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Time for capacitor charge to reach 99.9% maximum

Consider a series circuit containing a resistor of resistance R and a capacitor of capacitance C connected to a source of EMF E with negligible internal resistance. The wires are also assumed to have zero resistance. Initially, the switch is open and the capacitor discharged.
Let us try to understand the processes that take place after the switch is closed. The charge of the capacitor, the current in the circuit, and, correspondingly, the voltages across the resistor and the capacitor, will be changing. Note that at any moment in time during the life of our circuit, Kirchhoff's loop rule holds and indeed, it is helpful: E - V_R - V_C = 0, where V_R is the voltage across the resistor, and V_C is the voltage across the capacitor.

Given that:
q(t) = C*E(1-e^(-t/RC)

The Problem states:
Find the time t_2 it would take the charge of the capacitor to reach 99.9% of its maximum value given that R=12 Ohms and C = 500 * 10^ -6 F.

Express your answer numerically in seconds. Use three significant figures.

As a hint is given that:
1. find the time t_1 that it takes the charge of the capacitor to reach 99.9% of its maximum value.

To do 1. it says to:
2. What is the maximum charge on the capacitor? Set q(t) = to 99.9% of this value and solve for the corresponding time.

See attached image.


Solution Summary

It finds the time for the charge of the capacitor to reach 99.9% maximum in a RC circuit.