CASE STUDY CURRENTS AND VOLTAGES IN CIRCUITS CASE STUDY continued

8.3 CASE STUDY 223

8.3 CASE STUDY continued

The voltage or loop rule specifies that the algebraic sum of the potential differences i.e., voltage changes in any loop must equal zero. For a resistor circuit, this is expressed as ξ − iR = 0 8.9 where ξ is the emf electromotive force of the voltage sources, and R is the resistance of any resistors on the loop. Note that the second term derives from Ohm’s law Fig. 8.8b, which states that the voltage drop across an ideal resistor is equal to the product of the current and the resistance. Kirchhoff’s voltage rule is an expression of the conservation of energy . Solution. Application of these rules results in systems of simultaneous linear algebraic equations because the various loops within a circuit are interconnected. For example, con- sider the circuit shown in Fig. 8.9. The currents associated with this circuit are unknown both in magnitude and direction. This presents no great difficulty because one simply assumes a direction for each current. If the resultant solution from Kirchhoff’s laws is negative, then the assumed direction was incorrect. For example, Fig. 8.10 shows some assumed currents. Structure Machine Reactors Circuit ⫹ ⫺ a Chemical engineering b Civil engineering d Mechanical engineering c Electrical engineering i 1 i 3 i 2 V i V j R ij i ij a b FIGURE 8.8 Schematic representations of a Kirchhoff’s current rule and b Ohm’s law. FIGURE 8.7 Engineering systems which, at steady state, can be modeled with linear algebraic equations.

8.3 CASE STUDY continued

Given these assumptions, Kirchhoff’s current rule is applied at each node to yield i 12 + i 52 + i 32 = 0 i 65 − i 52 − i 54 = 0 i 43 − i 32 = 0 i 54 − i 43 = 0 Application of the voltage rule to each of the two loops gives −i 54 R 54 − i 43 R 43 − i 32 R 32 + i 52 R 52 = 0 −i 65 R 65 − i 52 R 52 + i 12 R 12 − 200 = 0 or, substituting the resistances from Fig. 8.9 and bringing constants to the right-hand side, −15i 54 − 5i 43 − 10i 32 + 10i 52 = 0 −20i 65 − 10i 52 + 5i 12 = 200 Therefore, the problem amounts to solving six equations with six unknown currents. These equations can be expressed in matrix form as ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 1 1 1 −1 1 −1 −1 1 1 −1 10 −10 −15 −5 5 −10 −20 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ i 12 i 52 i 32 i 65 i 54 i 43 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 200 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ R ⫽ 10 ⍀ V 1 ⫽ 200 V V 6 ⫽ 0 V 3 2 1 4 5 6 R ⫽ 5 ⍀ R ⫽ 5 ⍀ R ⫽ 10 ⍀ R ⫽ 15 ⍀ R ⫽ 20 ⍀ FIGURE 8.9 A resistor circuit to be solved using simultaneous linear algebraic equations. 3 2 1 4 5 6 i 12 i 65 i 52 i 32 i 54 i 43 FIGURE 8.10 Assumed current directions. 8.3 CASE STUDY 225

8.3 CASE STUDY continued