Question:
A capacitor of capacitance $12.0 \mu \mathrm{F}$ is connected to a battery of emf $6.00 \mathrm{~V}$ and internal resistance1.00 $\Omega$ through resistance less leads. $12.0 \mu \mathrm{s}$ after the connections are made, what will be
(a) The current in the circuit,
(b) The power delivered by the battery,
(c) The power dissipated in heat and
(d) The rate at which the energy stored in the capacitor is increasing.
Solution:
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.