This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: terms of the node voltages: Next, express the controlled voltage in terms of the node voltages: v 1v 2 = 3i x , or using v 1 =12V 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:...
View
Full Document
 Winter '10
 JosephFord
 Volt, Mesh Analysis, Voltage drop, node voltages

Click to edit the document details