An inductor coil stores `16` joules of energy and dissipates energy as heat at the rate of `320 W` when a current of 2 Amp is passed through it. When the coil is joined across a battery of emf `20V` and internal resistance `20Omega` the time constant for the circuit is `tau`. Find `100tau` ( in seconds).

Answer

8

Explanation

`1/2L(4)=16impliesL=8H`
`4R=320impliesR=80Omega`
When connected to the battery, `R_("net")=100Omega, L=8H`
`=tau=8/100implies100tau=8`