V=6(1-e^t/rc)
3.6/6=1-e^1/rc
take ln of both sides
ln(.6)=0+-1/rc
r=-ln(.6)/5E-6
A 6.0 V battery is being used to charge a 5.0x10^-6 F capacitor through a resistor R. The capacitor attains a potential difference of 3.6 V in 1 second after the charging begins. The value of the resistance R is?
Been stuck on this for a while and if anyone can point me into the right direction that will be great.
2 answers
Oh ok, thank you.
1 answer