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Unformatted text preview: terry (ect328) – homework 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 A 0 . 03 μ F capacitor is given a charge Q . After 9 s, the capacitor’s charge is 0 . 5 Q . What is the effective resistance across this capacitor? Correct answer: 432 . 809 MΩ. Explanation: Let : C = 0 . 03 μ F = 3 × 10 8 F and t = 9 s . The charge on the capacitor is Q ( t ) = Q e t/τ . e t/τ = Q Q t τ = ln Q Q τ = t ln Q Q . The effective resistance is R = τ C = t C ln Q Q = 9 s (3 × 10 8 F) ln Q . 5 Q · 1 MΩ 10 6 Ω = 432 . 809 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 = R 1 C 2. τ = 2 ( R 1 + R 2 ) C 3. τ = 1 ( R 1 + R 2 ) C 4. τ = 1 R 1 C 5. τ = 1 R 2 C 6. τ = ( R 1 + R 2 ) C correct 7. τ = R 1 + R 2 2 C 8. τ = R 2 C 9. τ = 1 √ R 1 R 2 C 10. τ = R 1 R 2 C Explanation: In charging an R C circuit, the characteris tic time constant is given by τ = R C , where in this problem R is the equivalent resistance, or R = R 1 + R 2 ....
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 Spring '10
 Turner
 Electric charge, Jaguar Racing, 770 W, 2676 J, 2676 W

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