(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

To measure the inductance and resistance of a real inductor, a physicist first connects the inductor to a 3V battery. At these conditions, the final, steady, current is 24 A. Then the physicist suddenly short-circuits the inductor with a thick (resistance-less) wire place across its terminals. The current then decreases from 24 A to 12 A in 0.22 s. Find the resistance and the inductance.

2. Relevant equations

ε = 3V

I_{1}= 24A

I_{2}= 12A

t = 0.22s

L = (ε_{L}X dt) / DI

R = L / time constant

3. The attempt at a solution

For L: (3v X .22s) / 12A = 0.0275H

For R: 0.0275H / 0.22s = 0.125Ω

Would that be right? It seems like what I did is so simple for a physics assignment, I get suspicious of it..

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# Homework Help: Find the resistance and inductance ensuing a short-circuit

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