Atkinson, Albert Algernon, 1867- [from old catalog] · New York, D. Van Nostrand company · leaf 281
R = 2.08 ohms. <£ = 7°28/. E = 20.8 volts. Solution. — Joint im-pedance of parallel part isobtained by finding thethird side of the triangle.1 R base and 2 ir/C = 6.28 X100 x 0.0002 = 0.1256forming the perpendicular.The hypotenuse is 0.127.¥i^- 40. Its reciprocal = 7.87 ohms. Now lay the latter… … 2TTfc=A256
Electricity
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