The time constant of an inductor is τ1. When a pure resistor of RΩ is connected in series with it, the time constant is found to decrease to τ2 .The internal resistance of the inductor is

Question

The time constant of an inductor is τ1. When a pure resistor of RΩ is connected in series with it, the time constant is found to decrease to τ2 .The internal resistance of the inductor is (A) (Rτ2/τ1-τ2) (B) (Rτ1/τ1-τ2) (C) (R(τ1-τ2)/τ1) (D) (R(τ1-τ2)/τ2)

Tardigrade Admin · Accepted Answer

τ1 = (L/r) ; τ2 = (L/R + r) Solve for r from te above equations.

Q. The time constant of an inductor is $τ_{1}$ . When a pure resistor of $R Ω$ is connected in series with it, the time constant is found to decrease to $τ_{2}$ .The internal resistance of the inductor is

$\frac{R τ _{2}}{τ _{1} - τ _{2}}$

$\frac{R τ _{1}}{τ _{1} - τ _{2}}$

$\frac{R ( τ _{1} - τ _{2} )}{τ _{1}}$

$\frac{R ( τ _{1} - τ _{2} )}{τ _{2}}$

$τ_{1} = \frac{L}{r}; τ_{2} = \frac{L}{R + r}$ Solve for r from te above equations.

Q. The time constant of an inductor is τ1​. When a pure resistor of RΩ is connected in series with it, the time constant is found to decrease to τ2​ .The internal resistance of the inductor is

τ1​−τ2​Rτ2​​

τ1​−τ2​Rτ1​​

τ1​R(τ1​−τ2​)​

τ2​R(τ1​−τ2​)​