This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: terry (ect328) oldhomework 16 Turner (59130) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points 2 . 6 10 6 1 . 9 F 10 V S An emf of 10 V is connected to a series RC circuit consisting of a 2 . 6 10 6 resistor and a 1 . 9 F capacitor. Find the time required for the charge on the capacitor to reach 88% of its final value. Correct answer: 10 . 4741 s. Explanation: Let : R = 2 . 6 10 6 C = 1 . 9 F = 1 . 9 10 6 F E = 10 V , and p = 0 . 88 . From Q = Q max (1 e t/ ), where = R C , we get p = Q Q max = 1 e t/ e t/ = 1 p t = ln(1 p ) t = ln(1 p ) = R C ln(1 p ) = ( 2 . 6 10 6 )( 1 . 9 10 6 F ) ln(1 . 88) = 10 . 4741 s . keywords: 002 (part 1 of 4) 10.0 points A typical television dissipates 284 W when plugged into a 115 V outlet. Find the resistance of the television. Correct answer: 46 . 5669 . Explanation: Let : P = 284 W and V = 115 V . Power is P = V 2 R so the television resistance is R TV = V 2 P TV = (115 V) 2 284 W = 46 . 5669 . 003 (part 2 of 4) 10.0 points The 2 . 41 fuse (integrated into the outlet) and wires connecting the outlet form a series circuit that works like a voltage divider. As sume there are no other appliances plugged into the outlet....
View
Full
Document
This note was uploaded on 02/17/2010 for the course PHY 59130 taught by Professor Turner during the Fall '09 term at University of Texas.
 Fall '09
 Turner

Click to edit the document details