## Help understanding RC circuits

1. The problem statement, all variables and given/known data
In a problem I have a circuit that I have not set up to know what happens as far as voltages across my battery(4v), two equal resistors and a capacitor.

2. Relevant equations

3. The attempt at a solution
Because i have not seen a circuit like this im thinking that this circuit is not ideal for a capacitor to collect charge. im thinking that the current will only go through resistor A and not but b or the capacitor. So Voltage bat= 4v Voltage resistor A = 4v and resistor b and capacitor = 0V
Attached Thumbnails

 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity

Recognitions:
Homework Help
 Quote by rh23 1. The problem statement, all variables and given/known data In a problem I have a circuit that I have not set up to know what happens as far as voltages across my battery(4v), two equal resistors and a capacitor. 2. Relevant equations 3. The attempt at a solution Because i have not seen a circuit like this im thinking that this circuit is not ideal for a capacitor to collect charge. im thinking that the current will only go through resistor A and not but b or the capacitor. So Voltage bat= 4v Voltage resistor A = 4v and resistor b and capacitor = 0V
Suppose, just for argument's sake, that resistor A was not there. So there is just the battery, resistor B, and the capacitor. Can you describe how the circuit would behave when the battery is first connected?
 well the resistor is actually a light bulb so when the circuit is first connected lets say with a switch the light bulb will light up but at time goes on the light bulb will go out. No more current will be flowing and the capacitor will be charged.

Recognitions:
Homework Help

## Help understanding RC circuits

 Quote by rh23 well the resistor is actually a light bulb so when the circuit is first connected lets say with a switch the light bulb will light up but at time goes on the light bulb will go out. No more current will be flowing and the capacitor will be charged.
Right. So current initially flows until the capacitor is charged up. When the capacitor is charged up, what is the voltage across the capacitor? How about the resistor?
 voltage across capacitor is 4v and resistor is 0v

Recognitions:
Homework Help
 Quote by rh23 voltage across capacitor is 4v and resistor is 0v
Excellent. That is correct.

Now place resistor A back in the circuit. Note that it is a parallel branch -- it is in parallel with the capacitor branch, which we just looked at, and the battery. Being a branch that is across the voltage source, it will behave independently of the other branch; Nothing that resistor B or the capacitor do will effect the voltage that appears across resistor A.
 So when are switch is closed bulb b will light but eventually turn off and the capacitor will then be charge and while this goes on bulb A will be lit continuously?

Recognitions:
Homework Help
 Quote by rh23 So when are switch is closed bulb b will light but eventually turn off and the capacitor will then be charge and while this goes on bulb A will be lit continuously?
Yes, that's right.
 when switch is closed and i wanted to rank my voltages immediately would they be Vbattery=Vbulb A=Vbulb B > V capacitor . V capacitor= 0 After switch is closed for a longtime current rankings I battery = I bulb A > I bulb B . I bulb B = 0 Is this true? If the switch is located next to the negative terminal of the battery and the switch was closed for a long time is the voltage across bulb A after the switch is opened again be 2v?

Recognitions:
Homework Help
 Quote by rh23 when switch is closed and i wanted to rank my voltages immediately would they be Vbattery=Vbulb A=Vbulb B > V capacitor . V capacitor= 0
Yes.
 After switch is closed for a longtime current rankings I battery = I bulb A > I bulb B . I bulb B = 0 Is this true?
Yes.
 If the switch is located next to the negative terminal of the battery and the switch was closed for a long time is the voltage across bulb A after the switch is opened again be 2v?
Yes, if both bulbs have the same resistance.
 Wow thanks so much for taking the time to help me understand these circuits better gneil. I can't thank you enough, but if it makes you feel better i also help younger students with their work when asked :) its very rewarding.

Recognitions:
Homework Help
 Quote by rh23 Wow thanks so much for taking the time to help me understand these circuits better gneil. I can't thank you enough, but if it makes you feel better i also help younger students with their work when asked :) its very rewarding.
You're welcome
 If you dont mind me asking what would happen if bulb B was placed next to the negative terminal and the switch was closed for a long time? would the current through A be equal to the current through B

Recognitions:
Homework Help
 Quote by rh23 If you dont mind me asking what would happen if bulb B was placed next to the negative terminal and the switch was closed for a long time? would the current through A be equal to the current through B
I'm not sure that I can picture the circuit that you have in mind... what happens to the capacitor in this scenario? Can you draw the circuit?
 in this circuit when the switch is closed for a long time... the capacitor will charge correct? Bulb A will continuously be lit correct? Im not sure what happens to bulb B, will it light then dim itself a little because of the capacitor or will it light as bright as A ( bulb A and B have the same current) ? Attached Thumbnails
 Sorry bulb A is the resistor parallel with the capacitor

Recognitions:
Homework Help
 Quote by rh23 in this circuit when the switch is closed for a long time... the capacitor will charge correct? Bulb A will continuously be lit correct? Im not sure what happens to bulb B, will it light then dim itself a little because of the capacitor or will it light as bright as A ( bulb A and B have the same current) ?
After the switch is closed for a long time both bulbs conduct the same current. The capacitor is charged (to half the battery voltage --- the same voltage that appears across bulb A).

When the switch is first closed the capacitor is uncharged so there is 0V across it. Bulb B then gets the full benefit of the battery potential for that initial instant, while bulb A has zero voltage (and thus no current). As the capacitor charges and its voltage rises, bulb A conducts current. Bulb A brightens while bulb B dims, and they "meet in the middle" with both bulbs running at half brightness and half the battery voltage across each.

 Tags capacitors, circuit, resistors, volatage

 Similar discussions for: Help understanding RC circuits Thread Forum Replies Quantum Physics 8 Electrical Engineering 4 Electrical Engineering 2 Electrical Engineering 14 Electrical Engineering 4