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Find the current amplitude if the self-inductance of a resistanceless inductor that is connected across the source of the previous problem is (a) $0.0100 \mathrm{H},$ (b) $1.00 \mathrm{H},(\mathrm{c}) 100 \mathrm{H}$ .

a) $2.5 \mathrm{A}$b) 0.025 $\mathrm{A}$c) $2.5 \times 10^{-4} \mathrm{A}$

Physics 102 Electricity and Magnetism

Chapter 22

Alternating Current

Current, Resistance, and Electromotive Force

Direct-Current Circuits

Electromagnetic Induction

University of Michigan - Ann Arbor

Simon Fraser University

McMaster University

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So for this, we're going to need the formula for the current amplitude in terms of the reactions of the induct er and it's that right there now, the reactant ce of the in doctor is equal to the angle of frequency times the induction ce and so playing this in here, get that The current amplitude is the potential over that. And now finally, Omega be equal to the lowest value, Which eyes looks like 1,000 radiance for second and l A is equal. Ope Sorry. Rather Ellis changing in this sordid Let lb its lowest value 0.1 Henry's. Then the current amplitude ends up being 2.5 and piers, and now we're going to change valuable and see what happens to the current Appleton. But in order to speed this up, notice that the current amplitude is proportional 21 over l. And so if we multiply, l buy 100 I gets divided by hundreds. And that's what happens now. We're multiplying l by 100. So l goes to one Henry. So for l is equal to one. Henry l hear had to get multiplied by 100. And so I was going to get divided by 100. And that is 25. No amperes. Now, if I multiply l by 100 again, I'm going to divide. I buy 100 again. So I get Isaac all to 0.250 1,000,000 years and

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