Capacitor

Hello to every one i just want to ask about capacitor time constant equation.. T=RC
what is a mean of this...

i am very much clear about dynamic equation of the capacitor
vc(t)= vin[1-e*-t/rc]

when vc(0)=0


or when vc(0)=>0 then
vc(t)= [vc(0)-vin] e*-t/rc+vin

through these we can check at which time how much voltage across the capacitor plates but i am not understanding time constant meaning...

kindly help me in this please...

It's a quick approximation of RC timing. One time constant is 1-e^-1/rc or 63% of vin. Two time constants is 1-e^-2/rc or 86.5% of vin. Five time constants, or 99.3%, is considered the time it takes to fully charge the capacitor (though, theoretically, it never really reaches full charge).

So, T=RC is the time it takes to achieve 63% of full charge. If the charging voltage is, say, 5 volts, then at RC, the capacitor will have around 3 volts across it. At 5RC it will have 5*.993 = 4.965 volts across it.

It's the same thing in reverse for a capacitor discharging.

Fun fact: in the equation 1-e^-τ/rc, it is triditional to use the lowercase Greek letter Tau "[*τ*]" which is the time constant.

I just learned that, also the time constant τ is related to the cutoff frequency fc, an alternative parameter of the RC circuit, by

τ = RC = 1(2pi*fc)

or, equivalently,

fc = 1/(2pi*τ)

where resistance in ohms and capacitance in farads yields the time constant in seconds or the frequency in Hz.

so it means Sir, let suppose we have a resistor value 10ohm and capacitor value 2f with voltage capacity 10 volts and voltage source is 10 volts
1-So time constant is 20sec means it will take 20 seconds to charge 63% of its full voltage and take 100 second to charge fully 10 volts... m i right????

and it means what ever the voltage level value of resistor and capacitor decide how much time this capacitor will take to fully charge?????

Looking forward your valuable reply sir

Actually if you plot the function, you will get a good visual representation. Or you could take a container poke a small hole in the bottom, fill i with water then, open the hole. At first the flow is great, then gradually as the container empties, the flow diminishes, same function.

As for the cutoff -3dB frequency it could as well be a high pass +3db frequency, although technically it would be in radians or frequency if you have 1/(2*pi*RC)

Yes, you are right!!!!!!!

At 20 sec, that resistor/capacitor combination will be at 63% of full charge, and at 100 seconds the capacitor will be at (roughly) full charge!!!!!!!

And, regardless the charging voltage, the value of the resistor and the capacitor determine the time it takes to fully charge the capacitor!!!!!!!!

Here's a site where the Time Constants are ploted (just as Earl Albin suggested):

http://www.tpub.com/neets/book2/3d.htm

2-now a same case let suppose a capacitor with 10 f with 25v capacity but i am giving only 12 volts my second question is will it charge fully?? because i think it will only charge upto 12 volts it will not depend on the time now.

By "25V capacity" do you mean a capacitor rated to 25V max? If so, then that has nothing to do with how the capacitor will charge. Not how high the voltage will go, nor how fast it will get there. That is merely the maximum voltage the capacitor is rated to withstand (it will probably withstand higher voltages, but the rated voltage is the highest the manufacturer can be held liable for).

A 10F capacitor in series with, say, a 5 ohm resistor, will take 250sec to reach 99.3% of the APPLIED voltage--which is completely different than the voltage printed on the side of the capacitor. If, 12Volts is applied to the RC circuit, then the capacitor will FULLY CHARGE to 12Volts (not 25Volts). Now, if, say, 40Volts is applied to the same RC circuit, then the capacitor will charge to 63% of 40 Volts, or 25.2Volts in 50 seconds. After that, there is no telling how much longer the capacitor will charge before failing (e.g. short out). If it doesn't fail, then it wll still take 250 seconds for it to reach 99.3% of full charge, at which point there will be 39.7Volts acoss the capacitor (very unlikely).

if that isn't what you mean by "25V capacity", then I don't know what you're talking about.

sir i mean let suppose we have one RC series circuit in which we have resistor of 2 ohm and capacitor with 4f with 25volts of with stand capacity. if we give more it will damage the insulation between the plates...


what actually mean by fully charged??? i think it means source voltage should be equal to the capacitor terminal voltage or the voltage across the capacitor plates.....

i am very clear about applied voltages...
but i am getting mix up with the with stand capacity voltages and terminal voltages of the capacitor...

The 25 volts (in your example) is nothing more than the maximum voltage that the capacitor is rated to withstand. Nothing more. It only means that if the capacitor charges to a voltage higher than 25 volts, it is not guaranteed to continue functioning properly (i.e. the insulation between the plates will likely be damaged, as you said).

Fully charged means, charged to the same voltage as the applied voltage (as you said).

The only relationship between the “withstand voltage” (the max voltage printed on the side of the capacitor) AND the “terminal voltage” is if the terminal voltage exceeds the withstand voltage, then the capacitor will probably fail.

ir i mean let suppose we have one RC series circuit in which we have resistor of 2 ohm and capacitor with 4f with 25volts of with stand capacity. if we give more it will damage the insulation between the plates…

what actually mean by fully charged??? i think it means source voltage should be equal to the capacitor terminal voltage or the voltage across the capacitor plates…..
i am very clear about applied voltages…
but i am getting mix up with the with stand capacity voltages and terminal voltages of the capacitor

The 25 volts (in your example) is nothing more than the maximum voltage that the capacitor is rated to withstand. Nothing more. It only means that if the capacitor charges to a voltage higher than 25 volts, it is not guaranteed to continue functioning properly (i.e. the insulation between the plates will likely be damaged, as you said).

Fully charged means, charged to the same voltage as the applied voltage (as you said).

The only relationship between the “withstand voltage” (the max voltage printed on the side of the capacitor) AND the “terminal voltage” is if the terminal voltage exceeds the withstand voltage, then the capacitor will probably fail.

Now i got it sir....

we can charge capacitor at any voltage level but it should be less then with stand voltage and a capacitor will fully charged when it equal to the applied voltages...


Big Bundle of thanks sir for you....
May God help you and give you more and more knowledge....

may i contact with you on this form. actually i m planning for master degree but i am working in Dubai in a construction company and out of touch with the books that's why i started to study again thank you once again sir for your help

Cool, I'm glad i was able to help.

If you have further questions, just submit them to EEWeb and I or someone else will likely assist you.

Time constant of capacitor means how long the time need to discharge or charging an capacitor. The value of capacitance and resistance also affect the duration of charging or discharging an capacitor. Another equation for time constant for capacitance is t = 1 / (2 * Fc), which Fc is resonant frequency.

No, the time constant is one segment of the time needed to charge or discharge a capacitor (not the entire time it takes to charge or discharge a capacitor).

Also, don't you mean t = 1/(2π * Fc) where Fc is the "cutoff frequency" not the "resonant frequency" (if the resonant frequency, then what is resonating?)

hi sir today i have a different question...

i have one source transformer with the capacity of 50KVA, connected with two load 10kw (resistive and inductive load)...


Let suppose transformer p.f is 0.8...


How can i find load in KVA,load power factor...

Ask it in a new thread. Otherwise, if it is answered, it won't be visible to others who might have the same question (now or in the future). This isn't just about you

Sounds like you might be taking college exams. First of all with a PF of 0.8 you are good to go.

So there are a number of ways but since PF is a ratio of KW / KVA and that = 0.8, and the total power in KVA. There is another term not mention and that one is KWR ( KW reactive), although you don't need that.

KVA = 10KW / 0.8 = 2.5KVA