An initially uncharged 3.29-muF capacitor and a 7.53-kOhm resistor are connected
ID: 2057891 • Letter: A
Question
An initially uncharged 3.29-muF capacitor and a 7.53-kOhm resistor are connected in series to a 1.50-V battery that has negligible internal resistance What is the initial current m 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 lot the current to decrease lo 3.67% of its initial value? Initial current: Time constant: Elapsed lime:Explanation / Answer
a>initially capacitor hav zero resisitance.so,initial current=1.5/(7.53*10^3)=0.2 milli amperes b>time constant=RC=(3.29*10^-6*7.53*10^3)=24.77 milli secs c>current at time t , i(t)=i(max) *e^(-t/time constant) i(max)=1.5/(7.53*10^3) (3.67/100) *i(max)=i(max) *e^(-t/24.77*10^-3) =>t= 81.86 milli seconds