ch27-p065 - is zero Then i 1 = i 2 and the loop rule yields...

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65 (a), (b), and (c) 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 i 1 = i 2 + i 3 , the loop rule applied to the left-hand loop produces E− i 1 R 1 i 2 R 2 =0 , and the loop rule applied to the right-hand loop produces i 2 R 2 i 3 R 3 =0 . Since the resistances are all the same, you can simplify the mathematics by replacing R 1 , R 2 , and R 3 with R . The solution to the three simultaneous equations is i 1 = 2 E 3 R = 2(1 . 2 × 10 3 V) 3(0 . 73 × 10 6 ) =1 . 1 × 10 3 A and i 2 = i 3 = E 3 R = 1 . 2 × 10 3 V 3(0 . 73 × 10 6 ) =5 . 5 × 10 4 A . (d), (e), and (f) At t = , the capacitor is fully charged and the current in the capacitor branch
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Unformatted text preview: is zero. Then i 1 = i 2 and the loop rule yields E − i 1 R 1 − i 1 R 2 = 0 . The solution is i 1 = i 2 = E 2 R = 1 . 2 × 10 3 V 2(0 . 73 × 10 6 Ω ) = 8 . 2 × 10 − 4 A . (g) and (h) The potential difference across resistor 2 is V 2 = i 2 R 2 . At t = 0 it is V 2 = (5 . 5 × 10 − 4 A)(0 . 73 × 10 6 Ω ) = 4 . × 10 2 V and at t = ∞ it is V 2 = (8 . 2 × 10 − 4 A)(0 . 73 × 10 6 Ω ) = 6 . × 10 2 V . (i) The graph of V 2 versus t is shown to the right. t V 2 E / 6 E / 3 E / 2 ....................................................................................................................................
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This test prep was uploaded on 04/15/2008 for the course PHYS 101 taught by Professor Reich during the Spring '08 term at Johns Hopkins.

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