# hw6.9 - R R V I b To find we can use the following formula...

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HW 6.9 Given: V F i n d : 11 V ? 0 I F C 6 ? 0 . v P ? 4200 1 R c 10200 2 R ? t Q Q t 5 / 4 In the circuit shown, the capacitor is originally uncharged and the switch is open. (a) The switch is closed at t = 0. Calculate the current through the battery immediately afterward. The arrow on the diagram indicates the direction of positive current. (b) Calculate the rate at which the EMF is supplying energy to the circuit immediately after the switch is closed. (c) What is the time constant for charging the capacitor with the switch closed? (d) Find the current I passing through the EMF a long time after the switch is closed. (e) Sometime later, the switch is again opened. How long t does it take after the switch is opened for the charge in the capacitor to decrease to 4/5 of its previous (steady-state) value? Solutions a.) To find we can use the following formula: 0 I since the capacitor has no voltage drop we end up having just a battery with two resistor in parallel  1 1 2 1 1 R R V R V I   1 2 1 1

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Unformatted text preview: R R V I b.) To find we can use the following formula: . v P V R R V IV P 1 2 1 1 2 1 2 1 1 V R R P c.) To find c we can use the following formula: RC C R c 1 d.) To find we can use the following formula: Q since the capacitor is charged i.e current equal zero we just have a battery and one resistor 2 R V R V I 2 R V I e.) To find we can use the following formula: t / t Qe Q / t t Qe Q / t t e Q Q e t e Q Q t t ln / ln ln / / ln t Q Q t Q Q t t ln C R R RC 2 1 C R R Q Q t t t t 2 1 ln C R R Q Q t t 2 1 ln...
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hw6.9 - R R V I b To find we can use the following formula...

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