HW3solutions

HW3solutions - so i x = 12/5 A = 2.4 A. P4.4-9 Apply KCL at...

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ECE35 Win’09 Homework 3 Solutions P4.2-4 Node equations: When v 1 = 1 V, v 2 = 2 V P4.3-2 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 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 terms of the node voltages: Next, express the controlled voltage in terms of the node voltages:
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Unformatted text preview: 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 . 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 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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This note was uploaded on 04/15/2010 for the course ECE ece35 taught by Professor Xiejiang during the Winter '10 term at UCSD.

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HW3solutions - so i x = 12/5 A = 2.4 A. P4.4-9 Apply KCL at...

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