Meaning of Time Constant T=RC in Capacitor Charging and Discharging Equations

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#1 21662470 05 Jun 2012 14:22

ASAD ALI ASAD ALI

Anonymous

Post #1
21662470 05 Jun 2012 14:22

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...

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#2 21662471 05 Jun 2012 14:33

Steve Lawson Steve Lawson

Anonymous

Post #2
21662471 05 Jun 2012 14:33

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.
#3 21662472 05 Jun 2012 14:38

Steve Lawson Steve Lawson

Anonymous

Post #3
21662472 05 Jun 2012 14:38

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.
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#4 21662473 05 Jun 2012 14:44

Steve Lawson Steve Lawson

Anonymous

Post #4
21662473 05 Jun 2012 14:44

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.
#5 21662474 05 Jun 2012 15:27

ASAD ALI ASAD ALI

Anonymous

Post #5
21662474 05 Jun 2012 15:27

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
#6 21662475 05 Jun 2012 15:40

Earl Albin Earl Albin

Anonymous

Post #6
21662475 05 Jun 2012 15:40

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)
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#7 21662476 05 Jun 2012 15:41

Steve Lawson Steve Lawson

Anonymous

Post #7
21662476 05 Jun 2012 15:41

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!!!!!!!!
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#8 21662477 05 Jun 2012 15:46

Steve Lawson Steve Lawson

Anonymous

Post #8
21662477 05 Jun 2012 15:46

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

http://www.tpub.com/neets/book2/3d.htm
#9 21662478 05 Jun 2012 16:06

ASAD ALI ASAD ALI

Anonymous

Post #9
21662478 05 Jun 2012 16:06

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.
#10 21662479 05 Jun 2012 16:22

Steve Lawson Steve Lawson

Anonymous

Post #10
21662479 05 Jun 2012 16:22

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.
#11 21662480 05 Jun 2012 17:37

ASAD ALI ASAD ALI

Anonymous

Post #11
21662480 05 Jun 2012 17:37

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...
#12 21662481 05 Jun 2012 17:44

Steve Lawson Steve Lawson

Anonymous

Post #12
21662481 05 Jun 2012 17:44

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.
#13 21662482 05 Jun 2012 17:45

ASAD ALI ASAD ALI

Anonymous

Post #13
21662482 05 Jun 2012 17:45

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
#14 21662483 05 Jun 2012 17:50

Steve Lawson Steve Lawson

Anonymous

Post #14
21662483 05 Jun 2012 17:50

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.
#15 21662484 05 Jun 2012 17:54

ASAD ALI ASAD ALI

Anonymous

Post #15
21662484 05 Jun 2012 17:54

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
#16 21662485 05 Jun 2012 17:57

Steve Lawson Steve Lawson

Anonymous

Post #16
21662485 05 Jun 2012 17:57

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.
#17 21662486 05 Jun 2012 21:58

Yun Siong Leong Yun Siong Leong

Anonymous

Post #17
21662486 05 Jun 2012 21:58

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.
#18 21662487 06 Jun 2012 01:38

Steve Lawson Steve Lawson

Anonymous

Post #18
21662487 06 Jun 2012 01:38

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?)
#19 21662488 07 Jun 2012 14:44

ASAD ALI ASAD ALI

Anonymous

Post #19
21662488 07 Jun 2012 14:44

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...
#20 21662489 07 Jun 2012 14:47

Steve Lawson Steve Lawson

Anonymous

Post #20
21662489 07 Jun 2012 14:47

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
#21 21662490 07 Jun 2012 15:05

Earl Albin Earl Albin

Anonymous

Post #21
21662490 07 Jun 2012 15:05

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
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Topic summary

✨ The time constant (τ = RC) in capacitor charging and discharging circuits represents the characteristic time it takes for the capacitor voltage to reach approximately 63% of the applied voltage during charging, or to decay to about 37% during discharging. It is calculated as the product of resistance (R) in ohms and capacitance (C) in farads, yielding time in seconds. The capacitor voltage over time follows the exponential equations vc(t) = Vin[1 - e^(-t/RC)] for charging from zero initial voltage, and vc(t) = [vc(0) - Vin]e^(-t/RC) + Vin for other initial conditions. Multiple time constants indicate further progression toward full charge: at 2τ about 86.5%, and at 5τ approximately 99.3% of the applied voltage, which is conventionally considered fully charged despite the theoretical asymptote. The time constant also relates inversely to the cutoff frequency (fc) of the RC circuit by τ = 1/(2πfc). The capacitor's voltage rating (withstand voltage) is the maximum voltage it can safely handle; the capacitor charges up to the applied voltage, not its rated voltage, and exceeding the rated voltage risks damage. The resistor and capacitor values determine the charging/discharging speed, independent of the applied voltage magnitude. Visualizing the charging curve or analogies like water draining from a container can aid understanding. The discussion clarifies the distinction between terminal voltage, applied voltage, and capacitor voltage rating, emphasizing that "fully charged" means the capacitor voltage equals the applied voltage.

FAQ

TL;DR: In an RC circuit, one time constant τ=RC brings a capacitor to 63% of its final voltage; “T=RC is the time it takes to achieve 63% of full charge.” [Elektroda, Steve Lawson, post #21662471]

Why it matters: These rules let you estimate charge/discharge timing, select parts, and avoid over‑voltage damage in real designs.

