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 potential difference across the capacitor `V=Q/C=(Q)/(C)(1-e^(-t//RC)`
Here, at t= 1 `mus,` the potential differences is 4V whereas the steady potential difference is `Q_(0)/C=12V`
So, `4V=12V (1-e^(-t//RC)`
`or 1-e^(-t//RC)=1/3`
`or, e^(-t//RC)=2/3`
`or, t//RC=ln (3/2)=0.405`
`or, RC=(t)/(0.405) =(1mus)/(0.405)=2.469mus`
`or, C=(2.469mus)/(10Omega)=0.25muF`