# hw6.7 - 3 2 1 I I I B 2 1 1 V V V C b To find P& we can...

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HW 6.7 Given: V F i n d : 6 1 V ? 0 . 1 I v 3 2 V ? 0 . 2 I ? 0 . 3 490 1 R I 800 2 R ? 0 . 1 P ? 0 . 2 P 290 3 R F C 10 ? . 1 I ? Q The capacitor C is initially uncharged and the switch S is closed at t = 0. Arrows in the diagram designate the direction of positive currents. (a) Calculate the currents just after the switch is closed. (b) What power is each EMF supplying to the circuit just after the switch is closed? (c) Find the current through E 1 a long time after the switch is closed. (d) Find the charge on the capacitor a long time after the switch is closed. Solutions a.) To find , & we can use the following formula: 0 . 1 I 0 . 2 I 0 . 3 I We need two loop equations and one current equation 3 2 1 I I I 0 0 3 3 1 1 1 I R I R V since 0 2 2 2 1 1 1 c V V I R I R V q 0 c V

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After algebra… you get 3 2 1 2 1 2 1 1 1 1 3 R R R R R V R V R V I 2 1 R R 1 3 2 1 2 1 2 1 2 1 1 1 1 3 1 1 R R R R R R R R V R V R V R V I 3 1 2 I I I Best way to solve the question is to use a Matrix 0 0 1 1 1 2 1 3 1 R R R R A
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Unformatted text preview: 3 2 1 I I I B 2 1 1 V V V C b.) To find P & we can use the following formula: . 1 . 2 P IV P 1 1 . 1 V I P 2 2 . 2 V I P c.) To find I we can use the following formula: . 1 R V I since the capacitor is charged 2 I 3 1 1 R R V I . 3 . 1 I I I 3 1 1 . 1 R R V I d.) To find we can use the following formula: Q c R V V V 2 3 3 1 1 3 3 3 3 R R V R I R V R 2 3 1 1 3 2 3 V R R V R V V V R c 2 3 1 1 3 V R R V R C CV CV Q c 2 3 1 1 3 V R R V R C Q Click here for fast calculation...
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hw6.7 - 3 2 1 I I I B 2 1 1 V V V C b To find P& we can...

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