A capacitor of capacitance 8.0 μF is connected to a battery of emf 6.0 V through a resistance of 24Ω. Find the current in the circuit

(a) just after the connections are made and


(b) one time constant after the connections are made.


Charging a capacitor:


A capacitor of capacitance C is being charged using a battery of emf ϵ through a resistance R . A switch S is also connected in series with the capacitor. The switch is initially open. The capacitor is uncharged at first. At t=0, the switch is closed. The current through the circuit at anytime t>0 is given by:



Where I0 is the initial current and is the time constant.


Given,


Capacitance,


EMF of battery,


Resistance,


(a)


Just after the connections are made, there is no potential difference across the capacitor and it acts as a short circuit; hence, the current can simply be calculated form Ohm’s law:





(b)


We know that,



Now, at t=τ,




Using the result from (a), we get,



1