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A parallel plate capacitor of 1 muF capa...

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

A

`(2)/(ln(2))MOmega`

B

`(4)/(ln(2))MOmega`

C

`(0)/(ln(2))MOmega`

D

`(16)/(ln(2))MOmega`

Text Solution

Verified by Experts

The correct Answer is:
A
`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`
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