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Unformatted text preview: nguyen (jmn727) – oldhomework 16 – Turner – (59070) 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 . 2 × 10 6 Ω 2 . 7 μ F 10 V S An emf of 10 V is connected to a series RC circuit consisting of a 2 . 2 × 10 6 Ω resistor and a 2 . 7 μ F capacitor. Find the time required for the charge on the capacitor to reach 96% of its final value. Correct answer: 19 . 1201 s. Explanation: Let : R = 2 . 2 × 10 6 Ω C = 2 . 7 μ F = 2 . 7 × 10 − 6 F E = 10 V , and p = 0 . 96 . From Q = Q max (1 e − t/τ ), where τ = R C , we get p = Q Q max = 1 e − t/τ e − t/τ = 1 p t τ = ln(1 p ) t = τ ln(1 p ) = R C ln(1 p ) = ( 2 . 2 × 10 6 Ω )( 2 . 7 × 10 − 6 F ) × ln(1 . 96) = 19 . 1201 s . keywords: 002 (part 1 of 4) 10.0 points A typical television dissipates 253 W when plugged into a 115 V outlet. Find the resistance of the television. Correct answer: 52 . 2727 Ω. Explanation: Let : P = 253 W and V = 115 V . Power is P = V 2 R so the television resistance is R TV = V 2 P TV = (115 V) 2 253 W = 52 . 2727 Ω . 003 (part 2 of 4) 10.0 points The 2 . 82 Ω 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....
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This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
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