Solution manual fundamentals of electric circuits

Chapter 1, Solution (a) q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C (b) q = 24x1018 x [-1.602x10-19 C] = -0.19865 C (c) q = 2.46x1019 x [-1.602x10-19 C] = -3.941 C (d) q = 1.628x1020 x [-1.602x10-19 C] = -26.08 C Chapter 1, Solution (a) (b) (c) (d) (e) i = dq/dt = mA i = dq/dt = (16t + 4) A i = dq/dt = (-3e-t + 10e-2t) nA i=dq/dt = 1200π cos 120π t pA i =dq/dt = − e −4t (80 cos 50 t + 1000 sin 50 t ) µ A Chapter 1, Solution (a) q(t) = ∫ i(t)dt + q(0) = (3t + 1) C (b) q(t) = ∫ (2t + s) dt + q(v) = (t + 5t) mC (c) q(t) = ∫ 20 cos (10t + π / ) + q(0) = (2sin(10t + π / 6) + 1) µ C (d) 10e -30t ( −30 sin 40t - 40 cos t) 900 + 1600 = − e - 30t (0.16cos40 t + 0.12 sin 40t) C q(t) = ∫ 10e -30t sin 40t + q(0) = Chapter 1, Solution q = ∫ idt = ∫ = 10 −5 5sin π t dt = cos 6π t 6π (1 − cos 0.06π ) = 4.698 mC 6π Chapter 1, Solution q = ∫ idt = ∫ = e dt mC = - e -2t (1 − e ) mC = 490 µC Chapter 1, Solution (a) At t = 1ms, i = dq 80 = = 40 mA dt (b) At t = 6ms, i = dq = mA dt (c) At t = 10ms, i = dq 80 = = - 20 mA dt Chapter 1, Solution 25A, dq  i= = - 25A, dt   25A, -2t 0[...]... (a) i= Chapter 1, Solution 37 q = 5 × 10 20 (− 1 602 × 10 −19 ) = −80 1 C W = qv = −80 1 × 12 = − 901.2 J Chapter 1, Solution 38 P = 10 hp = 7460 W W = pt = 7460 × 30 × 60 J = 13.43 × 106 J Chapter 1, Solution 39 p = vi → i = p 2 × 10 3 = = 16.667 A v 120 Chapter 2, Solution 1 v = iR i = v/R = (16/5) mA = 3.2 mA Chapter 2, Solution 2 p = v2/R → R = v2/p = 14400/60 = 240 ohms Chapter 2, Solution 3 R =... Solution 3 R = v/i = 120/(2.5x10-3) = 48k ohms Chapter 2, Solution 4 (a) (b) i = 3/100 = 30 mA i = 3/150 = 20 mA Chapter 2, Solution 5 n = 9; l = 7; b = n + l – 1 = 15 Chapter 2, Solution 6 n = 12; l = 8; b = n + l –1 = 19 Chapter 2, Solution 7 7 elements or 7 branches and 4 nodes, as indicated 30 V 1 20 Ω 2 3 ++++ - 2A 30 Ω 60 Ω 4 40 Ω 10 Ω Chapter 2, Solution 8 12 A a i1 b 8A i3 i2 12 A c At node a, At...Chapter 1, Solution 33 i= dq → q = ∫ idt = 2000 × 3 × 10 3 = 6 C dt Chapter 1, Solution 34 (b) Energy = ∑ pt = 200 x 6 + 800 x 2 + 200 x 10 + 1200 x 4 + 200 x 2 = 10,000 kWh (c) Average power = 10,000/24 = 416.67 W Chapter 1, Solution 35 ( a) W = ∫ p( t ) dt = 400 × 6 + 1000 × 2 + 200 × 12 × 1200 × 2 + 400 × 2 = 7200 + 2800 = 10.4 kWh ( b) 10.4 kW = 433.3 W/h 24 h Chapter 1, Solution 36 160A... = -3A Chapter 2, Solution 9 Applying KCL, i1 + 1 = 10 + 2 1 + i2 = 2 + 3 i2 = i3 + 3 i1 = 11A i2 = 4A i3 = 1A Chapter 2, Solution 10 2 4A 1 -2A i2 i1 3 3A At node 1, At node 3, 4 + 3 = i1 3 + i2 = -2 i1 = 7A i2 = -5A Chapter 2, Solution 11 Applying KVL to each loop gives -8 + v1 + 12 = 0 -12 - v2 + 6 = 0 10 - 6 - v3 = 0 -v4 + 8 - 10 = 0 v1 = 4v v2 = -6v v3 = 4v v4 = -2v