Question

The current in an $LR$ circuit builds up to ${\left(\frac{3}{4}\right)}^{\text{th}}$ of its steady state value in $4\text{s}$. The time constant of this circuit is

• A. $\frac{1}{\ln 2}\text{s}$
• B. $\frac{2}{\ln 2}\text{s}$
• C. $\frac{3}{\ln 2}\text{s}$
• D. $\frac{4}{\ln 2}\text{s}$

Answer 1

The correct option is B $\frac{2}{\ln 2}\text{s}$

We know that,

$i=i_0[1-e^{(-\frac{Rt}{L})}]$

$\Rightarrow \frac{3}{4}{i}_0=i_0[1-e^{(-\frac{Rt}{L})}]$

$\Rightarrow \frac{3}{4}=1-e^{(-\frac{Rt}{L})}$


$\Rightarrow e^{(-\frac{Rt}{L})}=\frac{1}{4}$

$\Rightarrow e^{(-\frac{Rt}{L})}=2^{-2}$

$\Rightarrow e^{\left(\frac{Rt}{L}\right)}=2^2$

Taking natural log on both sides, we get,

$\frac{Rt}{L}=2\ln 2$

At, $t=4\text{s}$

$\frac{4R}{L}=2\ln 2$

$\Rightarrow \tau =\frac{2}{\ln 2}$

Hence, $\left(B\right)$ is the correct option.