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.

Answer

`0.25 muF`

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 is ` V= ( Q)/(c ) =(Q_0)/( 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 = 12 V`(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)/( 0.105) =( 1mus )/( 0.405) = 2.469) mus `
Or, ` C= ( 2.469 mus )/( 10 Omega ) =0.25 mu F . `