# hw6.10 - since 1 1 1 V I R V 1 V 1 Q R V R V R V I 2 2 1 1...

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HW 6.10 Given: 22 R F i n d : ? L R F C C C 2 3 2 1 ? / 1 dt dQ ? V 54 V I ? . 1 Q ? . 2 Q ? . 3 Q All the resistors in the circuit shown in the diagram have a common value: R = 22 . Similarly, all the capacitances are the same: C 1 = C 2 = C 3 = 2 μF and they are initially uncharged. The EMF is E = 54 V. The switch is closed at t = 0. (a) Find the net (load) resistance R L seen by the battery at t = 0+. (b) At t = 0+ what is the rate at which charge is being stored in capacitor C 1 ? (c) Calculate the current flowing through the battery after the switch has been closed for a long time. (d) Calculate the charge stored in each of the capacitors after a long time. Solutions a.) To find R we can use the following formula: L

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& 0 5 4 RI RI V 0 3 4 V RI V & 5 3 4 I I I 0 since 3 V 0 3 Q R V I 4 0 4 5 R R V R V R RI V I R V R V I I I 0 5 4 3 0 2 2 V RI V 0 2 V since 0 2 Q R V I 2 R V R V R V I I I 2 4 2 6 1 1 1 1 1 2 R R R R   2 2 1 1 1 R R R since the two resisters were in parallel
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Unformatted text preview: since 1 1 1 V I R V 1 V 1 Q R V R V R V I 2 2 1 1 R V R V R V I I I 4 2 2 6 1 4 4 4 R V VR R V V R L 4 R R L b.) To find we can use the following formula: dt dQ / 1 R V I dt dQ 2 / 1 1 R V dt dQ 2 / 1 c.) To find we can use the following formula: I R V R R V I 2 3 2 1 I I I since the capacitors are charged R V I 2 d.) To find , & we can use the following formula: . 1 Q . 2 Q . 3 Q since they are in parallel with the battery V C Q Q 1 . 1 . 1 is in parallel with R 3 C 2 2 3 3 3 3 . 3 V C R V R C RI C V C Q R 2 3 . 3 V C Q...
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hw6.10 - since 1 1 1 V I R V 1 V 1 Q R V R V R V I 2 2 1 1...

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