Page 81 - Matrix Analysis & Applied Linear Algebra

74 Chapter 2 Rectangular Systems and Echelon Forms

Kirchhoff’s Rules

NODE RULE: The algebraic sum of currents toward each node is zero.
That is, the total incoming current must equal the total
outgoing current. This is simply a statement of conser-
vation of charge.

LOOP RULE: The algebraic sum of the EMFs around each loop must
equal the algebraic sum of the IR products in the same
loop. That is, assuming internal source resistances have
been accounted for, the algebraic sum of the voltage
drops over the sources equals the algebraic sum of the
voltage drops over the resistances in each loop. This is
a statement of conservation of energy.

Kirchhoff’s rules may be used without knowing the directions of the currents
and EMFs in advance. You may arbitrarily assign directions. If negative values
emerge in the final solution, then the actual direction is opposite to that assumed.
To apply the node rule, consider a current to be positive if its direction is toward
the node—otherwise, consider the current to be negative. It should be clear that
the node rule will always generate a homogeneous system. For example, applying
the node rule to the circuit in Figure 2.6.1 yields four homogeneous equations in
six unknowns—the unknowns are the I k ’s:

Node 1: I 1 − I 2 − I 5 =0,
Node 2: − I 1 − I 3 + I 4 =0,
Node 3: I 3 + I 5 + I 6 =0,
Node 4: I 2 − I 4 − I 6 =0.
To apply the loop rule, some direction (clockwise or counterclockwise) must
be chosen as the positive direction, and all EMFs and currents in that direction
are considered positive and those in the opposite direction are negative. It is
possible for a current to be considered positive for the node rule but considered
negative when it is used in the loop rule. If the positive direction is considered
to be clockwise in each case, then applying the loop rule to the three indicated
loops A, B, and C in the circuit shown in Figure 2.6.1 produces the three non-
homogeneous equations in six unknowns—the I k ’s are treated as the unknowns,
while the R k ’s and E k ’s are assumed to be known.
Loop A: I 1 R 1 − I 3 R 3 + I 5 R 5 = E 1 − E 3 ,
Loop B: I 2 R 2 − I 5 R 5 + I 6 R 6 = E 2 ,
Loop C: I 3 R 3 + I 4 R 4 − I 6 R 6 = E 3 + E 4 .