effect of adding a resistor in series with a capacitor?

[dtexb]

What would happen if you added a resistor in series to a circuit all ready having a capacitor? I think the capacitance would stay the same but the charge stored on the capacitor would go up. Is this true? Also how would this affect the amount of energy stored on the capacitor? thanks

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[Meekah]

nevermind

 

[Rabolisk]

I agree that the capacitance C will stay the same, as it is a property of the capacitor (C = eA/d for parallel plate). The charge stored, Q, should decrease though. Q = CV. The voltage across the capacitor is LOWERED when it is in series with a resistor. Because of this, Q and E (energy stored)should be lower as well, since E = 0.5QV

I don't think so. The voltage will only be lowered transiently. It will go right back up to its original value (9 V) soon.

 

[GreenRabbit]

What would happen if you added a resistor in series to a circuit all ready having a capacitor? I think the capacitance would stay the same but the charge stored on the capacitor would go up. Is this true? Also how would this affect the amount of energy stored on the capacitor? thanks

If the capacitor is fully charged, adding a resistor will do nothing. No effect on the voltage across the capacitor (no current, so the resistor would have no effect), the capacitance certainly isn't changing, and therefore the charge stored isn't changing either.

 

[Graffiti]

circuits bleh

 

[golgiapparatus88]

The only thing that will change is the amount of time it takes to charge the capacitor. If you have a capacitor in place after a junction of two resistors of different resistances, the capacitor will be charged by the resistor with less resistance (since more current flows through it). In series, the more resistance you have, the longer it will take to charge. Either way, capacitance will not change.

 

[flodhi1]

What would happen if you added a resistor in series to a circuit all ready having a capacitor? I think the capacitance would stay the same but the charge stored on the capacitor would go up. Is this true? Also how would this affect the amount of energy stored on the capacitor? thanks

Okay well check this out the RESISTOR is in SERIES with the capacitance. Now since this is a series the current will be the SAME. However there will be a Voltage drop. Increased in resistance will lead to a decrease in Voltage when it reaches the Capacitor.
based on V= IR , if I (Current is same which it is for series always) then the Voltage drop will be greater with a greater resistor) Thus the elelctrical potential energy that the Capacitor gains will be LESS than the Work done by the battery. However the capacitance will remain the same. The energy stored (Ie. number of electrons displaced) will increase. This was all assuming the Capacitor was not already fully charged.

 

[GreenRabbit]

Sigh, so many wrong and irrelevant answers. At t=infinity, current = 0. Therefore the voltage across the capacitor is not affected, the capacitance is not affected, and charge and energy in the capacitor is not affected.

All the resistor does is make it take longer to charge up the capacitor. This wasn't what OP was asking.

 

[Rabolisk]

Okay well check this out the RESISTOR is in SERIES with the capacitance. Now since this is a series the current will be the SAME. However there will be a Voltage drop. Increased in resistance will lead to a decrease in Voltage when it reaches the Capacitor.
based on V= IR , if I (Current is same which it is for series always) then the Voltage drop will be greater with a greater resistor) Thus the elelctrical potential energy that the Capacitor gains will be LESS than the Work done by the battery. However the capacitance will remain the same. The energy stored (Ie. number of electrons displaced) will increase. This was all assuming the Capacitor was not already fully charged.

Beautifully done.. If it was only anywhere close to being right...

 

[GreenRabbit]

At no point in time will adding a resistor "transiently decrease" the voltage on the capacitor.

 

[flodhi1]

Sigh, so many wrong and irrelevant answers. At t=infinity, current = 0. Therefore the voltage across the capacitor is not affected, the capacitance is not affected, and charge and energy in the capacitor is not affected.

All the resistor does is make it take longer to charge up the capacitor. This wasn't what OP was asking.

AAMC PRACTICE EXAM 7R Passage VII question 40)

As the capacitor is charged the electrical potential energy that is gains IS LESS than the work done by the battery throughout the charging process DUE TO THE ADDITION OF A RESISTOR. Since energy is lost by heating the resistor. Therefore with the addition of a resistor the battery supplies the energy for both the resistor and capacitor.

