The figure shows a graph of the current in a charging circuit of a capacitor through a resistor of resistance `10Omega`.

Options

A. the initial potential difference across the capacitor is `100V`
B. The capacitor of the capcitor is `1/(10ln2) F`
C. the total heat produced in the circuit will be `(500/ln2)J`
D. All of the above

Answer

D

Explanation

`V_0=i_0R=(10)(10)=100V`
after `2s` , current becoms `1/4` th. Therefore, after `1s`, current will remain half also called half life.
`t_(1/2)=(ln2)tau_C=(ln 2)CR`
`:. C=((t_(1/2)))/((ln 2)R)=1/(10 ln 2) F`
total heat `=1/2 CV_0^2`
`=1/2x1/(10 ln 2) (100)^2`
`=500/(ln 2) J`