Question

The time constant of an inductor is ${\tau }_1$. When a pure resistor of R$\Omega$ is connected in series with it, the time constant is found to decrease to ${\tau }_2$. The internal resistance of the inductor is :

• A. $\frac{R{\tau }_2}{{\tau }_1-{\tau }_2}$
• B. $\frac{R{\tau }_1}{{\tau }_1-{\tau }_2}$
• C. $\frac{R({\tau }_1-{\tau }_2)}{{\tau }_1}$
• D. $\frac{R({\tau }_1-{\tau }_2)}{{\tau }_2}$

Answer 1

Answer: (A)

$\frac{R{\tau }_2}{{\tau }_1-{\tau }_2}$

let $r$ be the internal resistance of the inductor and $L$ be the inductance.
so, ${\tau }_1=\frac{L}{r}$
after connecting resistance R with the inductor, time constant becomes,
${\tau }_2=\frac{L}{r+R}$
${\tau }_2=\frac{{\tau }_{1}r}{r+R}$
$r{\tau }_2+R{\tau }_2=r{\tau }_1$
$R{\tau }_2=r({\tau }_1-{\tau }_2)$
$r=\frac{R{\tau }_2}{{\tau }_1-{\tau }_2}$