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?

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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

C=.25^4/3*120 farad=1.08*10^-5 F =10.85micro Farad.(answer)

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