-------------Assigned January 25, 2013-------------
1.24 a) q = area under i vs. t plot = 123,000 C
b) w_T = w1 + w2 + w3 = 468.667 + 674.133 + 104.4 = 1247.2 kJ
1.28 a) The subordinate engineer is correct.
b) The sign of Pe is wrong.
-------------Assigned January 28, 2013-------------
2.1 The interconnection is valid.
P_dev = 300 W
2.5 The interconnection is invalid.
2.19 a) ia = 2 A
b) ib = 0.5 A
c) v0 = 40 V
d) 25 W, 80 W, 20 W
e) 125 W
-------------Assigned January 30, 2013-------------
2.21 a) i1 = 2 A
b) 100 W, 50 W, 260 W, 90 W, 400 W
c) 900 W = 900 W
2.28 a) vy = 4.5 V
b) 741 mW - 741 mW = 0
-------------Assigned February 1, 2013-------------
2.30 a) i_sigma = 2 A, i_delta = 5 A, v0 = 80 V
b) P_generated = P_dissipated = 4230 W
3.2 a) The 10 Ohm and 40 Ohm resistors are in parallel, as are the 100 Ohm and 25 Ohm resistors.
b) The 9 kOhm, 18 kOhm, and 6 kOhm resistors are in parallel.
c) The 600 Ohm, 200 Ohm, and 300 Ohm resistors are in parallel.
3.6 a) Rab = 64 Ohms
b) Rab = 30 Ohms
c) Rab = 20 Ohms
-------------Assigned February 4, 2013-------------
3.10 P_5_ohm = 80 W
3.15 a) v0 = 66 V
b) P_R1 = 1.88 W, P_R2 = 1.32 W
c) R1 = 17,672 Ohms, R2 = 12,408 Ohms
-------------Assigned February 6, 2013-------------
3.25 v1 = 5/9 V, v2 = 1/3 V
3.30 a) vg = 60 V
if you didn't get this, some intermediate steps you might want to check: v_9 = 18 V, i_9 = 2A, i_2 = 3A, v_24 = 24 V, i_24 = 1 A, i_3 = 4 A, v_72 = 36 V, i_72 = 0.5 A, v_108 = 54 V, i_108 = 0.5 A
b) P_20 = 16.2 W
-------------Assigned February 11, 2013-------------
3.33 % error = -0.35%
3.54 a) i0 = 96 mA
b) i1 = 48 mA
c) i2 = 600 mA
d) P = 72.96 W
3.55 Rab = 10 kOhms
-------------Assigned February 20, 2013-------------
4.6 v1 = 100 V
v2 = 20 V
4.11 a) v0 = 4 V
b) P = -80 mW
c) P = 4800 mW
d) 4800 mW = 4800 mW
e) v0 is independent of any finite resistance connected in series with the 40 mA current source
-------------Assigned February 22, 2013-------------
4.25 v0 = 1.5 V
-------------Assigned February 25, 2013-------------
4.27 v_delta = 5 V, v0 = -20 V
4.31 i1, i3, and i6 are the mesh currents you should have solved for first.
i1 = 23.76 A, i2 = 5.33 A, i3 = 18.43 A, i4 = 15.10 A, i5 = 9.77 A, i6 = 8.66 A
-------------Assigned February 27, 2013-------------
4.40 i1 = 110 A; i2 = 52 A; i3 = 60 A; i_delta = -8 A
P_dependent = 46,640 W
4.51 a) i1 = 4.6 A, i2 = 5.7 A, i3 = 0.97 A
ia = 5.7 A, ib = 4.6 A, ic = 0.97 A, id = -1.1 A, ie = 3.63 A
b) P_developed = P_dissipated = 1319.685 W
-------------Assigned March 1, 2013-------------
4.53 a) There are three unknown node voltages and only two unknown mesh currents. Use the mesh current method to minimize the number of simultaneous equations.
b) i1 = 6mA, i2 = 8mA, P_1k = 4 mW
c) No, the voltage across the 10 A current source is readily available from the mesh currents, and solving two simultaneous mesh-current equations is less work than solving three node voltage equations.
