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Unformatted text preview: so i x = 12/5 A = 2.4 A. P4.49 Apply KCL at node 2: but so Next apply KCL to the supernode corresponding to the voltage source. P4.52 Top mesh: so R = 12 Ω . Bottom, right mesh: so v 2 = − 28 V. Bottom left mesh so v 1 = − 4 V. P4.56 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 . P4.63 Express the current source current as a function of the mesh currents: Apply KVL to the supermesh: P4.65 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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 Winter '10
 xiejiang
 Volt, Mesh Analysis, Voltage drop, node voltages

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