A parallel plate capacitor of `1 muF` capacity is discharging thorugh a resistor. If its energy reduces to half in one second. The value of resistance will be

Options

A. `(2)/(ln(2))MOmega`
B. `(4)/(ln(2))MOmega`
C. `(0)/(ln(2))MOmega`
D. `(16)/(ln(2))MOmega`

Answer

A

Explanation

`theta=theta_(0)e^(-t//tau)` When energy is 50%
then `theta=(theta_(0))/(sqrt2)`
`(theta_(0))/(sqrt2)=theta_0e^(-t//tau)`
`(1)/(tau)=ln(sqrt2)" "tau=(t)/(ln(sqrt2))`
`R_(C)=(1)/(ln(sqrt2))`
`R=(1)/(Cln(sqrt2))=(1)/(10^(-6).ln(sqrt2))=(10^(6))/(ln(sqrt2))=(2)/(ln(2))=MOmega`