Hw14_pdf - mcconnell (kam2342) – homework14 – Turner...

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Unformatted text preview: mcconnell (kam2342) – homework14 – Turner – (60230) 1 This print-out should have 12 questions. Multiple-choice 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 . 7 μ F, C 2 = 4 . 46 μ F, and C 3 = 16 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 . 02636 μ F. Explanation: Let : C 1 = 14 . 7 μ F , C 2 = 4 . 46 μ F , and C 3 = 16 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 = (4 . 46 μ F) 2 + 4 (14 . 7 μ F) (4 . 46 μ F) (16 μ F) (14 . 7 μ F) + (16 μ F) = 156 . 568 ( μ F) 2 , so C ab = C eq =- 1 2 (4 . 46 μ F) + 1 2 radicalBig 156 . 568 ( μ F) 2 = 4 . 02636 μ F . 002 (part 2 of 4) 10.0 points A dielectric with constant 2 . 22 is only inserted into the first capacitor labeled C 2 and not into the others. What is C ab now? Correct answer: 4 . 94249 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 = (2 . 22)(4 . 46 μ F) = 9 . 9012 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 . 7 μ F)(9 . 9012 μ F + 4 . 02636 μ F)(16 μ F) 14 . 7 μ F + 9 . 9012 μ F + 4 . 02636 μ F + 16 μ F = 4 . 94249 μ F . 003 (part 3 of 4) 10.0 points mcconnell (kam2342) – homework14 – Turner – (60230) 2 A 4 . 44 V battery is placed between the termi- nals ab ....
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This note was uploaded on 11/22/2010 for the course PHYS 303 taught by Professor Turner during the Spring '10 term at University of Texas.

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Hw14_pdf - mcconnell (kam2342) – homework14 – Turner...

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