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: P2.25 (a) (b) Using the quadratic formula When v = 0.8 V then . When v = 1.0 V then . (c) Using the quadratic formula So there is no real solution to the equation. P2.45 P2.52 (a) (b) P2.61 (a) (b) The voltage, 12 V, and the current, 0.5 A, of the voltage source obey the passive convention so the power P = 12 (0.5) = 6 W is the power received by the source. The voltage source delivers 6 W. P2.65 a.) b.) Element Power supplied voltage source current source resistor total 0 P2.77 i d and v d adhere to the passive convention so is the power received by the dependent source. P2.93 (a) v = 12 V (b) (c) v = 0 V (d) (the box symbol was supposed to be ~, ‘approximately’) DP22 1.) 2.) Therefore 20 < R < 3.75 Ω . These conditions cannot be satisfied simultaneously....
View
Full Document
 Winter '07
 MassimoFranceschetti
 Volt, Quadratic equation, Current Source, total power, REFERENCE DIRECTION

Click to edit the document details