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HW3solutionsWIN10

# HW3solutionsWIN10 - terms of the node voltages Next express...

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ECE35 Winter 2010 Homework 3 Solutions Assignment: Problems 4.2-4, 4.3-2, 4.3-4, 4.3-12, 4.4-3, 4.4-5, 4.4-9, 4.5-2, 4.5-6, 4.6-3, 4.6-5 P4.2-4 Node equations: When v 1 = 1 V, v 2 = 2 V P4.3-2 Comment: can do this more easily by KCL at supernode, expressing voltages in terms on one variable (v b ) Express the branch voltage of each voltage source in terms of its node voltages to get: KCL at node b : KCL at the supernode corresponding to the 8 V source: so Consequently

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P4.3-4 Apply KCL to the supernode: Solving yields (checked using LNAP 8/13/02) P4.3-12 Express the voltage source voltages in terms of the node voltages: Apply KVL to the supernode to get Comment: the current source flows from one side of the supernode to the other, so the contribution cancels. So The node voltages are P4.4-3 First express the controlling current in terms of the node voltages: Write and solve a node equation: P4.4-5 First, express the controlling current of the CCVS in

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Unformatted text preview: terms of the node voltages: Next, express the controlled voltage in terms of the node voltages: v 1-v 2 = 3i x , or using v 1 =12V so i x = 12/5 A = 2.4 A. P4.4-9 Apply KCL at node 2: but so Next apply KCL to the supernode corresponding to the voltage source. P4.5-2 Top mesh: so R = 12 Ω . Bottom, right mesh: so v 2 = − 28 V. Bottom left mesh so v 1 = − 4 V. P4.5-6 so the simplified circuit is The mesh equations are or The power supplied by the 12 V source is . The power supplied by the 8 V source is . The power absorbed by the 30 Ω resistor is . P4.6-3 Express the current source current as a function of the mesh currents: Apply KVL to the supermesh: P4.6-5 Express the current source current in terms of the mesh currents: Supermesh: Lower, left mesh: Eliminating i 1 and i 2 from the supermesh equation: The voltage measured by the meter is:...
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HW3solutionsWIN10 - terms of the node voltages Next express...

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