{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solution_pdf16

# solution_pdf16 - tovar(jdt436 homework 16 Turner(59070 This...

This preview shows pages 1–2. Sign up to view the full content.

tovar (jdt436) – homework 16 – Turner – (59070) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 0 . 15 μ F capacitor is given a charge Q 0 . After 16 s, the capacitor’s charge is 0 . 5 Q 0 . What is the effective resistance across this capacitor? Correct answer: 153 . 887 MΩ. Explanation: Let : C = 0 . 15 μ F = 1 . 5 × 10 - 7 F and t = 16 s . The charge on the capacitor is Q ( t ) = Q 0 e - t/τ . e t/τ = Q 0 Q t τ = ln parenleftbigg Q 0 Q parenrightbigg τ = t ln parenleftbigg Q 0 Q parenrightbigg . The effective resistance is R = τ C = t C ln parenleftbigg Q 0 Q parenrightbigg = 16 s (1 . 5 × 10 - 7 F) ln parenleftbigg Q 0 0 . 5 Q 0 parenrightbigg · 1 MΩ 10 6 Ω = 153 . 887 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 + R 2 2 C 2. τ = ( R 1 + R 2 ) C correct 3. τ = 1 R 1 R 2 C 4. τ = R 2 C 5. τ = 2 ( R 1 + R 2 ) C 6. τ = 1 R 1 C 7. τ = 1 R 2 C 8. τ = 1 ( R 1 + R 2 ) C 9. τ = radicalbig R 1 R 2 C 10. τ = 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 is the equivalent resistance, or R = R 1 + R 2 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 4

solution_pdf16 - tovar(jdt436 homework 16 Turner(59070 This...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online