P28_089 - i 2 derives from a succession of symmetric...

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89. (a) The six resistors to the left of E 1 = 16 V battery can be reduced to a single resistor R =8 . 0Ω , through which the current must be i R = E 1 /R =2 . 0 A. Now, by the loop rule, the current through the 3 . 0Ωand1 . 0 Ω resistors at the upper right corner is i 0 = 16 . 0V 8 . 0V 3 . 0Ω+1 . 0Ω =2 . 0A in a direction that is “backward” relative to the E 2 =8 . 0 V battery. Thus, by the junction rule, i 1 = i R + i 0 =4 . 0A and is upward (that is, in the “forward” direction relative to E 1 ).
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Unformatted text preview: i 2 derives from a succession of symmetric splittings of i R (reversing the procedure of reducing those six resistors to Fnd R in part (a)). We Fnd i 2 = 1 2 µ 1 2 i R ¶ = 0 . 50 A and is clearly downward. (c) Using our conclusions from part (a) in Eq. 28-14, we obtain P = i 1 E 1 = (4)(16) = 64 W supplied. (d) Using results calculated in part (a) in Eq. 28-14, we obtain P = i E 2 = (2)(8) = 16 W absorbed....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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