Question

The current in an l-r circuit builds up to (3/4)th of its steady state value in 4- seconds. The time constant of the circuit is

Answer 1

we have to use formula,
$\boxed{\boxed{\bf{I=I_0(1-e^{-t/\tau})}}}$
where I is current after time t, $\tau$ is proportionality constant known as time constant and $I_0$ initial current.

given, $I=\frac{3}{4}I_0$ in t = 4sec

so, $\frac{3}{4}I_0=I_0(1-e^{-4/\tau})$

or, $\frac{1}{4}=e^{-4/\tau}$

taking log both sides,

or, $-ln4=-4/\tau$

or, $\tau=\frac{2}{ln2}$ sec

hence, time constant is 2/ln2