Chapter 7:
7.3
a) yes (KCL or some other reasonable explanation)
b) I2 = I - I1 = 5A - 2A = 3A
c) yes (KCL)
d) V2 = E - V1 = 10V - 6V = 4V
e) Rt = R1 || R2 + R3 || R4 = 2 || 3 + 1 || 4 = 6/5 ohm + 4/5 ohm = 2 ohm
f) I = E/Rt = 10V / 2 ohm = 5A
g) Pdeliv'd = E*I = (10V)(5A) = 50 W
P1 = (V1**2)/R1 = 36/3 = 12W
P2 = (V2**2)/R2 = 36/2 = 18W
7.4
a) Rt = 16 ohm
b) I = 4A; I1 = 1.33A
c) V3 = 48 V
7.5
a) Rt = 10 || 12 || (10+2) = 4 ohm
b) Is = V/Rt = 36/4 = 9 A; I1 = V/R = 36 / (10||12) = 6 A; I2 = I - I1 = 3 A
c) Va = V(R4) = 2 ohm * I2 = 2 * 3 = 6V
7.15
REDRAW THE CIRCUIT!
a) I = V/(R2 + R3) = 9/15 = 0.6A
b) V = V1-V2 = 9 - (-19) = 28V
7.17
Redraw the circuit if helpful - notice that the R8 is shorted; therefore, I = 0 and V = 0
Also, R6||R7, R4||R5; when these are replaced, the replacement is in series with R3, and so on...
Therefore, Rt = 20 ohm
a) I2 = 1.667A; I6 = 1.111A; I = 100V/20ohm = 5A; and I8=0
b) V4 = 10V; V8 = 0
8.1
Vab = 28 V