Electrical Physics Question

[Ryll]

A capacitor that is initially uncharged is connected in series with a resistor and an emf source with Є=120v and negligible internal resistance. Just after the circuit is completed, the current through the resistor is 8.00 mA, and the time constant for the circuit is 4.00 s

What is the current as time reaches infinity? prove mathmaticly

My questions is don't know where to start I know it has something to do with q=Q(1-e_-t/RC) but must be missing something

 

[Hootenanny]

Welcome to the forums Ryll,

I'll give you two hints. Firstly, current is the rate of flow of charge. And secondly, one could make good use of limits in this question

 

[Moetasim]

As time reaches infinity, the capacitor will be fully charged and there will be no more current flowing through the circuit. This can be mathematically proven by using the equation q=Q(1-e^(-t/RC)), where q is the charge on the capacitor, Q is the maximum charge on the capacitor, t is time, R is the resistance in the circuit, and C is the capacitance of the capacitor.

At infinity, t will be a very large number, so the term e^(-t/RC) will approach 0. Therefore, the equation becomes q=Q(1-0), which simplifies to q=Q. This means that at infinity, the charge on the capacitor will be equal to the maximum charge it can hold.

Since the current in the circuit is given by I=dq/dt, where I is current, q is charge, and t is time, we can see that as t approaches infinity, the current will approach 0. This is because the charge on the capacitor is constant at this point, so there is no change in charge over time and therefore no current.

In conclusion, at infinity, the current in the circuit will be 0 mA, as the capacitor will be fully charged and no more current will flow. This can be mathematically proven using the equation q=Q(1-e^(-t/RC)).