# Need help on a Calculus homework question about electrical current

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### Question Description

A capacitor has 100V of charge on it.  An electrical current is supplied to it, with .  After .25 seconds, the capacitor has charged to 220 V. What is its capacitance?

ipesuni
School: New York University

We have the equation as

i=C*(dV/dt)

i=current,Missing in the question,from different sources I have found the i should be given as i=.25t^2

dV=change in voltage

C=capacitance

It can be written as

idt=CdV

i (dt)=C int(dV)

Integrating both side

int.25t^2**dt=C*(V2-V1)

(.25*t^3)/3=120C*(V2-V1)

Solve

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Anonymous
awesome work thanks

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