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In the above, we wrote 0I1 and 0I3 merely to keep the equations lined up in a convenient manner. Step III. In this problem we are asked to nd the voltage drop across the 9-ohm resistor, that is, the value of 9I3 ; hence, in this problem, we must nd the value of I3 . The rst step in doing this is to nd the
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CHAPTER 4 Basic Network Laws and Theorems
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value D of the denominator determinant, which is the determinant formed from the coe cients of the unknowns; thus,    11 6 0     D  6 13 4     0 4 13  which, upon expanding in terms of the minors of the rst column (and factoring the determinants where possible) is equal to   13 D 11  4    4   12 3 2  13  0  1215 13 
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To nd the value of I3 we use the same determinant as above, except now the coe cients of I3 are replaced by the constant terms on the right-hand sides of the network equation found in step II. The value of I3 is then equal to   11    6   0     11 6 24  4       13 18  6 6 13 3      0 4 4 6 1  6 71 0:35062 amperes; approx: D 1215 1215 6
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and therefore Va 9I3 3:15556 volts; answer
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Problem 47 In Fig. 52, the symbol  is the capital Greek letter omega, which is often used to denote ohms. Using the method of loop currents, nd the current in the 7-ohm resistance. (Answer: 1.24138 amperes)
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Fig. 52
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Problem 48 In the bridge-type network in Fig. 53, the resistance values are in ohms. Find the voltage drop across the 4-ohm resistance. (Answer: 1.8462 volts)
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CHAPTER 4 Basic Network Laws and Theorems
Fig. 53
Problem 49 In Fig. 54, nd the potential at point a with respect to ground. Resistance values are in ohms. (Answer: 8.98294 volts)
Fig. 54
Problem 50 This problem is included here as an example of the very important PRINCIPLE OF SUPERPOSITION, which we rst met in section 1.3 (eq. (7)). As applied to electric networks, the principle of superposition can be stated as follows. In a network composed of linear* elements and several generators, the current at any point in the network is the sum of the currents due to EACH GENERATOR CONSIDERED SEPARATELY, the other generators being replaced by their internal resistances. In our applications here, we ll assume the generators (batteries) to have zero internal resistance. In problem 47, Fig. 52, we found the current in the 7-ohm resistor to be, to ve decimal places, equal to 1.24138 amperes. Verify this answer by applying the principle of superposition to Fig. 52. Problem 51 Explain why all actual resistances are non-linear to some degree. Problem 52 In Fig. 53, suppose it is desired to replace the 3-ohm resistor with a resistor of R ohms, such that the current through the 4-ohm resistor will be zero. What must be the value of R (Answer: R 0:5 ohm)
* A linear resistance is one whose resistance value is independent of the amount of current owing in the resistance. (A resistance is said to be non-linear if the resistance changes with changes in current.)
CHAPTER 4 Basic Network Laws and Theorems
Conductance. Millman s Theorem
In certain types of network it is more convenient to work with the RECIPROCAL of resistance instead of directly with resistance itself. The reciprocal of resistance is called conductance, and is denoted by G. Hence, by de nition, G 1 1=R R 58
Thus conductance is measured in units of reciprocal ohms, called mhos ( mho is ohm spelled backward). The term siemen is also used for reciprocal ohms and is the SI unit of electrical conductance. As eq. (58) shows, HIGH RESISTANCE means LOW CONDUCTANCE, and vice versa; for example, if or if R 1000 ohms then G 1=1000 0:001 mho; R 0:001 ohm then G 1=0:001 1000 mhos:
Thus the basic Ohm s law, I V=R, becomes, in terms of conductance, I GV showing that amperes equals mhos times volts. Next consider POWER, P. In section 2.3 we found that P V 2 =R, and thus P GV 2 watts 60 59
Conductance is especially convenient to use when dealing with purely parallel networks, as the following will show. In section 2.6 (eq. (32)), we found that the total resistance RT of n parallel-connected resistances is found by means of the relationship 1 1 1 1 1 RT R1 R2 R3 Rn which is not especially easy to use. Note, however, that by the de nition of eq. (58) we have that the total conductance of n parallel-connected resistances is equal to the simple sum of the individual conductances; thus GT G1 G2 G3 Gn 61
The use of conductance is especially helpful in dealing with parallel networks in which each branch is composed of a conductance in series with a battery, as in Fig. 55.
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