Electrical Engineering - Capacitors - Discussion

Discussion Forum : Capacitors - General Questions (Q.No. 11)
11.
A 4.7 M resistor is in series with a 0.015 F capacitor. The combination is across a 12 V source. How long does it take the capacitor to fully charge?
35 ms
352 ms
3.5 s
70.5 ms
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
20 comments Page 1 of 2.

Yashwanth said:   5 years ago
Time Constant of Capacitor = 5RC.

T = 5 * (4.7*10^-6)(0.015*10^-3).
T = 352.

Nitin Dhotre said:   6 years ago
Q = RC*60/V.
(1)

Dhiraj said:   7 years ago
It is fully charged so multiply by 5.

Lalithanandigam said:   8 years ago
As we know v =v(1-epower(-t/rc)).

So let us take t=Rc then v=v*0.63.
If u take t=5rc then v=v*0.993..
So during t=5rc capacitor is fully charged..
If you want to find the voltage across the capacitor after time t =5rc then we can use v=12*5*r*c.

Manisha said:   8 years ago
What is the role of voltage here?

Saurabh Kr. Verma said:   8 years ago
If t=5rc then it means there is no effect of 12V on it?
Is it take same time across any voltage?
Please tell me that.

Abhishek said:   8 years ago
@Murtadha

The time required to charge a capacitor to 100% is 5T. Where t=R*c. that u got. not multiply that T with 5 you will get the answer.

Murtadha said:   9 years ago
Q = C*V = 0.015*10^-6 *12 =1.8^-7 colmb.
I = V/R = ( 12 / 4.7*10^6) =2.55*10^-6 amper.
I = Q/t 2.55*10^-6 = 1.8*10^-6 /t.
t= 70.5 ms.

So I think the choice (D) correct not (b).

Apoorva said:   9 years ago
Please give full reason?

Kanwaljit singh said:   9 years ago
5RC, in formula 5 is include because capacitor takes 5 times more time to charge itself when it is in series with resistor than normal charging so that why we use 5 constant in the formula.


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