A solenoid has an inductance of 10 henty and a resistance of 2 ohm. It is connected to a 10 volt battery. How long will it take for the magnetic energy to reach `1//4` of its maximum value?

Answer

B, C, D

Explanation

`U=1/2Li^(2) i.e. U prop i^(2)`
`U` will reach `1/4th` of its maximum value when current is reached half of its maximum value.In `L-R` circuit, equation of current growth is written as
`i=i_(0)(1-e^(-t//tau))`
Here `i_(0)`=Maximum value of current
`tau`=Time constant =`L//R`
`tau=(10henry)/(2 ohm)=5s^(-1)`
Therefore, `i=i_(0)//2=i_(0)(i-e^(-t//5))`
or `1/2=1-e^(-t//5)` or `e^(-t//5)=1/2`
or `-t//5="In"(1/2)`
or `t//5="in"(2)=0.693`
`therefore t=(5)(0.693)s`
or `t=3.465s`