# Another DC Circuit (But more complicated with a SWITCH and Capacitor!)

by sweetdion
Tags: capacitor, circuit, complicated, switch
 P: 64 1. The problem statement, all variables and given/known data For the circuit shown below, the capacitors are initially uncharged. At t=0, the switch S is closed. a) Determine the current in each resistor immediately after the switch is closed. b) Determine the current in each resistor a very long time after the switch is closed. c) Determine the voltage across the capacitor a very long time after the switch is closed. After the switch has been closed for a very long time, it is reopened. d) Determine the initial current I0 through the 100 ohm resistor, as a function of time after the switch is reopened. e) Determine I(t), the current through the 100 ohm resistor, as a function of time after the switch is reopened. f) How long after the switch is reopened does the charge on the capacitor fall to 10% of its fully charged state? 2. Relevant equations V=IR I through a capacitor is I = Cdv/dt = Q/C Kirchoff's Loops Rules 3. The attempt at a solution a) Immediately after the switch is closed, the capacitor behaves ls a short circuit and hence no current flows through the 100 ohm resistor. I in the 200 ohm resistor = V/R=15V/200ohms=.075A b) long after the switch is closed, the capacitor is fully charged and behanves as an open circuit. All of the current passes through both resistors. Left Loop Clockwise: 15V-200ΩI1-100ΩI1=0 Right Loop Counter clockwise: Q/C-100ΩI1-0 I1 is the only current passing through both the resistors I= 20A as for the rest of the parts, I want to make sure I get A and B right before I move on.
HW Helper
Thanks
P: 9,267
 Quote by sweetdion [b]1. b) Determine the current in each resistor a very long time after the switch is closed. b) long after the switch is closed, the capacitor is fully charged and behanves as an open circuit. All of the current passes through both resistors. Left Loop Clockwise: 15V-200ΩI1-100ΩI1=0 Right Loop Counter clockwise: Q/C-100ΩI1-0 I1 is the only current passing through both the resistors I= 20A
I do not understand. What is that I=20 A? A few hundred ohm and 15 V emf, how can be the current 20 A?

ehild
P: 4,513
 Quote by sweetdion I through a capacitor is I = Cdv/dt = Q/C
I = C dv/dt is correct.

But I = Q/C is not. Don't use it. V = Q/C is the correct equation.

P: 64

## Another DC Circuit (But more complicated with a SWITCH and Capacitor!)

 Quote by ehild I do not understand. What is that I=20 A? A few hundred ohm and 15 V emf, how can be the current 20 A? ehild
I solved the top equation and that's what I got. Did I label my currents right?
 HW Helper Thanks P: 9,267 Do you mean 15V-200ΩI1-100ΩI1=0? That is 15 V = 300Ω * I1. Do you mean I1= 20 A? ehild
 P: 64 Yeah, that's what I meant.
 HW Helper Thanks P: 9,267 Amazing. Tell me please, how much is 300*20??? ehild
 P: 64 A lot, 6000. Which, is wrong. I'm asking for your help because I don't understand what I'm doing wrong, ehild.
 HW Helper Thanks P: 9,267 You have an equation. 300 I = 15. This equation is correct. Solve it for I. Is not it 15/300 A? ehild
P: 64
 Quote by ehild You have an equation. 300 I = 15. This equation is correct. Solve it for I. Is not it 15/300 A? ehild
Oh. Now I see what I was doing wrong. Simple Math error. I = 0.05A
 P: 64 Moving on to part C now. When the switch is closed for a very long time, the capacitor starts discharging through the resistors. After a long time, the voltage across the capacitor becomes equal to the voltage across the 100 Ω resistor. V= IR=(0.05A)(100Ω)=5V
P: 145
 Quote by sweetdion Moving on to part C now. When the switch is closed for a very long time, the capacitor starts discharging through the resistors. After a long time, the voltage across the capacitor becomes equal to the voltage across the 100 Ω resistor. V= IR=(0.05A)(100Ω)=5V
That's right. After a very long time, the capacitor becomes essentially an open circuit.
 P: 64 Thanks Melawrghk. :)

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