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Thus, when the switch is closed: for RL 2 ohms; IL 8:33102=4:05715 2:05342 amperes; answer; for RL 3 ohms; IL 8:33102=5:05715 1:64738 amperes; answer; for RL 4 ohms; IL 8:33102=6:05715 1:37540 amperes; answer: Without the use of Thevenin s theorem it would be necessary to rework problem 53 for each new value of RL . 56. The rst step is to remove RL and nd the open-circuit voltage between terminals a and b. One way to do this is to write the two loop-voltage equations (see gure below), thus 20I1 12I2 100 12I1 24I2 0 from which I2 25=7 amperes, and hence the open-circuit voltage at a, b is equal to 9 25=7 32:143 volts.
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The next step is to nd the resistance looking back into terminals a, b (keeping RL disconnected, as before). The 8-ohm and 12-ohm resistances are in parallel when looking back from terminals a, b and hence combine together (product over sum, eq. (34), Chap. 2) to give 12 8 =20 4:8 ohms. The 4.8 ohms is now in series with the 3-ohm resistance, giving us a total of 7.8 ohms in parallel with the 9-ohm resistance. Therefore, looking into a, b we see 7.8 ohms in parallel with 9 ohms, which, using the product of the two, over the sum, is equal to 7:8 9 =16:8 4:179 ohms, approximately. Thus the equivalent Thevenin generator for Fig. 59 is therefore as shown in the following gure, answer.
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57. First, Isc ows into a conductance equal to, by eq. (61), GT Gg GL , and thus, by eq. (59), Isc Gg GL VL also, IL GL VL thus VL IL =GL , and putting this value of VL into the rst equation and solving for IL gives the answer, eq. (67). 58. (a) The rst step is to nd the short-circuit current Isc , which is the current that ows into the short-circuited terminals a, b as shown in the gure to the left below, with the two loop-voltage equations to the right.
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5I1 3Isc 25 4I1 5Isc 0
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thus,
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   5 25     4 0  100  7:692308 amperes; approx: Isc   5 3  13    4 5
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We must next nd the value ofthe conductance component Gg ofthe Norton equivalentgenerator. To do this, we now remove the short-circuit from the terminals a, b in the pre ceding gure; Gg now equals the conductance seen looking to the left into the now open-circuited terminals a, b. From inspection of Fig. 59 (disregard the 9ohm branch for the moment), note that we look into one branch consisting of 3 ohms in series with the parallel combination of 8 ohms and 12 ohms, that is, into 3 8 12 =20 7:8 ohms. Since this resistance is in parallel with the 9ohm resistance, the total resistance seen looking into the open-circuited terminals a, b is Rg 9 7:8 =16:8 4:178571 ohms
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thus, Gg 1=Rg 0:239316 mhos; approx: and hence the answer is that the Norton equivalent generator for Fig. 59 is
The Thevenin equivalent of Fig. 59 appears in the solution to problem 56, and the Norton equivalent appears above. Setting RL 10 ohms in the Thevenin equivalent generator and applying Ohm s law, and setting GL 1=10 0:1 mho in eq. (67), we have that Thevenin case: IL 32:143=14:179 2:267 amperes; approx: Norton case: IL 0:7692308=0:339316 2:267 amperes; approx: there being some di erence beyond the third decimal place because the same degree of accuracy was not used in calculating in problem 56 as in problem 58.
59. Let a, b be the output terminals of any given network; to convert the network into the Thevenin equivalent generator we must take the following two steps. (a) (b) Find the open-circuit voltage at the terminals a, b; this is the generated voltage, Vg , of the equivalent generator (Fig. 57). Find the resistance looking into the open-circuited terminals a, b; this is the internal resistance, Rg , of the equivalent generator (Fig. 57). Here we use the basic Ohm s law formula V RI 1=G I, thus V (b) Isc 6:155 56:468 volts Vg Gg 0:109
Now apply the two steps to Fig. 60 (keeping the switch open), as follows. (a)
We must now replace all generators with their internal resistances. In this case we must remember that a constant-current generator has INFINITELY GREAT internal resistance. Thus, as far as resistance is concerned, the constant-current generator has no shunting e ect on Gg , and thus 1=Gg 1=0:109 9:174 ohms Rg hence the Thevenin equivalent generator is