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oldhw11 - oldhomewk 11 LEE BENJAMIN Due 4:00 am Question 1...

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oldhomewk 11 – LEE, BENJAMIN – Due: Sep 28 2007, 4:00 am 1 Question 1, chap -1, sect -1. part 1 of 4 10 points An infinite chain of capacitors is pictured below with C 1 = 8 μ F, C 2 = 4 . 37 μ F, and C 3 = 11 . 3 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: 2 . 83924 μ F (tolerance ± 1 %). Explanation: Let : C 1 = 8 μ F , C 2 = 4 . 37 μ F , and C 3 = 11 . 3 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 . 37 μ F) 2 + 4 (8 μ F) (4 . 37 μ F) (11 . 3 μ F) (8 μ F) + (11 . 3 μ F) = 100 . 972 ( μ F) 2 , so C ab = C eq = - 1 2 (4 . 37 μ F) + 1 2 radicalBig 100 . 972 ( μ F) 2 = 2 . 83924 μ F . Question 2, chap -1, sect -1. part 2 of 4 10 points A dielectric with constant 2 . 67 is only in- serted into the first capacitor labeled C 2 and not into the others.

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oldhw11 - oldhomewk 11 LEE BENJAMIN Due 4:00 am Question 1...

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