Register to reply 
Another DC Circuit (But more complicated with a SWITCH and Capacitor!) 
Share this thread: 
#1
Apr1010, 11:28 PM

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 I_{0} 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: 15V200ΩI_{1}100ΩI_{1}=0 Right Loop Counter clockwise: Q/C100ΩI_{1}0 I_{1} 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. 


#2
Apr1010, 11:59 PM

HW Helper
Thanks
P: 10,377

ehild 


#3
Apr1110, 02:39 AM

P: 4,513

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


#4
Apr1110, 11:01 AM

P: 64

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



#5
Apr1110, 12:01 PM

HW Helper
Thanks
P: 10,377

Do you mean 15V200ΩI1100ΩI1=0? That is 15 V = 300Ω * I1. Do you mean I1= 20 A?
ehild 


#6
Apr1110, 12:36 PM

P: 64

Yeah, that's what I meant.



#7
Apr1110, 01:38 PM

HW Helper
Thanks
P: 10,377

Amazing. Tell me please, how much is 300*20???
ehild 


#8
Apr1110, 01:53 PM

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. 


#9
Apr1110, 02:13 PM

HW Helper
Thanks
P: 10,377

You have an equation. 300 I = 15. This equation is correct.
Solve it for I. Is not it 15/300 A? ehild 


#10
Apr1110, 02:27 PM

P: 64




#11
Apr1110, 02:41 PM

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 


#12
Apr1110, 03:26 PM

P: 145




#13
Apr1110, 03:31 PM

P: 64

Thanks Melawrghk. :)



Register to reply 
Related Discussions  
Circuit with switch resistors and capacitor  Introductory Physics Homework  4  
A question with capacitor and a switch!  Advanced Physics Homework  2  
Charging a capacitor with a switch  Introductory Physics Homework  2  
Circuit with capacitor, current source, voltagr source... and a switch  Engineering, Comp Sci, & Technology Homework  1  
Why does a capacitor prevent the spark?  General Physics  8 