A capacitor is connected to a 12 V battery through a resistance of `(10Omega)`.It is found that the potential difference across the capacitor rises to 4.0 V in 1(mu)s. find the capacitance of the capacitor.

Explanation

The charge on the capacitor during charging is given by
`Q=Q_(0)(1-e^(-t//RC)`
Hence the poential differnce across the capacitor is
`V=Q//C=Q_(0)//C(1-e^(-t//RC)`.
Here at `t=1(mu)s,`the potential difference is 4Vwhereas the steady state potential difference is `Q_(0)//C=12 V,`so,
`4V=12V(1-e^(-t//RC)`
`(1-e^(-t//RC)=1/3
`(e^(-t//RC)=2/3`
or,(t)/(RC)=In((3)/(2))=0.405`
`or, C=(2.469 (mu)s)/(10 Omega)~~0.25(mu)F.`