Chapter 8:
8.7
a) I(L) = I * (Rt/RL) = 12 * (1.33/2) = 8A
b) 4 ohm resistor in series with a 48 V source: I = 48V / 6 ohm = 8A (same as before)
8.8
a) 6.8 ohm resistor in series with a 13.6 V source replacing the current source//resistor
b) I = V/R = 25.6V / (39+10+6.8 ohm) = 0.459 A
c) V = RI = 39 (.45) = 17.9 V
8.13
I) I1 "up" through the 5.6k ohm resistor; I2 "up" through the 30V source; I3 "down" through the middle branch
You need three equn's for the 3 unknowns, I1, I2, and I3
I1 + I2 = I3
-10 + 5.6kI1 + 2.2KI3 - 20 = 0
20 - 2.2kI3 - 3.3KI2 + 30 = 0
solve...
I1 = 1.445mA; I2 = 8.51mA; I3 = 9.96mA
II) I1 "right" through the 1.2k ohm resistor; I2 "up" through the 9.1k resistor; I3 "left" through the middle branch
You need three equn's for the 3 unknowns, I1, I2, and I3
I1 + I3 = I2
1.1kI2 - 6 + 9.1kI2 + 8.2kI3 = 0
-8.2kI3 + 1.2kI1 - 9 = 0
solve...
I1 = 2.03mA; I2 = 0.8mA; I3 = 1.23mA
8.18
I) Assign "currents" I1 and I2 from left to right
IR1 = I1 = 1.445 mA
IR2 = I2 = 8.513 mA
IR3 = 7.07 mA (in the direction of I2)
II) Assign I1 to the top loop, I2 to the bottom loop
IR1 = I1 = 2.034 mA
IR3 = IR4 = I2 = 1.231 mA
IR2 = 0.802 mA (in the direction of I1)
8.22
I) Assign "currents" I1, I2 and I3 from left to right
I1 = 1.206 mA
I2 = -0.481 mA
I3 = -0.621 mA
II) Assign I1 to the top loop, I2 to the bottom left loop, and I3 to the bottom right loop
I1 = -0.239 A
I2 = -0.517 A
I3 = -1.28 A