# A 6v battery charges up a 470 microfarad capacitor in 1 second

• lionman
In summary, this may seem a very simple question but i can't work it out and have no clue where to start.
lionman
Hello to all, this may seem a very simple question but i can't work it out and have no clue where to start.
My question is: A 6v battery charges up a 470 microfarad capacitor in 1 second, what is the final charge of the capacitor? and what is the current while the capacitor is charging?

i know capacitance is charge/potential or c = Q/V
i know charge is Q = CV

Is the final charge in coulombs, 470 microfarad x voltage? or should i convert microfarad to farad?
current i know means charge flowing per second but i do not know how to apply this to the question.
The final part of the question asks for the value of the resistor, is this just ohms law? R = V/I?
Any help would be much appreciated.

lionman said:
Hello to all, this may seem a very simple question but i can't work it out and have no clue where to start.
My question is: A 6v battery charges up a 470 microfarad capacitor in 1 second, what is the final charge of the capacitor? and what is the current while the capacitor is charging?

i know capacitance is charge/potential or c = Q/V
i know charge is Q = CV

Is the final charge in coulombs, 470 microfarad x voltage? or should i convert microfarad to farad?
current i know means charge flowing per second but i do not know how to apply this to the question.
The final part of the question asks for the value of the resistor, is this just ohms law? R = V/I?
Any help would be much appreciated.
Convert to farads if you want the result in coulombs. Otherwise you'll get your result in microcoulombs

The problem you've described appears to be an RC series circuit where a capacitor is being charged through some resistance. You will have to review your notes on RC circuits and in particular the equations that describe charging and discharging a capacitor (time constants, etc.).

I have reviewed my notes and there is little there to help me. I think time constant = resistance x capacitance? but i don't know the resistance.
looking at the question is 1 second the time constant? this doesn't sound right but I literally have no idea.

You have to remember that $i=\frac{\delta q}{\delta t}$ - so the current as you've put already is the rate of change of charge with respect to time... does this help?

Is this all the information that is given in the question?

lionman said:
I have reviewed my notes and there is little there to help me. I think time constant = resistance x capacitance? but i don't know the resistance.
looking at the question is 1 second the time constant? this doesn't sound right but I literally have no idea.

Yes, the time constant would be ##\tau = RC##. 1 second was given as the time for the capacitor to charge. A basic "rule of thumb" you can use for these exponential decays is to assume that all the action is essentially over after five time constants have elapsed

What's the equation that describes the current with respect to time for the circuit?

Im sorry but neither reply is making much sense to me.
gneill - do you mean after 5 constants the capacitor would be fully charged?
As for the equation i have tried to find this but a lot of what i see i don't understand and have never seen before.

After 5 constants the capacitor would be fully charged? - Yes, practically

I think the equation gneill means is $i=i_{o}\exp^{(\frac{-t}{\tau})}$ where t is the time, ... have you seen this? $\tau$ =RC and $i_{o}$ is the time at t=0 - I take it you've seen this before?

lionman said:
Im sorry but neither reply is making much sense to me.
gneill - do you mean after 5 constants the capacitor would be fully charged?
As for the equation i have tried to find this but a lot of what i see i don't understand and have never seen before.

After 5 time constants the charge on the capacitor will reach about 99.3% of its final value. Close enough to "done" for engineering purposes

Your notes should contain formulas involving ##e^{-\frac{t}{\tau}}##. This is the basic "decaying exponential" that governs the way voltages and currents will evolve in a circuit containing resistance and capacitance (or resistance and inductance).

If you can determine the time constant and hence resistor value, then you should be able to determine the initial current (based upon the characteristics of a capacitor at t=0) and to what value the current should eventually head as time goes on, and then select the appropriate equation to represent the current over time.

Thanks for the replies, the reason I am struggling because this is not even something i have been taught in class but it has come up in my coursework. I have no notes to call upon.

This equation does e stand for eulars constant? minus t which is 0 divided by T which is 1 second?

lionman said:
Thanks for the replies, the reason I am struggling because this is not even something i have been taught in class but it has come up in my coursework. I have no notes to call upon.

