BME211HW4 - P4.4-6 Pick a reference node and label the unknown node voltages Express the controlling current of the dependent source in terms of

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Unformatted text preview: P4.4-6 Pick a reference node and label the unknown node voltages: Express the controlling current of the dependent source in terms of the node voltages: va va i 4 = - . Then v b = 2 i 4 = - . 6 3 Apply KCL at node a: v a - 12 v a v a - v b + + =0 3 6 4 So: v va - a - v a - 12 v a 3 0 + + = 3 6 4 va 4 ( v a - 12 ) + 2 v a + 3 a + 0 v a = 4.8 V v = 3 12 - 4.8 7.2 = = 2.4 A The current in the 12-V voltage source is i = 3 3 So the power supplied by the voltage source is 12(2.4) = 28.8 W. P4.4-7 Label the node voltages: First, v2 = 10 V, due to the independent voltage source. Next, express va and ib, the controlling voltage and current of the dependent sources, in terms of the node voltages: ib = and v3 - v2 8 = v 3 - 10 8 v a = v1 - v 2 = v1 - 10 Next, express ib and 3va, the controlled voltages of the dependent sources, in terms of the node voltages: v 3 - 10 8i b = v 1 - v 3 8 = v1 - v 3 8 and 3v a = v1 3 ( v1 - 10 ) = v1 v1 = 15 V So Next v a = 15 - 10 = 5 V and ib = v 3 - 10 = 15 - v 3 v 3 = 12.5 V 12.5 - 10 = 0.3125 A 8 Finally, apply KCL to the top node to get ic = P4.6-5 va 2 + ib = 5 + 0.3125 = 2.8125 A 2 Express the current source current in terms of the mesh currents: i 3 - i1 = 2 i1 = i 3 - 2 Supermesh: 6 i1 + 3 i 3 - 5 ( i 2 - i 3 ) - 8 = 0 6 i1 - 5 i 2 + 8 i 3 = 8 Lower, left mesh: -12 + 8 + 5 ( i 2 - i 3 ) = 0 5 i 2 = 4 + 5 i 3 Eliminating i1 and i2 from the supermesh equation: 6 ( i 3 - 2 ) - ( 4 + 5 i 3 ) + 8 i 3 = 8 9 i 3 = 24 24 = The voltage measured by the meter is: 3 i 3 = 3 8 V 9 P4.6-6 Mesh equation for right mesh: 4 ( i - 2 ) + 2 i + 6 ( i + 3) = 0 12 i - 8 + 18 = 0 i = - P4.6-8 Use units of V, mA and k. Express the currents to the supermesh to get i1 - i 3 = 2 Apply KVL to the supermesh to get 4 ( i1 - i 3 ) + ( 1) i 3 - 3 + ( 1) ( i1 - i 2 ) = 0 *Correction:4(i3-i2) for the first term in KVL Apply KVL to mesh 2 to get 2i 2 + 4 ( i 2 - i 3 ) + ( 1) ( i 2 - i1 ) = 0 Solving, e.g. using MATLAB, gives 1 0 -1i1 2 -5 5 i 1 = 3 2 1 7 -4 i 3 - 0 P4.7-2 ib = 4ib - ia ib = 1 ia 3 i1 3 i2 = 1 i3 1 i1 - 5 i 2 + 5 i 3 = 3 * 10 5 A=- A 12 6 ( -1) i1 + 7i 2 - 4i 3 = 0 1 -100 ia 200ia + 8 = 0 + 3 ia = - 0.048 A P4.7-4 Express the controlling voltage of the dependent source as a function of the mesh current: vb = 100 (.006 - ia ) Apply KVL to the right mesh: -100 (.006 - ia ) + 3 [ 100(.006 - ia ) ] + 250 ia = 0 ia = -24 mA P4.7-8 Label the mesh currents: Express ia, the controlling current of the CCCS, in terms of the mesh currents i a = i 3 - i1 Express 2 ia, the controlled current of the CCCS, in terms of the mesh currents: i1 - i 2 = 2 i a = 2 ( i 3 - i1 ) Apply KVL to the supermesh corresponding to the CCCS: 80 ( i1 - i 3 ) + 40 ( i 2 - i 3 ) + 60 i 2 + 20 i1 = 0 Apply KVL to mesh 3 10 + 40 ( i 3 - i 2 ) + 80 ( i 3 - i1 ) = 0 -80 i1 - 40 i 2 + 120 i 3 = -10 100i1 + 100i 2 - 120i 3 = 0 3 i1 - i 2 - 2 i 3 = 0 These three equations can be written in matrix form -1 -2 i1 0 3 100 100 -120 i 2 0 = 80 -40 120 i 3 10 - - Solving, e.g. using MATLAB, gives i1 = -0.2 A, i 2 = -0.1 A and i 3 = -0.25 A Apply KVL to mesh 2 to get v b + 40 ( i 2 - i 3 ) + 60 i 2 = 0 v b = -40 ( -0.1 - ( -0.25 ) ) - 60 ( -0.1) = 0 V So the power supplied by the dependent source is p = v b ( 2i a ) = 0 W . P5.2-3 Source transformation at left; equivalent resistor for parallel 6 and 3 resistors: Equivalents for series resistors, series voltage source at left; series resistors, then source transformation at top: Source transformation at left; series resistors at right: Parallel resistors, then source transformation at left: Finally, apply KVL to loop - 6 + i (9 + 19) - 36 - vo = 0 i = 5 / 2 vo = -42 + 28 (5 / 2) = 28 V P5.2-5 -12 - 6 ia + 24 - 3 ia - 3 = 0 ia = 1 A ...
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This note was uploaded on 03/07/2008 for the course BME 211 taught by Professor Cain during the Winter '07 term at University of Michigan.

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