BME211HW2

# BME211HW2 - Homework#2 Solutions Ex 4.3-2 v 8 b 10 12 v b...

• Notes
• heavenlygrl
• 4

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Homework #2 Solutions Ex. 4.3-2 ( 29 ( 29 8 12 3 8 V and 16 V 10 40 v v b b v v b a + - - + = = = Ex. 4.4-1 Apply KCL at node a to express i a as a function of the node voltages. Substitute the result into 4 b a v i = and solve for v b . 9 6 4 4 4.5 V 8 12 12 v v b b i v i v a b a b + + = = = = Ex. 4.6-2 Mesh equation: ( 29 ( 29 ( 29 33 2 15 3 6 3 0 3 6 15 6 3 = 3 A 9 3 i i i i - + + + = + = - - = - P4.3-3 Apply KCL to the supernode: 10 8 .03 0 7 V 100 100 100 a a a a v v v v - - + + - = =

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P4.4-2 Write and solve a node equation: 6 4 0 12 V 1000 2000 3000 a a a a a v v v v v - - + + = = 4 12 mA 3000 a a b v v i - = = - P4.4-3 First express the controlling current in terms of the node voltages: 2 4000 b a v i - = Write and solve a node equation: 2 2 5 0 1.5 4000 2000 4000 b b b b v v v v - - - + - = = P4.4-5 First, express the controlling current of the CCVS in terms of the node voltages: 2 2 x v i = Next, express the controlled voltage in terms of the node voltages: 2 2 2 24 12 3 3 V 2 5 x v v i v - = = = so i x = 12/5 A = 2.4 A.
P4.6-5 Express the current source current in terms of the mesh currents: 3 1 1 3 2 2 i i i i - = = - Supermesh: ( 29 1 3 2 3 1 2 3 6 3 5 8 0 6 5 8 8 i i i i i i i + - - - = - + = Lower, left mesh: ( 29

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Unformatted text preview: i i i-+ +-= ⇒ = + Eliminating i 1 and i 2 from the supermesh equation: ( 29 ( 29 3 3 3 3 6 2 4 5 8 8 9 24 i i i i--+ + = ⇒ = The voltage measured by the meter is: 3 24 3 3 8 V 9 i = = P4.7-3 Express the controlling current of the dependent source as a function of the mesh current: .06 b a i i =-Apply KVL to the right mesh: a a a 100 (0.06 ) 50 (0.06 ) 250 10 mA a i i i i--+-+ = = Finally: o b 50 50(0.06 0.01) 2.5 V v i = =-= P4.7-15 Label the node voltages as shown. The controlling currents of the CCCS is expressed as a 28 v i = . The node equations are a a b a 12 28 4 14 v v v v-= + + and a b a b 4 14 8 v v v v-+ = Solving the node equations gives a 84 V v = and b 72 V v = . Then a 84 3 A 28 28 v i = = = ....
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• Winter '07
• Cain
• Mesh Analysis, Harshad number, Voltage drop, node voltages

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