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Note that since there is zero battery voltage in loops 2 and 3, the right-hand sides of the voltage equations for those loops is zero. Step III. In this problem we re asked to nd the voltage drop across the 4ohm resistance. From the above gure we see that, since I2 and I3 both ow in that resistor, we must nd the values of both I2 and I3 . The rst step in doing this is, as usual, to nd the value of delta, which is found from step II to be    3 1 2      D  1 8 4  65    2 4 7 Next you can verify that   3   6 1    2 I2  2    4    7  1   0  0
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90 1:384615 amperes; 65
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120 1:846154 amperes: 65
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From inspection of the gure, the net resultant current owing from left to right in the 4-ohm resistor is I3 I2 0:461539 amperes, and therefore the voltage drop across the 4-ohm resistor is IR 0:461539 4 1:8462 volts approx:; answer: 49. Going from left to right in Fig. 54, draw three loop currents, I1 , I2 , and I3 , which we ll assume to all be in the clockwise sense. Then the three voltage equations are, going from left to right in Fig. 54, 15I1 10I2 0I3 32 10I1 17I2 4I3 21 0I1 4I2 11I3 12 In this problem we must nd current I3 , because this is the current through the 7-ohm resistor. To do this we now form a determinant from the coe cients of the unknown currents; thus    15 10 0     D  10 17 4  1465    0 4 11 
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thus,   15    10   0 10 17 4 1465  32    21   12 
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I3 hence,
1880 1:283277 amperes 1465
Va 7I3 8:98294 volts; answer: 50. The procedure is to nd the current in the 7-ohm resistance in Fig. 52 due to each battery considered separately, the other batteries being replaced by their internal resistances (considered to be zero in this case). We must therefore, in this case, nd the current in the 7-ohm resistance in each of the following three seriesparallel networks. (Let us call down currents negative and up currents positive. )
The three separate current values below the gures were calculated using the procedures of section 2.7. The net current in the 7-ohm resistance is then the algebraic sum of the three currents, which is, in this case, to ve decimal places, 1.24138 amperes, the same value as found in problem 47. We might mention that, as was pointed out in section 2.5, current is assumed to ow out of a battery at the positive terminal and re-enter at the negative terminal. 51. A change in current causes the temperature of a physical resistor to change, which in turn causes the resistance to change (section 2.4). In our problems here we re assuming such changes in resistance are small enough to be disregarded. 52. This problem can be worked in two di erent ways, as follows. FIRST WAY: Note the gure to the right. If the ratio of 1 to 2 is equal to the ratio of R to 1, then zero potential di erence will exist between the terminals of the 4-ohm resistor, and thus zero current will ow in that resistor. That is, we must have 1 R ; hence R 1=2 ohm; answer: 2 1
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SECOND WAY: We can make use of the solution given with problem 48, as follows. Note that requiring zero current in the 4-ohm resistance is the same as requiring that currents I2 and I3 cancel each other out in the 4-ohm resistance, which, since I2 and I3 ow through the 4-ohm resistance in opposite directions, will be done only if I2 and I3 have equal magnitudes, that is, if I3 I2 . Therefore, in the network diagram given with the solution to problem 48, change 3 to R and I3 to I2 . Doing this, the three loop equations become I1 I2 2 I1 1 R I2 0 2I1 3I2 0 A B C
Now add (A) and (B) together to get I2 2=R. Now replace I2 with 2=R in (B) and (C) to get I1 1 R 2 0 R 3 I1 0 R
In these two equations, multiply through the rst equation by R and the second by R, then add the two together to get R 1=2 ohm, answer. 53. Figure 56 is of the form of Fig. 55, and thus eq. (63) applies. Let us agree to number the branches from 1 through 5, from left to right in Fig. 56. The rst step is to convert ohms to mhos, using the de nition of eq. (58). Doing this, and noting that, in Fig. 56, V2 V5 0, eq. (63) becomes Vo 0:08333 15 0:06666 22 0:11111 12 0:08333 0:12500 0:06666 0:11111 0:10000 4:04979 8:33102 volts; answer: 0:48611
54. We would not have V4 12 volts, and upon making this change in the above solution to problem 53 we nd that Vo 1:38315 2:84534 volts; answer: 0:48611
55. The rst step is to draw the Thevenin equivalent generator for Fig. 56. To do this, imagine that we are at the right-hand side of Fig. 56, looking to the left at the 10-ohm resistance. Then, from the solution to problem 53, we have Vo Vg 8:33102 volts; and Rg 1=0:48611 2:05715 ohms and thus, by Thevenin s theorem, we now have the simple series circuit condition shown as follows.
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