Quick Facts

τ (tau) equals R×C; at 1τ the capacitor reaches about 63% of the applied DC voltage. [Elektroda, Steve Lawson, post #21662471]
After 5τ, charge is about 99.3% complete; this is commonly treated as “near full.” [Elektroda, Steve Lawson, post #21662471]
Cutoff frequency for a first‑order RC is fc = 1/(2πRC). [Elektroda, Steve Lawson, post #21662473]
“Fully charged” means the capacitor voltage equals the applied source voltage. [Elektroda, Steve Lawson, post #21662481]
Voltage rating (withstand) is a maximum safe voltage; it does not change τ or charge speed. [Elektroda, Steve Lawson, post #21662479]

What does the time constant T=RC actually mean?

The time constant τ=RC is the characteristic time of a first‑order RC circuit. After one τ, the capacitor’s voltage has moved 63% of the way from its start value toward the final value set by the source. After two τ it’s about 86.5%. After five τ it’s roughly 99.3%. These percentages apply to charging and discharging. “It’s the same thing in reverse for a capacitor discharging.” [Elektroda, Steve Lawson, post #21662471]

When is a capacitor considered fully charged?

Fully charged means the capacitor’s terminal voltage equals the applied source voltage. The RC curve approaches this level asymptotically, so engineers treat about five time constants as effectively full for design timing. “Fully charged means, charged to the same voltage as the applied voltage.” [Elektroda, Steve Lawson, post #21662481]

How do I calculate capacitor voltage vs. time during charging?

Use vc(t) = vin·(1 − e^(−t/RC)) if the initial voltage is zero. For a non‑zero initial voltage vc(0), use vc(t) = [vc(0) − vin]·e^(−t/RC) + vin. These equations let you read the capacitor’s voltage at any moment and see how it approaches vin as t increases. [Elektroda, ASAD ALI, post #21662470]

How many time constants does it take to reach near full charge?

About five. At 1τ the capacitor is ~63% toward final; 2τ ≈ 86.5%; 3τ ≈ 95%; 4τ ≈ 98%; 5τ ≈ 99.3%. Designers commonly call 5τ “fully charged” for practical timing. This rule of thumb speeds estimates without full simulation. [Elektroda, Steve Lawson, post #21662471]

Does the capacitor’s voltage rating affect how fast it charges?

No. The printed voltage rating (withstand) is only the maximum safe operating voltage. It does not influence τ or the charge/discharge rate. Time behavior depends only on R and C. Change R or C to change timing; keep the applied voltage below the rating to avoid damage. [Elektroda, Steve Lawson, post #21662479]

What happens if I exceed the capacitor’s voltage rating?

Exceeding the withstand voltage risks dielectric breakdown and failure. The insulation between plates can be damaged, leading to leakage or a short. “If the terminal voltage exceeds the withstand voltage, then the capacitor will probably fail.” Always choose a rating with margin over the highest expected voltage. [Elektroda, Steve Lawson, post #21662481]

How does RC relate to cutoff frequency?

A first‑order RC has cutoff frequency fc = 1/(2πRC). Equivalently, τ = RC = 1/(2πfc). This ties time‑domain behavior (τ) to frequency‑domain behavior (−3 dB point) for the same network. It’s useful when swapping between timing specs and filter specs. [Elektroda, Steve Lawson, post #21662473]

Is the RC time constant the total time to charge a capacitor?

No. The time constant is one segment of the process, not the whole time. The capacitor asymptotically approaches its final voltage. Engineers often use 5τ as the practical endpoint. “The time constant is one segment of the time needed to charge or discharge.” [Elektroda, Steve Lawson, post #21662487]

How does discharging follow the time constant?

Discharge mirrors charge behavior. After one τ, the voltage has dropped by 63%, leaving 37% of the initial value. After five τ, about 0.7% remains. This symmetry lets you reuse the same exponential equation with a negative sign for the change. “It’s the same thing in reverse for a capacitor discharging.” [Elektroda, Steve Lawson, post #21662471]

Quick how‑to: estimate RC charge time without math?

Compute τ = R×C from your resistor (Ω) and capacitor (F).
Multiply τ by 5 for near‑full charge (~99.3%).
For checkpoints, use 1τ≈63%, 2τ≈86.5%, 3τ≈95%.
These milestones provide fast design estimates for microcontroller delays or analog settling. [Elektroda, Steve Lawson, post #21662471]

Why does the charging current slow down over time?

As the capacitor’s voltage rises, the voltage difference across the resistor shrinks, reducing current exponentially. A good analogy is a water tank draining through a small hole: fast at first, then slower as the level drops. This maps to the same exponential behavior in RC circuits. [Elektroda, Earl Albin, post #21662475]

Can an RC capacitor charge above the source voltage with DC?

No. In a simple RC with a DC source, the capacitor’s final voltage equals the applied source voltage. It cannot exceed that level without additional circuitry like charge pumps or inductive kickback paths. “Fully charged means… the same voltage as the applied voltage.” [Elektroda, Steve Lawson, post #21662481]

Does changing the source voltage change the time constant?

No. Time constant depends only on R and C. Changing the source voltage changes the final target voltage, not the speed of approach. “Regardless the charging voltage, the value of the resistor and the capacitor determine the time it takes.” [Elektroda, Steve Lawson, post #21662476]