An inductor of `3H` is connected to a battery of emf `6V` through a resistance of `100 Omega`. Calculate the time constant. What will be the maximum value of current in the circuit ?

Explanation

Give that `L = 3H, E = 6V, R = 100 Omega`
Time constant `tau_(L) = (L)/(R ) = (3)/(100) = 0.03 sec`
Maximum Current `I_(0) = (E)/(R ) = (6)/(100)amp = 0.06 amp`