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.
 

FAQ: Time for capacitor charging in RLC circuit.

How does a capacitor charge in an RLC circuit?

In an RLC circuit, a capacitor charges when current flows through it and builds up an electric field between its plates. This electric field stores energy in the capacitor, causing it to charge up to a certain voltage.

What factors affect the time it takes for a capacitor to charge in an RLC circuit?

The time it takes for a capacitor to charge in an RLC circuit is affected by the capacitance of the capacitor, the resistance of the circuit, and the inductance of the inductor. A higher capacitance or lower resistance will result in a faster charging time, while a higher inductance will slow down the charging process.

How can the time for capacitor charging be calculated in an RLC circuit?

The time for capacitor charging in an RLC circuit can be calculated using the equation t = RC, where t is the time in seconds, R is the resistance in ohms, and C is the capacitance in farads. This assumes that the inductance in the circuit is negligible.

Why is it important to consider the time for capacitor charging in an RLC circuit?

The time for capacitor charging is important because it affects the overall behavior and performance of the circuit. A longer charging time can result in delays or fluctuations in the circuit's output, while a shorter charging time may lead to overcharging and potential damage to the components.

How can the time for capacitor charging be reduced in an RLC circuit?

The time for capacitor charging in an RLC circuit can be reduced by using a capacitor with a lower capacitance or a circuit with a lower resistance. Additionally, using a higher voltage source can also decrease the charging time, as the capacitor will charge up to the desired voltage more quickly.

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