And I already said that THE CAPACITANCE WILL BE THE SAME.

And for GreenRabit "At no point in time will adding a resistor "transiently decrease" the voltage on the capacitor." I said the voltage reaching the capacitor NEVER SAID ON THE CAPACITOR. As the electrons flow through the resistor energy isw dissipated, and there is a voltage drop. The voltage drop is additive, that is for the series that the OP provided in the diagram. Therefore since there is a voltage drop due to the resistor from the source would it NOT make sense that the SAME exact Voltage from the source will NOT reach the capacitor?
I know I provided a little more information but this was ALL to address the concept of adding a resistor in series with respect to a capacitor.

 

[GreenRabbit]

AAMC PRACTICE EXAM 7R Passage VII question 40)

As the capacitor is charged the electrical potential energy that is gains IS LESS than the work done by the battery throughout the charging process. Since energy is lost by heating the resistor. Therefore with the addition of a resistor the battery supplies the energy for both the resistor and capacitor.

And I already said that THE CAPACITANCE WILL BE THE SAME.

And for GreenRabit "At no point in time will adding a resistor "transiently decrease" the voltage on the capacitor." I said the voltage reaching the capacitor NEVER SAID ON THE CAPACITOR.

Your caps are ridiculous. "Transiently decrease" was mentioned by Rabolisk, not you. Also note that I said as time goes to infinity, all properties of the capacitor are equivalent to an identical circuit without the resistor. Now note that the OP was referring to a circuit that already had a capacitor (t=infinity), not the same case as whatever your AAMC problem was referring to. You also said "The energy stored (Ie. number of electrons displaced) will increase." This is wrong. You don't need to quote explanations at me.

Edit: You do realize that as the capacitor is charged, the current drops to zero, and therefore the voltage "drop" by the resistor also drops the zero, which is why at t=infinity the capacitor is identical to a circuit without the resistor right?

 

[flodhi1]

Your caps are ridiculous. "Transiently decrease" was mentioned by Rabolisk, not you. Also note that I said as time goes to infinity, all properties of the capacitor are equivalent to an identical circuit without the resistor. Now note that the OP was referring to a circuit that already had a capacitor (t=infinity), not the same case as whatever your AAMC problem was referring to. You also said "The energy stored (Ie. number of electrons displaced) will increase." This is wrong. You don't need to quote explanations at me.

Edit: You do realize that as the capacitor is charged, the current drops to zero, and therefore the voltage "drop" by the resistor also drops the zero, which is why at t=infinity the capacitor is identical to a circuit without the resistor right?

Qualitatively, the current through the circuit at any time t would be smaller because of the higher resistance. The flow of charge to the capacitor is therefore slowed, and so it will take longer for the capacitor to charge up to a particular level. TIME WOULD BE EXTENDED. The problem I showed you from AAMC is the same exact problem as the OP posted it is with respect to a resistor in series with a Capacitor. Like I said before and I stand by what I said the potential energy that reaches the capacitor would be less than the work done by the battery/source. Why would you have the entire work/ potential energy of the battery reach the capacitor when the resistor is dissipating energy....And like Dpspanda said the Voltage that reaches the Capacitor will be lowered when a resistor is present. This is a series the Voltage/voltage drop is never the same in a series with a resistor.

 

[GreenRabbit]

Alright, if OP's diagram is from that question, I suppose he misstated his question. I agree with most of what you said. I also said that it would take longer for the capacitor to charge. This does imply what you keep saying: that at any finite point in time there would be less energy on the capacitor due to the voltage drop from the resistor.

At t=infinity, however, the entire voltage of the battery does reach the capacitor. As the capacitor is charging, the current decreases to zero, causing the voltage on the battery and the capacitor to be the same. Much of the battery's energy will have been released through the resistor, you are right. But as long as that battery is a solid 9 volts, the capacitor will reach that voltage.

 

[Rabolisk]

When I said transient loss, I meant any loss of charge of a nonideal capacitor that cannot hold charge definitely. The point was though, is that eventually, the resistor in series makes no difference with respect to the voltage, charge stored, or energy stores by the capacitor.

 

[dtexb]

thanks for the answers guys!