Time for capacitor charging in RLC circuit.

In summary, the time constant for a capacitor in a series RLC circuit is given by RC, where R is the resistance and C is the capacitance. However, there is some uncertainty about the exact formula to use, as some sources suggest 2L/R while others suggest 1/R*(L/C)^1/2. The circuit is connected to a battery through a switch, and the time it takes for the capacitor to charge to 0.63% of the source voltage depends on the values of R, L, and C and can result in oscillations. Solving the differential equation for the charge on the capacitor with initial conditions of zero charge and current can provide a more accurate estimation.
  • #1
Rts11
2
0

Homework Statement


In series RLC circuit how much time it takes for capacitor to charge to 0.63% of source voltage..?
time constant is the time taken by a capacitor to charge to 0.63% of source voltage and in rc circuit time constant is RC. but what about for capacitor in series RLC circuit..?

Homework Equations

The Attempt at a Solution


I searched allover the web and Wikipedia but i didn't find anything. i saw a formula saying that time constant is
2L/R and some where i saw time constant as 1/R*(L/C)1/2.but i don't know it is for inductor(L) or capacitor(C) and how to relate it to capacitor voltage.
.
 
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  • #2
Rts11 said:

Homework Statement


In series RLC circuit how much time it takes for capacitor to charge to 0.63% of source voltage..?

.
Copy the whole text of the original problem, please. Is the RLC circuit connected to a battery by a switch?
 
  • #3
ehild said:
Copy the whole text of the original problem, please. Is the RLC circuit connected to a battery by a switch?
Yes the RLC circuit is connected to a battery by closing the switch and how much time in takes only the capacitor to get 0.63% of source voltage.
 
  • #4
Is it 63% of the source voltages or 0.63 %?
There can be oscillations in the circuit, so the voltage across the capacitor can be a given value quite a few times, depending the actual values of R, L, C. Were they given?
You have two solve the differential equation for the charge on the capacitor, taking the initial conditions into account. In this case it is that both the charge and the current (time derivative of the charge) are zero. You get quite a complicated equation.
 

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