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An inductor is connected to the terminals of a battery that has an emf of 12.0 V and negligible internal resistance. The current is 4.86 mA at 0.700 ms after the connection is completed. After a long time the current is 6.80 mA.
What are
(a) the resistance R of the inductor and
(b) the inductance L of the inductor?

Respuesta :

Answer:

a) 1764.71 ohms

b) 1.73 H

Explanation:

From the question, we can identify the following parameters;

Vo =12 V , i = 4.86 mA, t =0.700 ms, io =6.80 mA

(a) Indcued emf V = L di/dt =0

From ohms law Vo = ioR

R = 12/6.80*0.001

R=1764.71 ohms

(b) For LR circuit

i =io (1-e^-t/T)

Time constant T = L/R

4.86 = 6.80 (1-e^-0.7*10^-3/T)

divide both side by 6.8

0.715 = 0.0007/T

L/R = 0.0007/0.715

L/R = 0.000979020979

Substitute R from above

L = 0.000979020979 * 1764.71

L =1.73 H

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