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?

Question

Alana Beltran

Physics

7 June, 21:43

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?

+3

Answers (1)

Know the Answer?

Samantha Cannon · 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