ch27-p065 - 65. At t = 0 the capacitor is completely...

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65. At t = 0 the capacitor is completely uncharged and the current in the capacitor branch is as it would be if the capacitor were replaced by a wire. Let i 1 be the current in R 1 and take it to be positive if it is to the right. Let i 2 be the current in R 2 and take it to be positive if it is downward. Let i 3 be the current in R 3 and take it to be positive if it is downward. The junction rule produces iii 123 = + , the loop rule applied to the left-hand loop produces ε = iR iR 11 2 2 0, and the loop rule applied to the right-hand loop produces 22 33 0 = . Since the resistances are all the same we can simplify the mathematics by replacing R 1 , R 2 , and R 3 with R . (a) Solving the three simultaneous equations, we find i R 1 3 6 3 2 3 212 10 3 0 73 10 11 10 == × × . . . V A ch c h , (b) () 3 4 2 6 1.2 10 V 5.5 10 A, 3 30 .7310 i R × = × ×Ω and (c) 4 32 5.5 10 A. ii == × At t = the capacitor is fully charged and the current in the capacitor branch is 0. Thus, i 1 = i 2 , and the loop rule yields = 12 0.
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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ch27-p065 - 65. At t = 0 the capacitor is completely...

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