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" " Let us now use the method of loop currents, and let I1 and I2 denote the two loop currents required for Fig. 142. In doing this, let us designate that positive current ows in the clockwise sense around each loop (as in Fig. 135). Thus (see discussion with Fig. 135) the simultaneous vector equations for Fig. 142 are " 15 j25 I1 " 15I2 30 " 3 j5 I1 " 3I2 6
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   3 j5 6     3 18 18 2 j 0 " I2 0:55385 j0:27692 A 13 2 j 13 2 j 65 " " Vy 10I2 5:5385 j2:7692 volts rect: form ; answer; or " Vy 6:19221= 26:5668 volts polar form ; answer:
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126. We must nd the TOTAL CURRENT owing through the 15-ohm resistance. Since " " we already know the value of I2 , this means we must now nd the value of I1 . Since D 13 2 j as before, we have that   6 3      " 0 5 j4 6 5 j4 6 5 j4 2 j 0:55385 j1:20000 amperes I1 13 2 j 13 2 j 65 Since our equations have been written for the case where positive current ows in the clockwise sense, the situation for the 15-ohm resistance is as shown to the right. From inspection we " see that the total resultant current IT in the 15-ohm resistance is equal to " " " IT I1 I2 j0:92308 A approx: and thus, by Ohm s law, " " Vx 15 IT j13:8462 volts; answer; or, in polar form " Vx 13:8462= 908; answer: Note: In problems 125 and 126 we chose the clockwise direction around the loop to be the positive direction. It should be noted, however, that the same rms values of currents and voltages will be obtained regardless of which direction, cw or ccw, is chosen to be the positive direction. Of course, once the positive direction is designated for each loop, in a given problem, that designation must not be changed during the writing of the network equations.
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127. The rst step is to nd the voltage between terminals a and b in Fig. 143; this is the " voltage V 0 of the equivalent generator, and is found by applying Ohm s law to Fig. 143 as follows. Note that the voltage between terminals a and b is the voltage drop across " capacitor C, which is equal to the current I times the reactance jXC of capacitor C; " 0 I jXC . However, from inspection of Fig. 143, " that is, we have V " I V= R jXC , and thus we have jVXC " V0 R jXC or, rationalizing, jVXC R jXC VXC XC jR " V0 ; answer; in terms of XC : 2 2 R2 XC R2 XC " Next, the impedance Z 0 of the equivalent generator is the impedance seen looking into terminals a, b in Fig. 143; thus, since R and jXC are in parallel, we have, using eq. (209), jRXC jRXC R jXC RXC XC jR " Z0 ; answer; in terms of XC : 2 2 R jXC R2 XC R2 XC Now setting XC 1=!C in the above answers, you can verify that V 1 jR!C " V0 1 R!C 2 and R 1 jR!C " Z0 ; answers: 1 R!C 2
Thevenin s theorem thus allows the replacement of a somewhat complicated network with a simple series circuit in the form of Fig. 144. This is useful when, given a complicated network, we wish to nd the current that would ow in a number of di erent loads when connected to the original complicated network. 128. (a) " First, XC 2 ohms and XL 3 ohms. Now, from left to right in Fig. 145, let I1 " and I2 denote the two loop currents, with the positive sense to be the clockwise direction. Then, noting that 5=908 j5 (note 16 in Appendix), the two simultaneous equations for Fig. 145 are " 2 1 j I1 " 2I2 5 2 j " " 2I1 4 j3 I2 j5 " " " Since, from inspection of Fig. 145, Vx 2I2 , let us solve for I2 , as follows. First,    1 j 2   2 5 j  D 2 1 4 j3  and hence    1 j 2 j    2 5  5 1 j  " I2 2 5 j 5 j
thus 10 10 5 j " " 1:9231 j0:3846 volts; answer: Vx 2I2 5 j 26 (b) " jVx j 1:9612 volts approx., answer.
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