Charging a capacitor through a resistor -- final voltage?

[Kajan thana]

Why does capacitor charges up to full voltage as the emf of the battery even if there is a resistor is connected in series? I know that having a resistor means it will take longer to charge the capacitor but shouldn't some energy need to be lost in the resistor as well?

 

[.Scott]

If the capacitor is at a different voltage, then there will be current across the resistor - so the voltage will still be changing.

 

[Kajan thana]

If the capacitor is at a different voltage, then there will be current across the resistor - so the voltage will still be changing.

Can we still get the capacitor to charge up to same PD as the battery even if there is a resistor? If we can, how is it possible?

 

[sophiecentaur]

Can we still get the capacitor to charge up to same PD as the battery even if there is a resistor? If we can, how is it possible?

It's the same as connecting two water tanks together with a very narrow pipe. Initially, there is a big pressure difference so the flow will be fast. As the levels become more equal, the pressure difference is less and the flow rate is less. In an ideal situation, the flow never stops - it just gets slower and slower and the levels are never exactly zero. An Exponential Decay curve describes what happens. It's the same with charging a capacitor through a resistor, the current drops ever slower and the voltage rises ever slower - also following an Exponential Decay curve. You could say that the capacitor never 'fully' charges - but pretty soon it gets near enough to be considered 'full'. For a 1MΩ resistor and a 1μF capacitor, the voltage on the capacitor will be (1/e)V away from being fully charged (V) after 1s. We call RC the Time Constant.

 

[anorlunda]

Why does capacitor charges up to full voltage as the emf of the battery even if there is a resistor is connected in series? I know that having a resistor means it will take longer to charge the capacitor but shouldn't some energy need to be lost in the resistor as well?

Energy is lost in the resistor. You are wrongly equating voltage and energy. In circuits, power is voltage times current. Power is the rate of energy delivery.

Always keep your units straight to avoid confusion.

 

[Kajan thana]

Energy is lost in the resistor. You are wrongly equating voltage and energy. In circuits, power is voltage times current. Power is the rate of energy delivery.

Always keep your units straight to avoid confusion.

I think I have not made my question clear, when energy is lost this also means voltage is used up as well, then how does the capacitor charge up to full PD as the battery. Considering about Kirchhoff's laws , the voltage of the resistor and the capacitor should equal to EMF of the battery

 

[sophiecentaur]

As with any two elements in series, the source voltage is divided between them. The PD across the R tends towards zero and the PD across the C tends to V.
Interestingly, the energy supplied by the battery is divided equally between what's stored in the C and what's dissipated in the R. Whatever the values of C and R.

 

[anorlunda]

I think I have not made my question clear, when energy is lost this also means voltage is used up as well, then how does the capacitor charge up to full PD as the battery. Considering about Kirchhoff's laws , the voltage of the resistor and the capacitor should equal to EMF of the battery

No, that is not correct. You can't "use up" voltage. In the steady state, capacitor voltage equals battery voltage, current is zero, power flow is zero, and the voltage difference across the resistor is zero.

It sounds like you are still trying to equate energy and voltage in your head.

 

[Kajan thana]

No, that is not correct. You can't "use up" voltage. In the steady state, capacitor voltage equals battery voltage, current is zero, power flow is zero, and the voltage difference across the resistor is zero.

It sounds like you are still trying to equate energy and voltage in your head.

The way I learned it; voltage is amount of joules per charge, if energy is used up then isn't the mean voltage also used up?
If I got the concept wrong, please correct me.
Thank you

 

[ZapperZ]

The way I learned it; voltage is amount of joules per charge, if energy is used up then isn't the mean voltage also used up?
If I got the concept wrong, please correct me.
Thank you

You need to understand the right concept for the right context. What you described in the energy gain or lost by a charge particle undergoing a change in potential.

A "battery" is a source that maintains a potential difference between its terminals, as stated by anorlunda.

Zz.

 

[anorlunda]

The way I learned it; voltage is amount of joules per charge, if energy is used up then isn't the mean voltage also used up?
If I got the concept wrong, please correct me.
Thank you

OK I think I see the source of your confusion. You can use up charge and use up energy (joules). Electric potential, measured in volts, is the ratio of the two. Voltage, also measured in volts, is a circuit is always the difference in potential between two points.

Edit: the last sentence should have said: Voltage (also measured in volts) in a circuit is always the difference in potential between two points.

 

[sophiecentaur]

It sounds like you are still trying to equate energy and voltage in your head.

That's not a heinous crime. At least he's not talking in terms of Forces! . . . and the Voltage is the Energy per unit Charge
The Electrical Potential Energy of each charge is 'lost' (aka Used Up) as it flows towards the negative terminal - I think that Kajan just needs to alter his wording a bit but I think he has got the idea correctly. Let's face it, we have read some much worse utter nonsense about Electrical Quantities elsewhere in this forum.

