Linear Algebra

74 Chapter 1. Linear Systems
We begin by labeling the branches of the network. Call the flow of current
coming out of the top of the battery and through the top wire i0, call the
current through the left branch of the parallel portion i1, that through the right
branch i2, and call the current flowing through the bottom wire and into the
bottom of the battery i3. (Remark: in labeling, we don’t have to know the
actual direction of flow. We arbitrarily choose a direction to establish a sign
convention for the equations.)
i0
The fact that i0 splits into i1 and i2, on application of Kirchhoff’s First Law,
gives that i1 + i2 = i0. Similarly, we have that i1 + i2 = i3. 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 and the voltage
drop is i1 ·12, so Kirchoff’s Second Law gives that 12i1 = 20. In the circuit from
the battery to the right branch and back to the battery there is a voltage rise of
20 and a voltage drop of i2 ·8, so Kirchoff’s Second law gives that 8i2 = 20. And
finally, in the circuit that just loops around in the left and right branches of the
parallel portion (taken clockwise), there is a voltage rise of 0 and a voltage drop
of 8i2 − 12i1 so Kirchoff’s Second Law gives 8i2 − 12i1 =0.
All of these equations taken together make this system.
i3
i1
i0 − i1 − i2 = 0
− i1 − i2 + i3 = 0
12i1 =20
8i2 =20
−12i1 +8i2 = 0
The solution is i0 =25/6, i1 =5/3, i2 =5/2, and i3 =25/6 (all in amperes).
(Incidentally, this illustrates that redundant equations do arise in practice, since
the fifth equation here is redundant.)
Kirchhoff’s laws can be used to establish the electrical properties of networks
of great complexity. The next circuit has five resistors, wired in a combination
of series and parallel. It is said to be a series-parallel circuit.
i2