algebra 4 - Chapter One Linear Systems 74 and also let the...

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74 Chapter One. Linear Systems and also let the current through the battery be i 0 . Note that we don’t need to know the actual direction of flow — if current flows in the direction opposite to our arrow then we will get a negative number in the solution. " i 0 i 1 # # i 2 The Current Law, applied to the split point in the upper right, gives that i 0 = i 1 + i 2 . Applied to the split point lower right it gives i 1 + i 2 = i 0 . In the circuit that loops out of the top of the battery, down the left branch of the parallel portion, and back into the bottom of the battery, the voltage rise is 20 while the voltage drop is i 1 · 12 , so the Voltage Law gives that 12i 1 = 20 . Similarly, the circuit from the battery to the right branch and back to the battery gives that 8i 2 = 20 . And, in the circuit that simply loops around in the left and right branches of the parallel portion (we arbitrarily take the direction of clockwise), there is a voltage rise of 0 and a voltage drop of 8i 2 - 12i 1 so 8i 2 - 12i 1 = 0 .
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