Chapter 2, Solution 12 + 15v... 4+6 Chapter 2, Solution 28 We first combine the two resistors in parallel 15 10 = 6 Ω We now apply voltage division, v1 = 14 (40) = 20 V 14 + 6 v2 = v3 = Hence, 6 (40) = 12 V 14 + 6 v1 = 28 V, v2 = 12 V, vs = 12 V Chapter 2, Solution 29 The series combination of 6 Ω and 3 Ω resistors is shorted Hence i2 = 0 = v2 v1 = 12, i1 = 12 = 3A 4 Hence v1 = 12 V, i1 = 3 A, i2 = 0 = v2 Chapter 2, Solution 30 8Ω... Chapter 2, Solution 39 (a) Req = R 0 = 0 (b) (c) (d) (e) R R + = R 2 2 Req = (R + R ) (R + R ) = 2R 2R = R Req = R R + R R = 1 Req = 3R (R + R R ) = 3R (R + R ) 2 3 3Rx R 2 =R = 3 3R + R 2  R ⋅ 2R  Req = R 2R 3R = 3R    3R  2 3Rx R 2 3 = 6R = 3R R= 2 11 3 3R + R 3 Chapter 2, Solution 40 Req = 3 + 4 (2 + 6 3) = 3 + 2 = 5Ω I= 10 10 = = 2A Re q 5 Chapter 2, Solution 41 Let R0 = combination of three... α R α = ⋅ = = 10 VS 2R 2 4 Chapter 2, Solution 25 V0 = 5 x 10-3 x 10 x 103 = 50V Using current division, I20 = 5 (0.01x50) = 0.1 A 5 + 20 V20 = 20 x 0.1 kV = 2 kV p20 = I20 V20 = 0.2 kW α = 40 Chapter 2, Solution 26 V0 = 5 x 10-3 x 10 x 103 = 50V Using current division, I20 = 5 (0.01x50) = 0.1 A 5 + 20 V20 = 20 x 0.1 kV = 2 kV p20 = I20 V20 = 0.2 kW Chapter 2, Solution 27 Using current division, i1... v2 = 0 v2 = 22V Chapter 2, Solution 17 + v1 - 24V + loop 1 - + v3 - - v2 + loop 2 -+ 12V It is evident that v3 = 10V Applying KVL to loop 2, v2 + v3 + 12 = 0 v2 = -22V Applying KVL to loop 1, -24 + v1 - v2 = 0 v1 = 2V Thus, v1 = 2V, v2 = -22V, v3 = 10V Chapter 2, Solution 18 Applying KVL, -30 -10 +8 + I(3+5) = 0 8I = 32 I = 4A -Vab + 5I + 8 = 0 Vab = 28V + - 10V Chapter 2, Solution 19 Applying KVL around... = 6 A 6 + 12 i1 = 9 − 6 = 3A, v = 4i1 = 4 x 3 = 12 V p6 = 12R = 36 x 6 = 216 W Chapter 2, Solution 31 The 5 Ω resistor is in series with the combination of 10 (4 + 6) = 5Ω Hence by the voltage division principle, v= 5 (20V) = 10 V 5+5 by ohm's law, i= v 10 = = 1A 4 + 6 4+ 6 pp = i2R = (1)2(4) = 4 W Chapter 2, Solution 32 We first combine resistors in parallel 20 30 = 20 x30 = 12 Ω 50 10 40 = 10x 40... sources: p12V = 12 (- -2) = 24W p10V = 10 (-2) = -20W p8V = (- -2) = -16W Chapter 2, Solution 20 Applying KVL around the loop, -36 + 4i0 + 5i0 = 0 i0 = 4A Chapter 2, Solution 21 Apply KVL to obtain 10 Ω -45 + 10i - 3V0 + 5i = 0 + v0 - But v0 = 10i, -45 + 15i - 30i = 0 P3 = i2R = 9 x 5 = 45W i = -3A 45V + + - 5Ω 3v0 Chapter 2, Solution 22 4Ω + v0 6Ω 10A 2v0 At the node, KCL requires that v0 + 10 + 2 v 0 = ... Chapter 2, Solution v = iR i = v/R = (16/5) mA = 3.2 mA Chapter 2, Solution p = v2/R → R = v2/p = 14400/60 = 240 ohms Chapter 2, Solution R = v/i = 120/(2.5x10-3) = 48k ohms Chapter 2, Solution. .. 20 mA dt Chapter 1, Solution 25A, dq  i= = - 25A, dt   25A, -2t 0