An initially uncharged 4.09 * 10^-6 F capacitor and a 7710 ohm resistor are conn

Question

An initially uncharged 4.09 * 10^-6 F capacitor and a 7710 ohm resistor are connected in series to a 1.5 V barrety that has negligible internal resistance...

An initially uncharged 4.09 * 10 6 F capacitor and a 7710 0 resistor are connected in series to a 1.50 V battery that has negligible internal resistance. What is the initial current in the circuit? Calculate the circuit's time constant. How much time must elapse from the closing of the circuit for the current to decrease to 3.69% of its initial value?

Explanation / Answer

a)

Initial current

Io=Vo/R =1.5/7710

Io=1.9455*10^-4 A

b)

Time constant

T=RC=7710*4.09*10^-6

T=0.03153 sec

c)

I=Io*e^(-t/T)

0.0369*Io=Io *e*(-t/0.03153)

ln(0.0369)=-t/0.03153

t=0.104 sec

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