Chapter 8:
8.21a
I1 on the left, I2 on the right
I1 = I(5 ohm resistor) = 72.16 mA
Va= -4 -(72.16mA)(6 ohm) = -4.433 V
8.24
[a]: I1 in the larger loop (bottom); I2 in the small loop (top)
(Remember: I1 - I2 = 6A)
I1 = 3A; I2 = -3A
[b]: see notes from class
8.34
I) V1 on the left node, V2 on the right
V1 = -2.653V; V2 = 0.952V
II)V1 is left; V2 is right; V3 in the middle
V1 = 8.877V; V2 = 9.831V; V3 = -3.005V
8.35
I) V1 on top; V2 on left (at the top of the current source); V3 on the right
V1 = 7.238V; V2 = -2.453V; V3 = 1.405V
II) V1 left; V2 center; V3 right
V1 = -6.642V; V2 = 1.293V; V3 = 10.664V
8.37
I) V1 left; V2 center
(Notice: you didn't really have to use the supernode approach on this one)
V1 = 10.083V; V2 = 6.944V
the voltage between the 24V source and the 12 ohm resistor is V2 - 24 = -17.056V
II) V1 on the left; V2 on the right.
1st eqn: Supernode eqn: V2 - V1 = 16
2nd eqn: add all currents leaving nodes V1 and V2: V1/20 + 3 - 3 + v2/40 - 4 = 0
V1 = 48V; V2 = 64V