Time it takes for a RC circuit to fully charge

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SUMMARY

The time it takes for an RC circuit to fully charge is defined by the equation Q = CV(1 - e^(-t/RC). Initially, when the capacitor is uncharged (Q = 0), the time T can be expressed as T = -ln(0 - 1)RC, which is undefined. However, as the capacitor charges, the current through the resistor decreases, leading to a theoretical scenario where the capacitor never fully charges. Practically, after five time constants (5 * RC), the voltage across the capacitor reaches 99% of the supply voltage.

PREREQUISITES
  • Understanding of RC circuit components: resistance (R), capacitance (C), and voltage (V)
  • Familiarity with the exponential function and natural logarithm
  • Knowledge of capacitor charging behavior in electrical circuits
  • Ability to interpret voltage-time graphs for RC circuits
NEXT STEPS
  • Study the mathematical derivation of the RC charging equation Q = CV(1 - e^(-t/RC))
  • Learn about the concept of time constants in RC circuits and their practical implications
  • Explore the effects of varying resistance and capacitance on charging time
  • Investigate real-world applications of RC circuits in timing and filtering
USEFUL FOR

Electrical engineering students, hobbyists working with circuits, and professionals designing timing circuits will benefit from this discussion on the charging behavior of RC circuits.

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Homework Statement


How much time does an RC circuit require to become fully charged assuming that the capacitor was initially uncharged.
resistance = R
capacitance = C
Voltage = V

Homework Equations


Q=CV(1-e^(-t/RC)

The Attempt at a Solution


so T = -ln(Q/(CV) - 1)RC
but Q = 0 since it's initially uncharged?
so T=-ln(0-1)RC

I'm not quite sure what to do for this problem because this error isn't in the domain of the natural log. Honestly, I don't even understand the formula, if CV is the final possible charge, then is Q=CV?
 
Last edited:
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Consider the current through the resistor...

I = VR/R

where the voltage VR across the resistor = Source Voltage - Capacitor voltage

So as the capacitor voltage rises the current through the resistor falls. This reduces the rate at which the capacitor charges. In theory the capacitor never becomes fully charged because the current falls towards zero.

The voltage on the capacitor looks like this..

http://www.interfacebus.com/RC-Time-Constant-Rising-Voltage-Chart.jpg

After "5 time constants" (eg 5 * RC) the voltage will be within 1% of the supply voltage.
 
Last edited:

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