Solutions manual for electrical engineering principles and applications 5th edition by hambley download

Solutions Manual for Electrical Engineering Principles and Applications 5th Edition by Allan R.Hambley CHAPTER Resistive Circuits 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 + 1/R2 + 1/1R3 + 1/R4 = Ω (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: Req= 1/R1 +1/[R2 +11/(1/R3 +1/R4 )] = Ω (c) R1 and R2 are in parallel Furthermore, R3, and R4 are in parallel Finally, the two parallel combinations are in series Req= 1/R1 +11/R2 + 1/R3 +1 1/R4 = 52.1 Ω (d) R1 and R2 are in series Furthermore, R3 is in parallel with the series combination of R1 and R2 Req= 1/R3 +11/(R1 +R2 ) = 1.5 kΩ 24 E2.2 (a) First we combine R2, R3, and R4 in parallel Then R1 is in series with the parallel combination 1/ 20 V 20 Req = / = 10 9.231 =1.04 A / =9 231 Ω i = R +Req + R 2+ R3 + R4 veq=Reqi1 = 9.600 V i2 =veq/R2 = 0.480A i3 =veq/R3 = 0.320 A i4 =veq/R4 = 0.240 A (b) R1 and R2 are in series Furthermore, R3, and R4 are in series Finally, the two series combinations are in parallel Req =R +R =20 Ω Req =R +R =20 Ω Req = 1 /Req +1 /Req =10 Ω veq= 2×Req= 20 V i1 =veq/Req1 = 1A i2 =veq/Req2 = 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 25 R eq v =20 Ω i = s =1 A = eq 1 /Req +1 /R R +Req = 20 V i2 =v2 /R2 = 0.5A i3 =v2 /Req1 = 0.5 A =R +R = 40 Ω R v2 =i1Req2 R R E2.3 (a) v1 =vsR1 +R2 +1R3 +R4 = 10 V v2 =vsR1 +R2 +2R3 +R4 = 20 V Similarly, we find v3 = 30Vand v4 = 60V (b) First combine R2 and R3 in parallel: Req= (1/R2 +1R3) = 2.917 Ω Then we have v1 =vsR1 +R R eq +R4 = 6.05 V Similarly, we find 26 E2.4 (a) First combine R1 and R2 in series: Req = R1 + R2 = 30 Ω Then we have i1 =isR3R+3Req= 1515+30 = A and i3 =isR3R+eqReq= 1530+30 = A (b) The current division principle applies to two resistances in parallel Therefore, to determine i1, first combine R2 and R3 in parallel: Req = R eq 1/(1/R2 + 1/R3) = Ω Then we have i1 =isR1 + Req= 10 + = A Similarly, i2 = A and i3 = A E2.5 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 E2.6 Node 1: v1 −1v3 +v1 −2v2 =ia R Node 2: v2R−2v1 +vR23 +v2R−4v3 = R Node 3: v3 +v3R−4v2 +v3R−1v1 +ib= R5 E2.7 Following the step-by-step method in the book, we obtain 1+1 −1 R1 R R2 v1 vv23 −R12 R12 +R13 +R14−R141 1 − R4 is R4 R5 27 i0s E2.8 Instructions for various calculators vary The MATLAB solution is given in the book following this exercise eq R v2 =vsR +Req+R4 = 5.88 Vand v4 = 8.07 V 28 E2.9 (a) Writing the node equations we obtain: Node 1: v1 −v3 +v +v1 −v2 = 2010 Node 2: v2 −v1 +10 +v2 −v3 = 10 Node 3: v3 −v1 +v3 +v3 −v2 = 20 10 (b) Simplifying the equations we obtain: 0.35v1 − 0.10v2 − 0.05v3 = − 0.10v1 + 0.30v2 − 0.20v3 =−10 − 0.05v1 − 0.20v2 + 0.35v3 = (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: − 0.2 0.5 3+ 0.2 = 10.32 V v1 = 0.7 − 0.2 = 0.35 − 0.04 − 0.2 0.5 0.7 29 0.7 +1.2 − 0.2 = 6.129 V v2 = 0.7 − 0.2 = 0.35 − 0.04 − 0.2 0.5 Many other methods exist for solving linear equations E2.11 First write KCL equations at nodes and 2: Node 1: v1 −10 +v +v1 −v2 = 210 Node 2: v2 −10 +v2 +v2 −v1 = 10 10 Then, simplify the equations to obtain: 8v1 −v2 = 50 and −v1 + 4v2 = 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 30 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: −v1 +10 +v2 = Then write KCL equations First for a supernode enclosing the 10-V source, we have: v1 +v1 −v3 +v2 −v3 = R1 R2 R3 Node 3: v +v −v +v −v = R4 R2 Reference node: R3 v1 +v3 = 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 10 v1 +v1 − +1 = R1 R2 Substituting values for the resistances and solving, we find v1 = 3.33 V 31 Then we have ia = 10 −v =1.333 A R2 (b) Select the reference node and assign node voltages as shown Then write KCL equations at nodes and v1 −25 +v1 +v1 −v2 = R2 