Answer :

Concepts/Formula used:


Current when capacitor is charging:


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.


Note that is known as time constant


Power supplied by the battery:


If a battery of emf ϵ gives a current I, then the power supplied by the battery is given by:



Energy dissipated by a resistor :


A resistor of resistance R with current I through it, dissipates energy U given by:



in time Δt.


Its power is given by:



The capacitor is being charged


Given,


Capacitance,


Resistance,


Emf of the battery,


Time,


Now, time constant,


(a)


The initial current is :



Now,



At t = 12.0μs





(b)


The power supplied by the battery is:



At t = 12.0μs,




(c)


The power dissipated as heat:



At t = 12.0μs,




(d)


By conservation of energy,


Energy supplied by battery = Energy stored by capacitor + Energy dissipated as heat.


Dividing by time, gives us


Power supplied by battery = Power dissipated as heat + rate at which energy is stored in the capacitor.


Hence, using the previous results, we have,


Rate at which energy is stored in the capacitor:



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