RECITATION-CH 27

RECITATION-CH 27 - RECITATION 7 9. (a) and (b) Assume i1...

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RECITATION 7 9. (a) and (b) Assume 1 i travels to the right, 2 i up and 3 i down. We use the loop rule twice, clockwise around the outer loop and clockwise around the right loop: ( 29 1 1 3 3 10 V 0 V; i R i R - - = ( 29 2 2 3 3 5.0 V 0 V. i R i R - - = We use the junction rule ( 29 3 1 2 i i i = + to eliminate 3 i in the loop rule equations. Substituting the given data into the loop rule equations yields ( 29 ( 29 1 2 10 V 8.0 4.0 ; i i = Ω + ( 29 ( 29 2 1 5.0 V 8.0 4.0 . i i = Ω + Solving for 1 2 and , i i we find 1 1.25 A. i = and 2 0 A i = . Therefore 3 1 2 1.25 A. i i i = + = As assumed above, 3 i is traveling down in the picture. 11. (a) The resistors R 2 , R 3 and R 4 are in parallel with each other, and that combination is in series with R 1 . The equivalent resistance for this network is 1 1 eq 1 2 3 4 1 1 1 1 1 1 100 118.75 120 . 50 50 75 R R R R R - - = + + + = Ω + + + = Ω ≈ (b) Based on our result from part (a), the current through the battery must be
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RECITATION-CH 27 - RECITATION 7 9. (a) and (b) Assume i1...

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