R4 R3 v2 −25 +v2 −v1 +v2 = R1 R3 R5 Substituting values for the resistances and solving, we find v1 = 13.79 V and v E2.15 v −v = 18.97 V Then we have ib= 1R = -0.259 A (a) Select the reference node and node voltage as shown Then write a KCL equation at node 1, resulting in v +v1 −5 10 −2ix= Then use ix= (10 −v1)/5 to substitute and solve We find v1 = 7.5 V Then we have ix= 10 −v = 0.5 A (b) Choose the reference node and node voltages shown: 32 Then write KCL equations at nodes and 2: v +v v +v −2iy+3 = −2iy = 10 Finally use iy=v2 /5 to substitute and solve This yields v2 = 11.54V and iy= 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) 33 P2.87 The equivalent circuit with a load attached is: For a load of 2.2 kΩ, we have iL= 4.4/2200 = 2mA, and we can write vL=Vt−RtiL Substituting values this becomes 4.4 =Vt− 0.002Rt (1) Similarly, for the 10-kΩ load we obtain =Vt− 0.0005Rt (2) Solving Equations (1) and (2), we find Vt= 5.2V andRt= 400 Ω P2.88 Open-circuit conditions: ix = 15 −vx v x 10 + 10 −ix+ 0.5ix = Solving, we find vx = 10V and then we have Vt =voc=vx10 Short-circuit conditions: 10 + 10 = V 79 15 ix = we have i sc = vx −v x v x 10 −ix+ 0.5ix = Solving, we find vx = 7.5 V and then 10 t equivalents are: = 0.75 A Then, we have R=vocisc= 6.67 Ω Thus the P2.89 As is Problem P2.80, we find the Thévenin equivalent: Then maximum power is obtained for a load resistance equal to the Thévenin resistance Pmax =v(tRt2)2 = 3.333 W P2.90 As in Problem P2.82, we find the Thévenin equivalent: 80 Then, maximum power is obtained for a load resistance equal to the Thévenin resistance Pmax =V(tRt2)2 = 7.2 W P2.91* To maximize the power to RL, we must maximize the voltage across it Thus, we need to have Rt = The maximum power is Pmax = 2052 = 80 W P2.92 The circuit is By the current division principle: iL=InRLR+tRt The power delivered to the load is PL= ( )iL 2RL= (In)2 (R( )RLt+2RRtL)2 Taking the derivative and setting it equal to zero, we have 81 dRdPLL = = (In)2 ( )Rt − (Rt+RL(R) t + R( ) R ) RL(Rt+RL ) Lt which yields RL=Rt The maximum power is PLmax = (In)2Rt P2.93 For maximum power conditions, we have RL=Rt The power taken from the voltage source is Ps =RtV( )t+R2L =V( )2tRt2 Then, half of Vt appears across the load and the power delivered to the load is ( PL= 0.R5Vtt)2 Thus, the percentage of the power taken from the source that is delivered to the load is L η=P ×100% = 50% Ps On the other hand, for RL= 9Rt, we have Ps =RtV( )t+R2L =V10( )tRt2 ( PL= 09.9RVtt)2 L η=P ×100% = 90% Ps Design for maximum power transfer is relatively inefficient Thus, systems in which power efficiency is important are almost never designed for maximum power transfer 82 P2.94* First, we zero the current source and find the current due to the voltage source iv =30 15 =2 A Then, we zero the voltage source and use the current-division principle to find the current due to the current source 10 =2 A +10 Finally, the total current is the sum of the contributions from each source i c =3 i=iv+ic= A P2.95* The circuits with only one source active at a time are: Req = =3.75 Ω +1 15 i s ,c = −1 10 = −0 667 A 10 +5 is,v=−10Req V =−2.667 A Then the total current due to both sources is is =is,v+is,c=−3.333A 83 P2.96 Zero the 2-A source and use the current-division principle: i1,1A= 11520+ 20 = 0.5714 A Then zero the A source and use the current-division principle: i 1,2A = −2 30 = −0.2857 A + Finally, i =i 1,1A +i 1,2A = 2857 A P2.97 The circuits with only one source active at a time are: i 1.4A = × =3 A 15 i 1,2A = −2 × 15 +5 = −1.5 15 15 +5 Finally, we add the components to find the current with both sources active 84 i1 =i1,4A +i1,2A = 1.5A P2.98 The circuit, assuming that v2 = 1V is: i (2 ) = v = 0.2 A v = 30 = V i i 10 =v 10 = 0.6 A i 30 =v 