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




Emf of the battery,


Now, time constant,


The initial current is :


At t = 12.0μs


The power supplied by the battery is:

At t = 12.0μs,


The power dissipated as heat:

At t = 12.0μs,


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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