The time constant of an LR circuit is 40 ms. The circuit is connected at t= 0 and the steady state current is found ot be 2.0 A. Find the current at (a) t= 10 ms (b) t =20ms, (c ) t =100ms and (d) t = 1 s.

Explanation

Here `tau= 40ms , i_0 = 2A `
` (a) t = 10ms,`
` Since I = i_0 (1-e^(-t/tau)) `
` = 2 (1- e(-10/40 ) ) `
`=2((1-e^(-1//4)`
`=2(1-0.7788)=2(0.2211)`
`(b)`When t=20 ms`
`t=i_0(1-e^(-t/tau)) `
` = 2 (1- e(-100/40 ) ) `
` = 2 (1- e(-10/4 ) ) `
` = 2 (1-0.082)`
`=1.835~= 1.8A`
(c ) When t=1s
i=i_0(1-e^(-t//tau)`
= 2 (1- e(-10/40 xx10^(-3)) ) `
` = 2 (1- e(-10/400) ) `
` = 2 (1- e(-25 ) )=2xx1=2A. `
`.