d) P_10mA = -200 mW
4.54 a) There are three unknown node voltages and three unknown mesh currents, so the number of simultaneous equations required is the same for both methods. The node voltage method has the advantage of having to solve the three simultaneous equations for one unknown voltage provided the connection at either the top or bottom of the circuit is used as the reference node. Therefore recommend the node voltage method.
b) v1 = 6.67 V, v2 = 13.33 V, v3 = 5.33 V
4.60 a) i0 = -0.85 A
b) i0 = -0.85 A
-------------Assigned March 4, 2013-------------
4.65 a) Vth = Voc = 30 V, Isc = 1.5 A, Rth = 20 Ohms
b) Rth = 20 Ohms
4.74 v_th = v_oc = -160 V, i_sc = -2837 uA, Rth = -160/2837/1000000 = 56.4 kOhms
-------------Assigned March 6, 2013-------------
4.87 a) Open-circuit: i1 = 99.6 A, i2 = 78 A, i3 = 100.8 A, iB = 21.6 A; V_th = -24 V
Short-circuit: i1 = 92 A, i2 = 73.33 A, i3 = 96 A, iB = 18.67 A; I_sc = -4 A
R_th = 6 Ohms
b) P_max = 24 W
-------------Assigned March 8, 2013-------------
4.91 a) v = v' + v'' = 770/13 + 120/13 = 50 V
b) 250 W
4.100 P = 23.09 W
4.101 P = 1.45 W
-------------Assigned March 13, 2013-------------
Midterm Voc = 184.8 V, Isc = 60 A, Rth = 3.08 Ohms
5.1 d) vo = 9 V
5.2 vo = -5 V, io = -1 mA
5.4 vo = -1 V, iL = -250 uA
-------------Assigned March 15, 2013-------------
An inverting amplifier is an op amp circuit producing an output voltage that is an inverted, scaled replica of the input. Derive equation 5.10 (pg. 150) using node-voltage (KCLs).
A summing amplifier is an op amp circuit producing an output voltage that is a scaled sum of the input voltages. Derive equation 5.14 (pg. 152) using node-voltage (KCLs).
A noninverting amplifier is an op amp circuit producing an output voltage that is scaled replica of the input voltage. Derive equation 5.18 (pg. 153) using node-voltage (KCLs).
A difference amplifier is an op amp circuit producing an output voltage that is a scaled replica of the input voltage difference. Derive equation 5.24 (pg. 155) using node-voltage (KCLs).
-------------Assigned March 18, 2013-------------
5.18 a) vo = 10.54 V
b) -4.55 V <= vg <= 4.55 V
c) Rf = 181.76 kOhms
5.21 a) This is a non-inverting summing amplifier.
b) v0 = 11(0.675va + 0.325vb) = 7.37 V
c) ia = 10 uA, ib = -10 uA
d) For va: weighting factor = 11 * 0.675 = 7.425
For vb: weighting factor = 11 * 0.325 = 3.575
-------------Assigned March 20, 2013-------------
5.28 a) vo = 16 V
b) -4.2 V < vc < 3.8 V
-------------Assigned April 3, 2013-------------
6.5 a) 0 < t < 1 s i = -10t^2 A
1 < t < 3 s i = 10t^2 - 40t + 20 A
3 < t < 5 s i = 20t - 70 A
5 < t < 6 s i = -10t^2 + 120t - 320 A
6 < t < infinity i = 40 A
b) v = 0 at t = 2 s and t = 6 s
i(2) = -20 A, i(6) = 40 A
c) Sketch
-------------Assigned April 5, 2013-------------
6.10 power = -339.57 W delivering
6.17 0 < t < 0.5: ic = 1.2t mA
0 < t < 1: ic = 1.2 (t-1) mA
Sketch of above equations
6.20 15 H
6.21 8 H
-------------Assigned April 8, 2013-------------
6.26 Equivalent capacitance is 2 uF with an initial voltage drop of +25 V.
6.34 v1(0+) = 400 V, v2(0+) = 0