BME211HW1 - HW#1 Solutions P1.2-1 i t d 41 e dt 5t 20 e 5t...

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HW#1 Solutions P1.2-1 ( 29 ( 29 5 5 4 1 20 t t d i t e e dt - - = - = A P1.2-2 ( 29 ( 29 ( 29 ( 29 5 5 5 0 0 0 0 4 4 0 4 1 0 4 4 4 5 5 t t t t t q t i d q e d d e d t e τ τ τ τ τ τ τ - - - = + = - + = - = + - C P1.2-6 5 C = 600 A = 600 s C s mg Silver deposited = 600 20min 60 1.118 = 8.05 10 mg=805 g s min C i d P2.4-5 1 2 1 2 1 1 1 1 1 150 V; 50 ; 25 and adhere to the passive convention so 150 3 A 50 s v v v R R v i v i R = = = = = = = = 2 2 2 2 2 150 and do not adhere to the passive convention so 6 A 25 v v i i R = - = - = - 1 1 1 1 The power absorbed by is 150 3 450 W R P v i = = = 2 2 2 2 The power absorbed by is 150( 6) 900 W R P v i = - = - - = P3.2-15 We can label the circuit as shown. The subscripts suggest a numbering of the circuit elements. Apply KVL to node the left mesh to get 1 1 1 20 15 25 20 0 0.5 A 40 i i i + - = = = Apply KVL to node the left mesh to get ( 29 2 1 2 1 25 0 25 25 0.5 12.5 V v i v i - = = = = Apply KCL to get m 2 i i = . Finally, apply Ohm’s law to the 50 resistor to get 2 m 2 12.5 0.25 A 50 50 v i i = = = =
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P4.2-1 KCL at node 1: 4 4 2 1 1 2 0 1.5 1.5 A 8 6 8 6 v v v i i i i - - - - = + + = + + = - + = (checked using LNAP 8/13/02) P4.2-2 KCL at node 1: 1 2 1 1 0 5 20 1 2 20 5 v v v v v - + + = - = - KCL at node 2: 1 2 2 3 2 3 2 40 1 2 3 20 10 v v v v v v v - - + = - + - = KCL at node 3: 2 3 3 1 3 5 30 2 3 10 15 v v v v v - + = - + = Solving gives v 1 = 2 V, v 2 = 30 V and v 3 = 24 V. P4.2-3 KCL at node 1: 4 15 4 1 2 1 2 A 1 1 5 20 5 20 v v v i i - - + = = + = - KCL at node 2: 1 2 2 3 2 5 15
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  • Winter '07
  • Cain
  • Resistor, Harshad number, 3 ohm, passive convention, V. P4.2-3 KCL, 8.05 105 mg

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