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: nguyen (jmn727) – homework 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 A 0 . 14 μ F capacitor is given a charge Q . After 14 s, the capacitor’s charge is 0 . 5 Q . What is the effective resistance across this capacitor? Correct answer: 144 . 27 MΩ. Explanation: Let : C = 0 . 14 μ F = 1 . 4 × 10 7 F and t = 14 s . The charge on the capacitor is Q ( t ) = Q e t/τ . e t/τ = Q Q t τ = ln parenleftbigg Q Q parenrightbigg τ = t ln parenleftbigg Q Q parenrightbigg . The effective resistance is R = τ C = t C ln parenleftbigg Q Q parenrightbigg = 14 s (1 . 4 × 10 7 F) ln parenleftbigg Q . 5 Q parenrightbigg · 1 MΩ 10 6 Ω = 144 . 27 MΩ . 002 10.0 points The switch S has been in the position “a” for a long time. Then at t = 0, it is moved from “a” to “b”. C R 1 R 2 E S b a Find the time when the charge in the ca pacitor is reduced to 1 e of its value at t = 0. 1. τ = radicalbig R 1 R 2 C 2. τ = R 1 C 3. τ = 1 ( R 1 + R 2 ) C 4. τ = ( R 1 + R 2 ) C correct 5. τ = 1 √ R 1 R 2 C 6. τ = 1 R 2 C 7. τ = R 2 C 8. τ = R 1 + R 2 2 C 9. τ = 2 ( R 1 + R 2 ) C 10. τ = 1 R 1 C Explanation: In charging an R C circuit, the characteris tic time constant is given by τ = R C , where in this problem R...
View
Full
Document
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 at Austin.
 Spring '08
 Turner
 Charge, Work

Click to edit the document details