(a) At `t=oo` , inductor offers zero resistance
`i_(0)=(E)/(R_(1)R_(2)//(R_(1)+R_(2)))=(E(R_(1)+R_(2)))/(R_(1)R_(2))`
`(i_(1))=(E)/(R_(1))`
`tau=(L)/(R_(1)+R_(2))`
(c) In discharging circuit
`i=(i_(1))_(0)e^(-t//tau)=(E)/(R_(1))e^(-t//tau)`
When `t=tau`
`i=(E)/(R_(1))e^(tau//tau)=(E)/(R_(1))e^(-1)=(E)/(eR_(1))`
![image](https://learnqa.s3.ap-south-1.amazonaws.com/images/16122510598674305761612251059.png)