An inductor is connected to the terminals of a battery that has an emf of 12.0V and negligible internal resistance. The current is 4.80mA at 0.745 ms after the connections is completed. After a long time the current is 6.40 mA. What is a) The resistance R of the inductor? b) The inductance L of the inductor?

We get the resistance of the inductor from the long-term [equilibrium] current and the voltage, as the current has stopped changing, so the
L(di/dt) contribution is zero. R=V/I= 12v/.0064amp= 1875 ohm.

The current approaches equilibrium as i/i' = 1-exp(-Rt/L), and i/i'=4.8/6.4=.75, which allows calculation of Rt/L, knowing t and R, as we do. I got L=1h, but I have not the time to re-check my calculations.

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aristeus aristeus · Answer 2 · 2020-01-12T23:32:29-05:00

Answer:

(a) Resistance will be 2500 ohm

(b) Inductance will be 0 Henry

Explanation:

We have given emf of the battery V = 12 volt

Current at time t = 0.745 sec is 4.80 mA

After a long time current is 6.40 mA

After a long time means time tends to infinite

At infinite time inductor behaves as short circuit, so there is only resistance in the circuit

So resistance \(R=\frac{V}{i}=\frac{12}{4.8\times 10^{-3}}=2500ohm\)

Current at time in inductive circuit is given by

\(i=\frac{V}{R}e^\frac{-t}{\tau }\)

\(4.8\times 10^{-3}=\frac{12}{2500}e^\frac{-0.745\times 10^{-3}}{\tau }\)

\(e^\frac{-0.745\times 10^{-3}}{\tau }=1\)

\(\frac{-0.745\times 10^{-3}}{\tau }=ln1\)

\(\tau =0sec\)

\(\tau =\frac{L}{R}\)

L = 0 Henry