HW1-8-11

# HW1-8-11 - 1 R alone is 29 W 5 367 6 131 V 220 2 = Ω = P...

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MAE 3780 MECHATRONICS FALL 2009 1 HOMEWORK 1 SOLUTIONS 8. a. By KCL, the current through element B is 5 A to the right. By KVL, 0 V 5 V 10 3 = + + - - V V A , so V 12 = A V (positive at the top of element A). Thus, A generates ( 29 ( 29 W 60 A 5 V 12 = . B generates ( 29 ( 29 W 15 A 5 V 3 = . C dissipates ( 29 ( 29 W 25 A 5 V 5 = . D dissipates ( 29 ( 29 W 30 A 3 V 10 = . E dissipates ( 29 ( 29 W 20 A 2 V 10 = . b. Conservation of energy is satisfied since total power generated equals total power dissipated. 9. a. For the parallel connection: ( 29 ( 29 2 2 1 2 V 220 V 220 W 2000 R R + = For the series connection: ( 29 2 1 2 V 220 W 300 R R + = Solving these two equations gives = 68 . 131 1 R , = 64 . 29 2 R (Note that the order of these two is not important.) b. The power dissipated by

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Unformatted text preview: 1 R alone is ( 29 W 5 . 367 6 . 131 V 220 2 = Ω = P The power dissipated by 2 R alone is ( 29 W 5 . 1632 7 . 29 V 220 2 = Ω = P MAE 3780 MECHATRONICS FALL 2009 2 10. MAE 3780 MECHATRONICS FALL 2009 3 11. Let A I be the current through the 1k Ω resistor. Then, by the current divider rule: I R R I x x A Ω + = k 1 , where x = 1,2,3 depending on the position of the switch. Solving for x R gives -Ω = A A x I I I R k 1 . Setting A 30 μ = A I , gives the following results: Ω = ⇒ = 3 R mA 10 1 I Ω = ⇒ = 3 . R mA 100 2 I Ω = ⇒ = 03 . R A 1 3 I...
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HW1-8-11 - 1 R alone is 29 W 5 367 6 131 V 220 2 = Ω = P...

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