here,
using KVL in the upper loop
-12 + 1 * I2 + 11 * I2 + 12 * I1 - 12 + ( 22 + 1)* I1 = 0
-12 + 1 * I2 + 11 * I2 + 12 * I1 - 12 + ( 22 + 1)* I1 = 0 ...(1)
using KVL in the lower loop
- 6 + 1 * I3 + 18 * I3 - (11 + 1) * I2 + 12 + 15 * I3 = 0
- 6 + 1 * I3 + 18 * I3 - (11 + 1) * I2 + 12 + 15 * I3 = 0 ...(2)
and
I1 = I2 + I3 ...(3)
from equation(1) , (2) and (3)
I1 = 0.51 A , I2 = 0.51 A
I3 = 0.003 A
B)
the terminal voltage , V = 6 - I3 * r
V = 5.997 V
the terminal voltage is 5.997 V
Determine the currents I_1, I_2, and I_3 in the figure. Assume the internal resistance of each...
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