HW 9 - Answers -

due Wednesday, 11/12/97

9.5
a) 6 ohm, 6 V
b) 2 ohm, 0.75 A
30 ohm, 0.1667A
100 ohm, 0.0566 A

 9.7
(I) 2 ohm, 84 V
(II) 1.579 kohm, -1.149 V

 9.8
a) R_th = 25 || 16 = 9.756 ohm
V_th (use superposition) = -20V + 29.268V = 9.268V
b) R_th = 4 || (2+6||3) = 2 ohm
for V_th:
I_t = 72 / [6+3||(2+4)] = 9A
I₂ = 3A (current division)
V_th = V_{6 ohm} + V_{2 ohm} = 6 I_t + 2 I ₂ = 60 V

9.9
(I) 45 ohm, -5 V
(II) 2.055 kohm, 16.772 V

 9.15
(a) 9.756 ohm, 0.95A
(b) 2 ohm, 30 A

 9.17
(a) 10 ohm, 0.2A
(b) 4.033 k ohm, 2.976 mA

 9.21 on your own. If this seems difficult, read about maximum power transfer...