# RC Time Constant

Discussion in 'Electronics Homework Help' started by Bluepoint SEO, Aug 14, 2016.

1. ### Bluepoint SEO

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Jul 28, 2016
Hi there,

I keep getting the wrong results for the calculation I do as follows:

5(470x10^-6F)(10000Ω)

Basically 470μF capacitor and a 10K resister in an RC circuit. I do the math on my calculator and keep ending up with 23.5 seconds when the answer I believe should be 4.7 seconds.

Maybe I'm just not using the calculator properly? Its driving me nuts. Thanks.

2. ### Alec_t

3,000
813
Jul 7, 2015
You forgot the '5'.

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3. ### Bluepoint SEO

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Jul 28, 2016
Hi Alec)t the 5 is there right before the first opening bracket. I keep getting 23.5!! Driving me nuts lol.

4. ### Bluepoint SEO

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Jul 28, 2016
I just did it without the 5 and got the right answer... hmmm.

5,164
1,087
Dec 18, 2013
The RC time constant is just that 1 RC not 5. 5 RC is when the capacitor is considered fully energized. 1 RC is when the capacitor is energized to 63% of maximum. This is because when the formula for the energizing of the capacitor = 1-e^-1 = 0.63 the -1 is - t / RC so when RC is the same as your time constant you are left with just -1.

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6. ### Bluepoint SEO

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Jul 28, 2016
Thank you so much for clearing that up! Okay my conclusion is I need to buy some good books when the cash is here, but hey I'm learning !

5,164
1,087
Dec 18, 2013
Or you save some money and ask us It's free lol

8. ### Bluepoint SEO

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Jul 28, 2016
Back to my blog to correct a few things Adam.

5,164
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Dec 18, 2013
Go for it. Where is your blog? Do you want us to check it out for you?

10. ### Bluepoint SEO

88
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Jul 28, 2016
So let me get this straight, the time constant is RC. 5RC is the length of time it takes to full charge the capacitor. When the capacitor is charged no more DC will flow?

5,164
1,087
Dec 18, 2013
1 RC is the time it takes to reach 63% of the applied voltage. I'll do you an example, give me a mo.

5,164
1,087
Dec 18, 2013
This shows a 1K and a 10uF cap, see if you can work out the voltage on the cap after 1 RC...is the graph correct?

hevans1944 likes this.
13. ### hevans1944Hop - AC8NS

4,630
2,159
Jun 21, 2012
Well, pretty much. The capacitor charges exponentially through a series resistor toward the applied voltage, theoretically never actually reaching the applied voltage. The use of five time constants to represent "fully charged" is just for convenience. The difference in capacitor voltage after ten time constants, versus the voltage after five time constants, is very small and can usually be ignored in practical circuit design.

There is no such thing as a "full charge" on a capacitor. Barring electrical breakdown, a capacitor can be charged to any voltage and the charge on it will be Q = CV. The method of "charging" the capacitor will affect how the voltage across the capacitor terminals changes. Charging through a fixed resistor from a constant voltage source is just one method, the description of which gives rise to the concept of time constant and exponential time functions.

In practice, there is always some parallel leakage resistance, Rleak, across the capacitor. This leakage path forms a voltage divider with the R in series with the capacitor. Therefore, if V is the charging voltage applied to capacitor C through resistor R, the final voltage will be V * (Rleak)/(Rleak+R). If Rleak is MUCH greater than R, the fraction approaches unity so the "final" voltage on the capacitor approaches V. Note that leakage resistance does modify the exponential increase and, because of the leakage resistance, all capacitors charged from a constant voltage source through a resistor will reach a constant steady-state voltage eventually, not just theoretically, in a finite amount of time.

Adam could modify his simulation to show the effect of leakage resistance, which often occurs as a result of trying to use the voltage across the capacitor to drive another circuit, say, an analog-to-digital converter or a voltage comparator. This is what happens with the ubiquitous 555 timer where internal leakage limits the maximum value of R that can be used for timing. The maximum value of C that can be used with the 555 is limited by the chips ability to discharge the capacitor when the threshold voltage is reached.

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14. ### Bluepoint SEO

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Jul 28, 2016
Well, 1 RC of that is 0.01 seconds. To work out the voltage what do I do haha! I feel some algebra coming on give me a mo.

Okay I know 1 Farad is 1 Coulomb at 1 Volt so 1 Farad's RC at a resistance of 1KΩ would be 1000 seconds. Oh my goodness this is hard. V = IR, resistance is 1000 x I, I is V/R voltage is unknown arrr, I don't about this so I'll Google some.

Right Charge is the product of Capacitance and Voltage we've got voltage and we know the value of 1 Farad.

Voltage = Charge / Capacitance

Is the voltage 1V?

If that's wrong don't tell me I've had some beers and will try again tomorrow. Or later tonight and I'll be gutted if someone answers it before me.

88
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Jul 28, 2016

16. ### Bluepoint SEO

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Jul 28, 2016
Thank you so much for this help.

17. ### Bluepoint SEO

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Jul 28, 2016
Hi Adam, yes it's just like a place for me to revise jot down notes etc, then when I find out I'm wrong ha I go back and edit the content so it's right but I never delete anything if it's incorrect I just strike it though and leave it there. http://electronics-blog.bluepointseo.co.uk

5,164
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Dec 18, 2013
First off, find the correct formula. You will need to search for capacitor charging formula. You don't need to worry about Coulombs and all that.

19. ### Bluepoint SEO

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Jul 28, 2016
Okay I'm back sorry been off sick! Infected elbow has left me on antibiotics. That's what I get for falling off a kids scooter after a few bevies. Any how:

After 1RC = Charge is 0.01μF