 

[sophiecentaur]

You can use up charge and use up energy (joules)

I think the idea of Using Up Charge could be a dodgy one as it is one of those things that is conserved. 'Using it up' can apply when it 'goes elsewhere' but we are discussing a series circuit here.

 

[anorlunda]

think the idea of Using Up Charge could be a dodgy one as it is one of those things that is conserved. 'Using it up' can apply when it 'goes elsewhere' but we are discussing a series circuit here.

Fair point, but exactly the same words can be said about energy. It is conserved but it can "go elsewhere".

 

[sophiecentaur]

Fair point, but exactly the same words can be said about energy. It is conserved but it can "go elsewhere".

Oh yes - but it does 'go elsewhere' from the circuit in the form of heat and mechanical in a motor. The total energy 'consumed equals the total energy supplied. But we all acknowledge KCL and KVL. We only fight about them when someone else asks for some actual words to express them. Words are the weak link and Maths is wonderful.

 

[David Lewis]

...energy supplied by the battery is divided equally between what's stored in the C and what's dissipated in the R.

Some energy will be dissipated as heat because of loss resistance, and some as electromagnetic radiation because of changing current.

 

[sophiecentaur]

Oh yes. That acknowledges the parasitic inductance that has to be there due to the finite size of the circuit. You can ignore that for the first pass through the problem, though.

 

[David Lewis]

Would circuit inductance cause the current to change more slowly?

 

[davenn]

Some energy will be dissipated as heat because of loss resistance,

... because of the losses IN the resistance

and some as electromagnetic radiation because of changing current.

No, a changing current doesn't cause EM radiation. Accelerating electrons ( charges) cause E/M radiation as in an oscillator,
which isn't the case in the above DC circuit

 

[sophiecentaur]

There need be no confusion if you realize that the RC circuit in the classic diagram cannot exist in practice. Along with the Inductance there will be a Radiation Resistance. But this idealised diagram is just one example of modelling in Science.

 

[sophiecentaur]

If you want a mechanical description of the energy radiated when accelerating electrons (charges), think of Force times Distance. For a circuit with no size, the Distance is zero so no radiation.

 

[David Lewis]

I use the term loss resistance to distinguish it from radiation resistance.

 

[davenn]

I use the term loss resistance to distinguish it from radiation resistance.

then that is just confusing the issue as there is no radiation resistance
stick with what I said previously

 

[sophiecentaur]

then that is just confusing the issue as there is no radiation resistance
stick with what I said previously

There is only No Radiation Resistance if the circuit has no volume. But the radiation resistance of a circuit with an RC time constant of, say 1ms and a size of a few cm would be extremely low, so you could call it zero. My opinion is that we should tackle Science one layer at a time. Radiation resistance in a case like this should not be allowed to spoil your day because there is an explicit resistive circuit element involved.
On the other hand, in an apparently very similar case - the 'paradox' of the missing energy when you discharge one capacitor into a similar, parallel capacitor - the radiation resistance is very relevant.
You just have to be reasonable and fairly well informed about these things if you want to avoid endless hassle.

 

[davenn]

. But the radiation resistance of a circuit with an RC time constant of, say 1ms and a size of a few cm would be extremely low, so you could call it zero. My opinion is that we should tackle Science one layer at a time. Radiation resistance in a case like this should not be allowed to spoil your day because there is an explicit resistive circuit element involved.

agreed

 

[Mark Harder]

Why does capacitor charges up to full voltage as the emf of the battery even if there is a resistor is connected in series? I know that having a resistor means it will take longer to charge the capacitor but shouldn't some energy need to be lost in the resistor as well?

It's not a complete answer, but you might want to know that there is a voltage drop across the resistor only when a current is flowing through the resistor, as given by Ohm's law: V = I R. I.e. I=0 implies V = 0. So Kirchoff's voltage law requires that the voltage across the resistance and a fully charged capacitor in series = 0 + V(source) = V(source). For a capacitor in the real world, a small current leaks across its dielectric gap. In which case, a little current always flows through the resistor and it dissipates a little energy, drops the voltage across it and limits the voltage of the fully charged capacitor to a tad less than the source voltage.

 

[sophiecentaur]

It's not a complete answer, but you might want to know that there is a voltage drop across the resistor only when a current is flowing through the resistor, as given by Ohm's law: V = I R. I.e. I=0 implies V = 0. So Kirchoff's voltage law requires that the voltage across the resistance and a fully charged capacitor in series = 0 + V(source) = V(source). For a capacitor in the real world, a small current leaks across its dielectric gap. In which case, a little current always flows through the resistor and it dissipates a little energy, drops the voltage across it and limits the voltage of the fully charged capacitor to a tad less than the source voltage.

Yes. And there is no limit to where you can take the effects of practicalities. It is essential to have the basic equation for the idealised circuit fully appreciated before starting down the real world road.