A capacitor of capacitance 12.0 μF is connected to a battery of emf 6.00 V and internal resistance 1.00 Ω through resistanceless leads 12.0 μs 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.


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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