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**Unformatted text preview: **i 3 from the equations by writing 2 1 3 i i i − − = . Plugging this into the other two equations we obtain 2 1 2 1 2 2 1 11 6 ) ( 6 5 2 5 6 2 i i i i i i i − − = − − + − = + − = We now have two equations and two unknowns, so we’re heading in the right direction. Now rewrite the upper equation to 1 2 6 2 5 i i + = and substitute this into the other: A. 33 . 3 1 2 . 19 4 . 6 2 . 13 4 . 4 6 5 6 2 11 6 2 1 1 1 1 1 − = − = − = ⇒ − − − = + − − = i i i i i (a) and (b) We see that i 1 = 0.33 A and it’s directed down . (c) and (d) The current i 2 is actually zero: ( ) ( ) 5 / ) 3 / 1 ( 6 2 5 / 6 2 1 2 = − + = + = i i . (e) and (f) The current i 3 is given by A 33 . 2 1 3 + = − − = i i i , so it’s directed up . (g) To find V a – V b we walk from point b to a : V. . 5 5 5 2 2 2 2 = − = − i R i...

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