hw8 - wilson(kw22588 – HW8 – Mackie –(20224 1 This...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: wilson (kw22588) – HW8 – Mackie – (20224) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. warning: this includes capacitors with di- electrics, which is material from before break. 001 (part 1 of 7) 10.0 points A capacitor network is shown in the following figure. 19 . 5 V 5 . 6 μ F 4 . 3 μ F 13 . 3 μ F a b What is effective capacitance C ab of the entire capacitor network? Correct answer: 15 . 7323 μ F. Explanation: Let : C 1 = 5 . 6 μ F , C 2 = 4 . 3 μ F , C 3 = 13 . 3 μ F , and E B = 19 . 5 V . E B C 1 C 2 C 3 a b C 1 and C 2 are in series with each other, and they are together are parallel with C 3 . So C ab = C 1 C 2 C 1 + C 2 + C 3 = (5 . 6 μ F) (4 . 3 μ F) 5 . 6 μ F + 4 . 3 μ F + 13 . 3 μ F = 15 . 7323 μ F . 002 (part 2 of 7) 10.0 points What is the voltage across the 4 . 3 μ F upper right-hand capacitor? Correct answer: 11 . 0303 V. Explanation: Since C 1 and C 2 are in series they carry the same charge C 1 V 1 = C 2 V 2 , and their voltages add up to V , voltage of the battery V 1 + V 2 = V C 2 V 2 C 1 + V 2 = V C 2 V 2 + C 1 V 2 = V C 1 V 2 = V C 1 C 1 + C 2 = (19 . 5 V)(5 . 6 μ F) 5 . 6 μ F + 4 . 3 μ F = 11 . 0303 V . 003 (part 3 of 7) 10.0 points If a dielectric of constant 2 . 39 is inserted in the 4 . 3 μ F top right-hand capacitor (when the battery is connected), what is the elec- tric potential across the 5 . 6 μ F top left-hand capacitor? Correct answer: 12 . 6221 V. Explanation: Let : κ = 2 . 39 . When the dielectric is inserted, the capaci- tance formerly C 2 becomes C ′ 2 = κC 2 , and the new voltage across C 1 is V ′ 1 = V C ′ 2 C 1 + C ′ 2 = κV C 2 C 1 + κC 2 = (2 . 39)(19 . 5 V)(4 . 3 μ F) 5 . 6 μ F + (2 . 39)(4 . 3 μ F) = 12 . 6221 V . wilson (kw22588) – HW8 – Mackie – (20224) 2 004 (part 4 of 7) 10.0 points If the battery is disconnected and then the dielectric is removed, what is the charge on 5 . 6 μ F top left-hand capacitor? Correct answer: 51 . 0255 μ C. Explanation: Immediately before the battery was discon- nected the charges on the capacitors had been Q ′ 3 = C 3 V = (13 . 3 μ F)(19 . 5 V) = 259 . 35 μ C Q ′ 1 = Q ′ 2 = C ′ 12 V = (3 . 62482 μ F)(19 . 5 V) = 70 . 6839 μ C . When we remove the dielectric, the sum of the charges stays the same, and the voltages on C 3 and on C 12 (where C 12 is the equivalent capacitance of C 1 and C 2 in series) are equal to each other Q ′′ 1 + Q ′′ 3 = Q ′ 1 + Q ′ 3 . (1) Since the potential drop across the top and bottom are equal, we have V ab = Q ′′ 1 C 12 = Q ′′ 3 C 3 , so Q ′′ 3 = C 3 C 12 Q ′′ 1 ....
View Full Document

This note was uploaded on 02/17/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

Page1 / 11

hw8 - wilson(kw22588 – HW8 – Mackie –(20224 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online