This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Kapoor (mk9499) – homework14 – Turner – (60230) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points An infinite chain of capacitors is pictured be low with C 1 = 14 . 9 μ F, C 2 = 7 . 91 μ F, and C 3 = 11 . 1 μ F. C 1 a b C 2 C 3 C 1 C 2 C 3 C 1 C 2 C 3 What is C ab ? Correct answer: 4 . 1665 μ F. Explanation: Let : C 1 = 14 . 9 μ F , C 2 = 7 . 91 μ F , and C 3 = 11 . 1 μ F . The capacitance C eq = C ab . Imagine points a ′ and b ′ in the circuit just past the first capacitor C 2 . The equivalent circuit to the right of points a ′ and b ′ is C eq , as shown in the figure below. C 1 a a' b' b C 2 C 3 C eq That is, we are replacing the circuit to the right of points a ′ and b ′ with the circuit we are trying to resolve. Solving for C eq 1 C eq = 1 C 1 + 1 C 2 + C eq + 1 C 3 1 C eq 1 C 2 + C eq = 1 C 1 + 1 C 3 C 2 C eq ( C 2 + C eq ) = C 1 + C 3 C 1 C 3 C 1 C 2 C 3 C 1 + C 3 = C 2 eq + C 2 C eq C 2 eq + C 2 C eq C 1 C 2 C 3 C 1 + C 3 = 0 C eq = C 2 + radicalbigg C 2 2 + 4 C 1 C 2 C 3 C 1 + C 3 2 . Under the radical, C 2 2 + 4 C 1 C 2 C 3 C 1 + C 3 = (7 . 91 μ F) 2 + 4 (14 . 9 μ F) (7 . 91 μ F) (11 . 1 μ F) (14 . 9 μ F) + (11 . 1 μ F) = 263 . 835 ( μ F) 2 , so C ab = C eq = 1 2 (7 . 91 μ F) + 1 2 radicalBig 263 . 835 ( μ F) 2 = 4 . 1665 μ F . 002 (part 2 of 4) 10.0 points A dielectric with constant 3 . 49 is only inserted into the first capacitor labeled C 2 and not into the others. What is C ab now? Correct answer: 5 . 30003 μ F. Explanation: After the dielectric is added denote symbols with primes, C 2 → C ′ 2 , C ab → C ′ ab , etc. C ′ 2 = κC 2 = (3 . 49)(7 . 91 μ F) = 27 . 6059 μ F . C ′ ab = 1 1 C 1 + 1 C ′ 2 + C eq + 1 C 3 = C 1 ( C ′ 2 + C eq ) C 3 C 1 + C ′ 2 + C eq + C 3 = (14 . 9 μ F)(27 . 6059 μ F + 4 . 1665 μ F)(11 . 1 μ F) 14 . 9 μ F + 27 . 6059 μ F + 4 . 1665 μ F + 11 . 1 μ F = 5 . 30003 μ F . 003 (part 3 of 4) 10.0 points Kapoor (mk9499) – homework14 – Turner – (60230) 2 A 7 . 38 V battery is placed between the termi nals ab ....
View
Full
Document
This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Physics, Work

Click to edit the document details