Explanation Consider system circuits, + i 1 + i 3 = i 2 (1) − 8 − 5 i 1 + 8 i 3 = 0 (2) − 48 + 6 i 2 + 8 i 3 = 0 (3) Rewrite the above equations as, i 1 − i 2 + i 3 = 0 (3) − 5 i 1 + 8 i 3 = 8 (4) 6 i 2 + 8 i 3 = 48 (5) Multiply the equation (3) by 5, then add the equation (3) and (4). 5 i 1 − 5 i 2 + 5 i 3 = 0 − 5 i 1 + 8 i 3 = 8 ________________ − 5 i 2 + 13 i 3 = 8 (6) Multiply the equation (5) and (6) by 6 and 5, respectively, then add the equation (5) and (6). 30 i 2 + 40 i 3 = 240 − 30 i 2 + 78 i 3 = 48 ________________ + 118 i 3 = 288 That is, i 3 = 144 59 . Substitute i 3 = 144 59 in the equation (6). − 5 i 2 + 13 ( 144 59 ) = 8 − 5 i 2 + 1872 59 = 8 − 5 i 2 = 472 − 1872 59 i 2 = − 1400 59 ⋅ 5 That is, i 2 = 280 59 . Substitute the values i 3 = 144 59 in the equation (4). − 5 i 1 + 8 ( 144 59 ) = 8 − 5 i 1 + 1152 59 = 8 − 5 i 1 = 8 − 1152 59 − 5 i 1 = 472 − 1152 59 On further simplification, − 5 i 1 = 472 − 1152 59 − 5 i 1 = − 680 59 i 1 = 136 59 Therefore, the solution of the system circuit is i 1 = 136 59 , i 2 = 280 59 and i 3 = 144 59

3rd Edition

ISBN: 9780321880116

Author: Michael Sullivan III, Katherine R. Struve, Janet Mazzarella

Publisher: PEARSON

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Chapter C.2, Problem 46E

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To solve: The system circuit { +i1+ i3=i2−8−5i1+8i3=0−48+6i2+8i3=0 and find the currents i1,i2 and i3.

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Chapter C.2, Problem 45E

Chapter C.2, Problem 47E

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The system circuit { + i 1 + i 3 = i 2 − 8 − 5 i 1 + 8 i 3 = 0 − 48 + 6 i 2 + 8 i 3 = 0 and find the currents i 1 , i 2 and i 3 .

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To solve: The system circuit { +i1+ i3=i2−8−5i1+8i3=0−48+6i2+8i3=0 and find the currents i1,i2 and i3.

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Chapter C Solutions