Question

An initially uncharged 4.85-?F capacitor and a 7.69-k? 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 2.31% of its initial value? Show in detail how you find elapsed time(t) please!

Answer 1

Answer:

0.000140530327543 s

Explanation:

V = Voltage = 1.5 V

r = Resistance = $7.69\ \Omega$

c = Capacitance = $4.85\ \mu F$

Initial current

$i_i=\dfrac{V}{r}\\\Rightarrow i_i=\dfrac{1.5}{7.69}\\\Rightarrow i_i=0.195058517555\ A$

Time constant

$t_1=rc\\\Rightarrow t_1=7.69\times 4.85\times 10^{-6}\\\Rightarrow t_1=0.0000372965\ s$

The current is

$i=i_ie^{\dfrac{-t}{t_1}}\\\Rightarrow -\ln\dfrac{i}{i_i}=\dfrac{t}{t_1}\\\Rightarrow t=-t_1\times \ln\dfrac{i}{i_i}\\\Rightarrow t=-0.0000372965\times \ln 0.0231\\\Rightarrow t=0.000140530327543\ s$

The elapsed time is 0.000140530327543 s