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Q. 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

Electromagnetic Induction

Solution:

$\tau_1 = \frac{L}{r} ; \tau_2 = \frac{L}{R + r}$ Solve for r from te above equations.