In the circuit diagram shown in the figure the switches `S_(1)` and `S_(2)` are closed at time `t=0`.After time `t=(0.1)ln 2sec`, switch `S_(2)` is opened.The current in the circuit at time `t=(0.2)/n 2sec` is equal to `x/32amp`.Find out value of `x`.

Answer

`67//32 A`

Explanation

When `S_(1)` and `S_(2)` are closed time constant
`tau_(1)=L/R_(1)=1/10 sec`
at `t=0.1 ln2 sec` awitch `S_(2)` is opened
so `tau_(2)=L/((R_(1)+R_(1)))=1/50 sec`.
Just before switch `S_(2)` is opened current in circuit
`i_(1)=100/10 (1-e^(-t//tau))`
`=10(1-e^(-ln2))=10(1-1/e^(ln2))=5 Amp`.
since current in inductor can not be changed instantionusly so this current will decay with time constant `tau_(2)`
current at `t=(0.2ln2-0.1lml)=0.1 ln2`
`i=5e^(t/tau_(2))+2(1-e^(-t//tau_(2)))=2+3e^(-5lm2)=2+3/32`
`=67/32`