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Network SolutionsAuthor: E.E. Kimberly A circuit containing other than components in simple series or simple parallel is called a network. An example of a network is shown in Fig. 36. The currents and voltages in them are determined by the simultaneous applications of Kirchhoff's voltage and current laws. In writing the path or loop equations for a network, such as that in Fig. 36, it is necessary that the directions taken around all paths be the same. After all the voltages have been found for all parts of the circuit, the voltage loop equations can be written for any closed traverse of the circuit. At any junction point in the circuit the currents in the branches connected to that point can be calculated by Kirchhoff's current law by socalled node equations.
In Fig. 36 there are only three unknown currents designated as I_{1}, I_{2}, and I_{3}, since I_{5} = I_{1} and I_{4} = I_{3}. Therefore, there must be three equations. These may consist of two voltage equations and one current equation and are as follows:
Also, It should be noted that the resistance voltage drop in a circuit is in the same direction as the current in it. Example 36.  In Fig. 36, let E_{1} = 100, E_{2} = 90, R_{1} = 10, R_{2} = 8, R_{3}=15, R_{4} = 20, and R_{5} = 30. Calculate the currents in and the voltage drops across all resistances. Solution.  By Kirchhoff's voltage laws,
100+10I_{1}+8I_{2}+30I_{5} = 0 After simplification by substituting equivalent currents, these equations become:
48I_{1}8I_{3} = 100 By the solution of these simultaneous equations,
I_{1} = +1.79 amp (I_{dc}) Then, I_{2} = I_{cf} = 1.79  (1.76) = 3.55 amp The negative sign before 1.76, the value of I_{3}, means that I_{3} is flowing in R_{3} in the direction opposite to that assumed when the loop B was traversed in a clockwise direction. Positive signs before the answers for I_{1} and I_{2} mean that the actual currents are in the same sense around the circuit as was assumed for loop A.


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