Question

An initially uncharged 3.47-μF capacitor and a 6.43-kΩ resistor are connected in series to a 1.50-V battery that has negligible internal resistance. What is the initial current in the circuit, expressed in milliamperes? Calculate the circuit\'s time constant in milliseconds. How much time, in milliseconds, must elapse from the closing of the circuit for the current to decrease to 2.57% of its initial value?

Answer 1

Answer: a) io=233.28 A ( initial current); b) τ=R*C= 22.31 ms; c) 81.7 ms

Explanation:  In order to explain this problem we have to use, the formule for the variation of the current in a RC circuit:

I(t)=io*Exp(-t/τ)

and also we consider that io=V/R=(1.5/6.43*10^3)

=233.28 A

then the time constant for the RC circuit is τ=R*C=6.43*10^3*3.47*10^-6

=22.31 ms

Finally the time to reduce the current to 2.57% of its initial value is obtained from:

I(t)=io*Exp(-t/τ)  for I(t)/io=0.0257=Exp(-t/τ) then

ln(0.0257)*τ =-t

t=-ln(0.0257)*τ=81.68 ms