This equation does e stand for eulars constant? minus t which is 0 divided by T which is 1 second?

$$e^{\left(-\frac{t}{\tau}\right)}$$
e is Euler's constant, e = 2.718...
##\tau## is the time constant
t is time

You can also write this as

$$exp(-\frac{t}{\tau})$$

thanks for the help so far but I am still none the wiser. Is there anyway that you can help me further without providing the answer?

lionman said:
thanks for the help so far but I am still none the wiser. Is there anyway that you can help me further without providing the answer?

How about you first describe, in your own words, how you think the current will behave in the circuit from the moment that the battery is first connected?

There will be a current flow until the capacitor has reached its full charge, at this point it will act as an open circuit - there will be no electron flow.
current flow should decrease over time?

lionman said:
There will be a current flow until the capacitor has reached its full charge, at this point it will act as an open circuit - there will be no electron flow.
current flow should decrease over time?

Yes, that is the general picture

So, the current must start out with some initial value and then taper off as the capacitor charges and the potential difference across it reaches its maximum value-- the same as that of the battery which is driving that current. The current will have the form:
$$i(t) = i_o e^{-t/\tau}$$
The trick, then, is to determine values for io and ##\tau##.

If you happened to know the values of all the components in the circuit, how would you determine the initial current?

if i knew all the values i would use I = V/R

lionman said:
if i knew all the values i would use I = V/R

Right. So it would appear that the only required information you are missing is a value for R (since you are given the battery voltage V).

Go back to the post I made about determining ##\tau## from the time to completion of charging. You should be able to find suitable values for ##\tau## and R.

The way I interpret the question is that the capacitor reaches the battery voltage in one second and that it is not a basic RC circuit but it is a circuit which charges at constant current.

The way I interpret the question is that the capacitor reaches the battery voltage in one second and that it is not a basic RC circuit but it is a circuit which charges at constant current.

What sort of circuit components might that comprise for an introductory electronics course? How would a student answer the question, "what is the current while the capacitor is charging?"

Firstly my interpretation of the question is due to the way the question is worded for example with the wording "what is the current" the implication is(at least to me)that the current is steady.In the UK it is an A level physics experiment to charge a capacitor at constant current where the student adjusts a variable resistor.The current cannot be held very steady,it is a crude experiment but it demonstrates a principle.Students also investigate exponential charging/discharging.The current is,of course,the rate of flow of charge(Q/t)
I may well be wrong in how I interpret the question but I suspect that the OP has summarised the question and may have left out some important details.

## 1. How does a 6v battery charge up a 470 microfarad capacitor in 1 second?

The 6v battery provides the necessary voltage to push electrons into the capacitor, while the 470 microfarad capacitance allows for a larger amount of charge to be stored in a short amount of time. Essentially, the battery supplies the energy and the capacitor stores it quickly due to its high capacitance.

## 2. What is the significance of the 1 second charging time?

The 1 second charging time indicates the speed at which the capacitor can be charged. In this case, it is a relatively short amount of time due to the combination of the 6v battery and the 470 microfarad capacitance. This can be useful in certain applications where a quick charge and discharge of energy is needed.

## 3. Can a different battery or capacitor be used to achieve the same charging time?

Yes, a different battery or capacitor can be used to achieve the same charging time. The key is to maintain the same voltage and capacitance values. For example, a 12v battery and a 235 microfarad capacitor could also charge up in 1 second, as long as all other factors remain constant.

## 4. What happens to the stored charge in the capacitor after it is fully charged?

After the capacitor is fully charged, it will maintain its stored charge until it is discharged. This means that the capacitor can hold onto the energy for an extended period of time, making it useful for applications such as flash photography or powering electronic devices.

## 5. Is it safe to touch the capacitor after it has been charged by the 6v battery?

It is generally not recommended to touch a charged capacitor, as it can still hold a dangerous amount of energy. It is important to discharge the capacitor before handling it, either by using a discharge tool or by letting it naturally dissipate its charge over time.

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