A capacitor is connected to a 36 V battery through a resistance of `20 Omega`. It is found that the potential difference across the capacitor rises to 12.0 V in `2 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//2RC))`.
Hence, the potential difference across the capacitor is `V=Q//C=Q_(0)//C(1-e^(-t//2RC))`.
Here, at ` t=2 mu s`, the potential difference is 12V whereas the steady potential difference is
`Q_(0)//C=36V. "So", rArr 12V=36V(1-e^(-t//2RC))`
or, `1-e^(-t//2RC)=(1)/(3) " or ", e^(-t//2RC)=(2)/(3)`
or, `(t)/(RC)=l n ((3)/(2))=0.405 " or, " RC=(t)/(0.405)=(2 mu s)/(0.45)=4.936 mu s`
or, `C=(4.936 mu s)/(20 Omega)=0.25 mu F`.