Inductors & AC Circuits (Homework) (RL Circuit)

AI Thread Summary
In a series RL circuit with a resistance of 7.0 kΩ, the inductance can be calculated using the time constant equation, Tau = L/R. Given that the current reaches half of its final value in 26 µs, the time constant Tau is determined to be 26 µs multiplied by 2 and then multiplied by 0.63212. This leads to the calculation of inductance (L), which is found to be approximately 0.2301 H. The discussion confirms the use of the time constant in solving for inductance in an RL circuit.
harkirat2009
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Homework Statement


What is the inductance in a series RL circuit in which R = 7.0 kΩ if the current increases to one half of its final value in 26 µs?



Homework Equations



am i correct in assuming that i should be using the time constant equation: Tau=L/R


The Attempt at a Solution



if i was correct in using this equation then:
Tau = ((26E^-6)*2))*.63212
R = 7000 Ohms

Tau*R=L
L=.23009168 H
 
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e^{-t/\tau}=\frac{1}{2} for t = 26 µs.
 
thank you for your help
 
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