Linear Algebra

72 Chapter One. Linear SystemsKirchhoff’s Current Law, applied to the top node, the left node, the right node,and the bottom node gives these.i 0 = i 1 + i 2i 1 = i 3 + i 5i 2 + i 5 = i 4i 3 + i 4 = i 0Kirchhoff’s Voltage Law, applied to the inside loop (the i 0 to i 1 to i 3 to i 0 loop),the outside loop, and the upper loop not involving the battery, gives these.5i 1 + 10i 3 = 102i 2 + 4i 4 = 105i 1 + 50i 5 − 2i 2 = 0Those suffice to determine the solution i 0 = 7/3, i 1 = 2/3, i 2 = 5/3, i 3 = 2/3,i 4 = 5/3, and i 5 = 0.We can understand many kinds of networks in this way. For instance, theexercises analyze some networks of streets.Exercises1 Calculate the amperages in each part of each network.(a) This is a simple network.3 ohm9 volt2 ohm2 ohm(b) Compare this one with the parallel case discussed above.3 ohm9 volt2 ohm 2 ohm(c) This is a reasonably complicated network.2 ohm3 ohm3 ohm9 volt3 ohm 2 ohm4 ohm2 ohm2 ohm2 In the first network that we analyzed, with the three resistors in series, we justadded to get that they acted together like a single resistor of 10 ohms. We can do