Discussion :: Capacitors - General Questions (Q.No.9)
After which time constant can a capacitor be considered to be fully charged?
Answer: Option C
No answer description available for this question.
|Balaji G said: (Feb 26, 2011)|
|Time constant is the time taken to charge the capacitance for 67% of the final voltage which denotes one 1RC. Five consecutive RC are required to the charge the capacitance fully :)|
|Vinod said: (Oct 2, 2011)|
|Any element to come to its original position takes 5 time constant.|
|Kishor said: (Feb 11, 2012)|
|Can't we get it mathematically?|
|Sarathy said: (Apr 5, 2012)|
|Its possible for all elements.|
|Ranjith Garuda said: (Apr 18, 2012)|
|(Voltage across capacitor) Vc = Vin[1-e^ (-t/rc) ].
Where Vin is input voltage applied, t = the elapsed time since the application of the supply voltage, rc is nothing but time constant.
For 1t (time) and 1rc we get 63. 2% of the input voltage across the capacitor.
How did you get 63. 2%?
Well take small t, means response time =1sec and rc time constant as 1 in the above equation. You will arrive to an equation like this.
Vc=Vin[1-0. 367879441].[Equation1] which will lead you to this.
Vc=Vin[0. 632120559]. [Equation2].
[Equation2] tells us that, at 1sec and 1rc of time constant, my input voltage is only '0.632Vin' which means 63% of my input has appeared across the capacitor at 1rc and 1 sec of time.
If we want to say a capacitor is fully or 100% charged, means. The voltage across the capacitor should be equal to the input voltage applied to it, in other words if I want 100% of my input to appear across the capacitor, I should make Vc=Vin in the above formula.
You can see from equation2 this can be acheived only if that 0. 6321 number should raise to 1. Vc=Vin*1. Means from the formula the part [1-e^ (-t/rc) ] should be made 1, now how do we acheive this.
At first we acheived 63% of output by making the term (-t/rc) as 1.
Now make it 2 you will get 0.864664717.
For 3 it is 0.950212932.
For 4 it is 0.981684361.
At 5 it is 0.993262053 means at 5T you will achieve 99. 3% of input across the capacitor. Which is approximately considered as 1means at 5T we got 99. 3% of input across the capacitor. Goahead and check for 6, 7, 8, n. You will never achieve answer as 1 which will result you 100%. Therotically its infinite. But practically you get full voltage by 5T.
By varying the resistor are and capacitor c value one can decide the % of charge in capacitor at different instant of time t. Decide the time you wanted to charge the capacitor by, and vary rc value to get it fully charged. :) thank you.
|Rajneesh Kumar said: (May 23, 2013)|
|Capacitor 67% charged in 1CR. (TIME CONSTANT).
& Capacitor are fully charged on 5CR.
|Md Asif said: (Jun 21, 2016)|
|From control system theory of to reach the stable position with approx 1% error, and their time will be 5 times.|
|Samir said: (Jun 27, 2016)|
|After about 5 time constant periods (5CR) the capacitor voltage will have very nearly reached the value E.
Because the rate of charge is exponential, in each successive time constant period Vc rises to 63.2% of the difference in voltage between its present value, and the theoretical maximum voltage (VC = E).
|Pallavi said: (Apr 25, 2017)|
|Settling time in contral system is 5t.|
|Hani said: (Mar 19, 2019)|
Thank you for the perfect explanation.
|Sravan said: (Feb 13, 2020)|
|@ Ranjith. Thank you very much for the elaborated explanation.|
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