Question

1. 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?

Answer 1

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.

Answer 2

(a) Resistance will be 2500 ohm

(b) Inductance will be 0 Henry

Step-by-step 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=(V)/(i)=(12)/(4.8* 10^(-3))=2500ohm`

Current at time in inductive circuit is given by

`i=(V)/(R)e^(-t)/(\tau )`

`4.8* 10^(-3)=(12)/(2500)e^(-0.745* 10^(-3))/(\tau )`

`e^(-0.745* 10^(-3))/(\tau )=1`

`(-0.745* 10^(-3))/(\tau )=ln1`

`\tau =0sec`

`\tau =(L)/(R)`

L = 0 Henry