30 = 0.2 A i s i i i 30 = A v 12i v 6i 24 V = + + = s s s = + 10 + We have established that for vs= 24V, we have v2 = V Thus, for vs= 12 V, we have: v2 = 1× = 0.5 V P2.99 We start by assuming i2 = A and work back through the circuit to determine the value of vs The results are shown on the circuit diagram 85 However, the circuit actually has vs = 10 V, so the actual value ofi2 is ×(1 A) = 0.5 A P2.100 We start by assuming i6 = A and work back through the circuit to determine the value of Is This results in Is = -4.5 A However, the circuit actually has Is = 10 A, so the actual value ofi6 is ×(1 A) = − 2.222 A P2.101 (a) With only the 2-A source activated, we have i2 = andv = 3( )i2 = 12 V (b) With only the 1-A source activated, we have i1 = A andv = 3( )i1 = V (c) With both sources activated, we have i= A andv = 3( )i = 27 V Notice that i≠i1 +i2 Superposition does not apply because device A has a nonlinear relationship between v and i P2.102 From Equation 2.90, we have (a) Rx=RR21 R3 = 101 k kΩΩ×3419 = 341.9 Ω (b) Rx=RR21 R3 = 10010 k kΩΩ×3419 = 34.19 kΩ 86 P2.103* (a) Rearranging Equation 2.90, we have R3 =RR21 Rx= 101044 ×5932 = 5932 Ω (b) The circuit is: The Thévenin resistance is 1 Rt= 1R3 +1R1 + 1R2 + 1Rx = 7447 Ω The Thévenin voltage is vt=vsR1R+3R3 −vsRxR+xR2 = 0.3939 mV Thus, the equivalent circuit is: Vt i detector =R t +R detector = 31 65 10 −9 A × Thus, the detector must be sensitive to very small currents if the bridge is to be accurately balanced 87 P2.104 If R1 and R3 are too small, large currents are drawn from the source If the source were a battery, it would need to be replaced frequently Large power dissipation could occur, leading to heating of the components and inaccuracy due to changes in resistance values with temperature If R1 and R3 are too large, we would have very small detector current when the bridge is not balanced, and it would be difficult to balance the bridge accurately P2.105 With the source replaced by a short circuit and the detector removed, the Wheatstone bridge circuit becomes The Thévenin resistance seen looking back into the detector terminals is Rt= 1R3 1+1R1 + 1R2 +11Rx The Thévenin voltage is zero when the bridge is balanced 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: Req=R1 + 1/R2 + 1/1R3 + 1/R4 = 16 Ω 88 is=Rveqs = A Then, using the current division principle, we have i4 =G2 +GG34 +G4 is= 1/48 +11//6016 + 1/60 = A T2.3 Writing KCL equations at each node gives v +v1 −v2 +v1 −v3 = v2 −v1 +v2 = 10 v3 +v3 −v1 =−2 In standard form, we have: 0.95v1 − 0.20v2 − 0.50v3 = − 0.20v1 + 0.30v2 = − 0.50v1 + 1.50v3 =−2 In matrix form, we have GV = I 0.95 − 0.20 − 0.50 v1 89 0.20 0.30 0.50 0 1.50 v2 v3 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 =Is 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 Thus, the short-circuit current is A flowing out of b and into a Zeroing the sources, we have 90 Thus, the Thévenin resistance is Rt= 1/40 + 1/(1 30 + 30) = 24 Ω and the Thévenin voltage is Vt=IscRt= 24V 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 91 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=Ri2 ), the superposition principle does not apply to power 92 ... Solve for the currents and voltages in the final circuit Transfer results back along the chain of equivalent circuits, solving for more currents and voltages along the way Check to see that KVL and. .. circuit, and its current is zero Thus, we can consider the 8-Ω and the 7-Ω resistances to be in series The current circulating clockwise in the left-hand loop is given by i1 15 = + = A, and we... for the voltage for both sources, thus P=−vifor both sources Pcurrent−source=−3 A×20V =−60W Pvoltage−source=−20i1 = 40 W Power is delivered by the current source and absorbed by the voltage source