Question

A series RL circuit is initially relaxed. A step voltage is applied to the circuit. If $\tau$ is the time constant of the circuit, the voltage across R and L will be the same at time t equal to

• A. $\tau In2$
• B. $\tau In\left(\frac{1}{2}\right)$
• C. 1/$\tau In2$
• D. 1/$\tau In\left(\frac{1}{2}\right)$

Answer 1

The correct option is A $\tau In2$

$i\left(t\right)=1-e^{-t/\tau }$

Where $\tau =\frac{L}{R}$

$v_R\left(t\right)=i\left(t\right)R$

$v_R\left(t\right)=(1-e^{-t/\tau })R$

$v_L\left(t\right)=L\frac{dt}{dt}=L.\frac{1}{\tau }{e}^{-t/\tau }$

$v_L\left(t\right)={\mathrm{Re}}^{-t/\tau }$

Let $t=t_{1}when{v}_R\left(t\right)=v_L\left(t\right)$

$\Rightarrow (1-e^{-t_1/\tau })R={\mathrm{Re}}^{-t_1/\tau }$

$\Rightarrow 2e^{-t1/\tau }=1$

$\Rightarrow e^{-t1/\tau }=1$

$\Rightarrow -\frac{t_1}{\tau }=ln\frac{1}{2}=-ln2$

$\Rightarrow t_1=\tau In2$