the output shorted, the A divides between the two resistances in parallel The short-circuit current is the current through the 40-Ω 10 = A Then, the Thévenin resistance is resistance, which is isc = 10 + 40 Rt = v oc / isc = 50 Ω E2.27 Choose the reference node at the bottom of the circuit as shown: Notice that the node voltage is the open-circuit voltage Then write a KCL equation: v oc − 20 v oc + =2 20 Solving we find that voc = 24 V which agrees with the value found in Example 2.17 E2.28 To zero the sources, the voltage sources become short circuits and the current sources become open circuits The resulting circuits are : 13 (a) Rt = 10 + (c) Rt = E2.29 = 14 Ω / + / 20 + 10 + 1 (b) Rt = 10 + 20 = 30 Ω =5Ω (1 / + / 20) (a) Zero sources to determine Thévenin resistance Thus Rt = = 9.375 Ω / 15 + / 25 Then find short-circuit current: I n = isc = 10 / 15 + = 1.67 A 14 (b) We cannot find the Thévenin resistance by zeroing the sources, because we have a controlled source Thus, we find the open-circuit voltage and the short-circuit current v oc − 2v x v oc =2 v oc = 3v x 10 30 Solving, we find Vt = v oc = 30 V + Now, we find the short-circuit current: 2v x + v x = ⇒ vx = Therefore isc = A Then we have Rt = v oc / isc = 15 Ω E2.30 First, we transform the 2-A source and the 5-Ω resistance into a voltage source and a series resistance: 15 10 + 10 = 1.333 A 15 From the original circuit, we have i1 = i2 − 2, from which we find i1 = −0.667 A Then we have i2 = The other approach is to start from the original circuit and transform the 10-Ω resistance and the 10-V voltage source into a current source and parallel resistance: = 3.333 Ω / + / 10 The current flowing upward through this resistance is A Thus the voltage across Req referenced positive at the bottom is 3.333 V and i1 = −3.333 / = −0.667 A Then from the original circuit we Then we combine the resistances in parallel Req = have i2 = + i1 = 1.333 A, as before E2.31 Refer to Figure 2.62b We have i1 = 15 / 15 = A Refer to Figure 2.62c Using the current division principle, we have i2 = −2 × = −0.667 A (The minus sign is because of the reference + 10 direction of i2.) Finally, by superposition we have iT = i1 + i2 = 0.333 A E2.32 With only the first source active we have: Then we combine resistances in series and parallel: Req = 10 + = 13.75 Ω / + / 15 Thus, i1 = 20 / 13.75 = 1.455 A, and v = 3.75i1 = 5.45 V 16 With only the second source active, we have: Then we combine resistances in series and parallel: Req = 15 + = 18.33 Ω / + / 10 Thus, is = 10 / 18.33 = 0.546 A, and v = 3.33is = 1.818 V Then, we have i2 = ( −v ) / 10 = −0.1818 A Finally we have vT = v + v = 5.45 + 1.818 = 7.27 V and iT = i1 + i2 = 1.455 − 0.1818 = 1.27 A Answers for Selected Problems P2.1* (a) Req = 20 Ω P2.2* Rx = Ω P2.3* Rab = 10 Ω (b) Req = 23 Ω P2.4* 17 P2.5* Rab = 9.6 Ω P2.23* i1 = A i2 = 0.5 A P2.24* v1 = V v = V P2.25* v = 140 V; i = A P2.34* i1 = 1.5 A i2 = 0.5 A P4A = 30 W delivering P2A = 15 W absorbing P5 Ω = 11.25 W absorbing P15 Ω = 3.75 W absorbing P2.35* i1 = 2.5 A i2 = 0.8333 A P2.36* v1 = V v2 = V P2.37* i1 = A i2 = A P2.38* v = 3.333 V P2.43* Rg = 25 mΩ P2.48* v = 14.29 V v = 11.43 V i1 = 0.2857 A P2.49* v = 6.667 V v = −3.333 V is = −3.333 A P2.56* v1 = V v2 = V ix = 0.4 A P2.57* v = 5.405 V v = 7.297 V P2.65* i1 = 2.364 A i2 = 1.818 A P2.66* v = 0.500 V P =6 W v = 13 V P = 4.471 W 18 P2.67* i1 = 0.2857 A P2.80* P2.81* Rt = 50 Ω P2.91* Rt = Pmax = 80 W P2.94* iv = A ic = A P2.95* is = −3.333 A P2.103* R3 = 5932 Ω i = iv + ic = A idetector = 31.65 × 10 −9 A Practice Test T2.1 (a) 6, (b) 10, (c) 2, (d) 7, (e) 10 or 13 (perhaps 13 is the better answer), (f) or (perhaps is the better answer), (g) 11, (h) 3, (i) 8, (j) 15, (k) 17, (l) 14 T2.2 The equivalent resistance seen by the voltage source is: = 16 Ω Req = R1 + / R2 + / R3 + / R4 is = vs = 6A Req Then, using the current division principle, we have G4 / 60 i4 = is = 6=1 A G2 + G + G / 48 + / 16 + / 60 T2.3 Writing KCL equations at each node gives 19 v1 + v1 − v2 v2 − v1 v3 5 + v2 10 + v1 − v3 =0 =2 v3 − v1 = −2 In standard form, we have: 0.95v − 0.20v − 0.50v = − 0.20v + 0.30v = − 0.50v + 1.50v = −2 + In matrix form, we have GV = I  0.95 − 0.20 − 0.50 v     − 0.20 0.30  v  =         − 0.50 1.50  v   − 2 The MATLAB commands needed to obtain the column vector of the node voltages are G = [0.95 -0.20 -0.50; -0.20 0.30 0; -0.50 1.50] I = [0; 2; -2] V = G\I % As an alternative we could use V = inv(G)*I Actually, because the circuit contains only resistances and independent current sources, we could have used the short-cut method to obtain the G and I matrices T2.4 We can write the following equations: KVL mesh 1: R1i1 −Vs + R3 (i1 − i3 ) + R2 (i1 − i2 ) = KVL for the supermesh obtained by combining meshes and 3: R4i2 + R2 (i2 − i1 ) + R3 (i3 − i1 ) + R5i3 = KVL around the periphery of the circuit: R1i1 −Vs + R4i2 + R5i3 = Current source: i2 − i3 = I s A set of equations for solving the network must include the current source equation plus two of the mesh equations The three mesh equations are dependent and will not provide a solution by themselves T2.5 Under short-circuit conditions, the circuit becomes 20 Thus, the short-circuit current is A flowing out of b and into a Zeroing the sources, we have Thus, the Thévenin resistance is Rt = = 24 Ω / 40 + /(30 + 30) and the Thévenin voltage is Vt = I sc Rt = 24 V The equivalent circuits are: Because the short-circuit current flows out of terminal b, we have oriented the voltage polarity positive toward b and pointed the current source reference toward b T2.6 With one source active at a time, we have 21 Then, with both sources active, we have We see that the 5-V source produces 25% of the total current through the 5-Ω resistance However, the power produced by the 5-V source with both sources active is zero Thus, the 5-V source produces 0% of the power delivered to the 5-Ω resistance Strange, but true! Because power is a nonlinear function of current (i.e., P = Ri ), the superposition principle does not apply to power 22