Linear Algebra

74 Chapter One. Linear Systems↑ i 0 i 1 ↓ ↓ i 2The Current Law, applied to the point in the upper right where the flow i 0meets i 1 and i 2 , gives that i 0 = i 1 + i 2 . Applied to the lower right it givesi 1 + i 2 = i 0 . In the circuit that loops out of the top of the battery, down theleft 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 givesthat 12i 1 = 20. Similarly, the circuit from the battery to the right branch andback to the battery gives that 8i 2 = 20. And, in the circuit that simply loopsaround in the left and right branches of the parallel portion (arbitrarily takenclockwise), there is a voltage rise of 0 and a voltage drop of 8i 2 − 12i 1 so theVoltage Law gives that 8i 2 − 12i 1 = 0.i 0 − i 1 − i 2 = 0−i 0 + i 1 + i 2 = 012i 1 = 208i 2 = 20−12i 1 + 8i 2 = 0The solution is i 0 = 25/6, i 1 = 5/3, and i 2 = 5/2, all in amperes. (Incidentally,this illustrates that redundant equations do indeed arise in practice.)Kirchhoff’s laws can be used to establish the electrical properties of networksof great complexity. The next diagram shows five resistors, wired in a seriesparallelway.10 volt5 ohm 2 ohm50 ohm10 ohm 4 ohmThis network is a Wheatstone bridge (see Exercise 4). To analyze it, we canplace the arrows in this way.i 1 ↙ ↘ i 2i 5 →↑ i 0i 3 ↘ ↙ i 4