1.

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 isA. `(R tau_(2))/(tau_(1)-tau_(2))`B. `(R tau_(1))/(tau_(1)-tau_(2))`C. `(R(tau_(1)-tau_(2)))/(tau_(1))`D. `(R(tau_(1)-tau_(2)))/(tau_(2))`

Answer» Correct Answer - A
`tau_(1) = (L)/(r )," "tau_(2) = (L)/(R+r)` Solve for from the above equations


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