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problem26_89

# problem26_89 - A 300 1 = I A 500 2 = I A 200 3 = I just as...

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26.89: a) Using Kirchhoff’s Rules on the circuit we find: Left loop: . 0 210 140 147 0 55 210 140 92 2 1 2 1 = - - = + - - I I I I Right loop: . 0 35 210 112 0 55 210 35 57 3 2 2 3 = - - = + - - I I I I Currents: . 0 3 2 1 = + - I I I Solving for the three currents we have: A, 300 . 0 1 = I A, 500 . 0 2 = I A. 200 . 0 3 = I b) Leaving only the 92-V battery in the circuit: Left loop: . 0 210 140 92 2 1 = - - I I Right loop: . 0 210 35 2 3 = - - I I Currents: . 0 3 2 1 = + - I I I Solving for the three currents: A, 541 . 0 1 = I A, 077 . 0 2 = I . A 464 . 0 3 - = I c) Leaving only the 57-V battery in the circuit: Left loop: . 0 210 140 2 1 = + I I Right loop: . 0 210 35 57 2 3 = - - I I Currents: . 0 3 2 1 = + - I I I Solving for the three currents: A, 287 . 0 1 - = I , A 192 . 0 2 = I A. 480 . 0 3 = I d) Leaving only the 55-V battery in the circuit: Left loop: . 0 210 140 55 2 1 = - - I I Right loop: . 0 210 35 55 2 3 = - - I I Currents: . 0 3 2 1 = + - I I I Solving for the three currents: A, 046 . 0 1 = I , A 231 . 0 2 = I A. 185 . 0 3 = I e) If we sum the currents from the previous three parts we find:
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Unformatted text preview: A, 300 . 1 = I A, 500 . 2 = I A, 200 . 3 = I just as in part (a). f) Changing the 57-V battery for an 80-V battery just affects the calculation in part (c). It changes to: Left loop: . 210 140 2 1 = + I I Right loop: . 210 35 80 2 3 =--I I Currents: . 3 2 1 = +-I I I Solving for the three currents: , A 403 . 1-= I A, 269 . 2 = I A. 672 . 3 = I So the total current for the full circuit is the sum of (b), (d) and (f) above: A, 184 . 1 = I A, 576 . 2 = I A. 392 . 3 = I...
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