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60. First note that Fig. 63 contains 7 node points (including the zero-volt reference node at ground potential), but note that only two of the node voltages, at nodes 1 and 2, are unknown. Let us therefore begin by writing the Kirchho current law equations for nodes 1 and 2; if we take the current direction as they happen to be drawn in Fig. 63 we have at node 1: I1 I2 I3 I4 0 A at node 2: I4 I5 I6 0 B
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Now apply eq. (68) to each current at nodes 1 and 2, then substitute into (A) and (B). Doing this (paying careful attention to battery polarities), eqs. (A) and (B) become 8 V1 V1 6 V1 V1 V2 0 6 15 5 8 and V1 V2 V2 12 10 V2 0 8 7 9 Now multiply the rst equation by 240 and the second by 504; this should give you the two simultaneous equations 67V1 15V2 304 63V1 191V2 48 the solutions of which are (using determinants is probably easiest) V1 4:9598 volts; answer; and V2 1:8873 volts; answer; both voltages with respect to the zero-volt reference node. 61. Since the trigonometric functions are de ned in terms of the ratios of the lengths of the sides of a right triangle to one another, let us, for simplicity, use the triangle below.
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p (By the Pythagorean theorem, the length of the vertical dashed line is equal to 3, as shown.) The answers are now found as follows. First, if we take the 608 angle as the reference angle, we have (see the de nitions just prior to eq. 69) p sin 608 3=2 0:866025 approx: and cos 608 1=2 0:500000
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Next, taking the 308 angle as the reference angle, we have sin 308 1=2 0:500000 and cos 308 p 3=2 0:866025; approx:
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66. Let us make use of the right triangle of Fig. 64 and eqs. (69) and (70) in section 5.2, as follows.
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