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The battery terminal voltage in Figure $\mathrm{P} 20.43$ is $\boldsymbol{\varepsilon}=9.00 \mathrm{V}$ and the current I reaches half its maximum value of 2.00 $\mathrm{A}$ at $t=0.100 \mathrm{s}$ after the switch is closed. Calculate $(\mathrm{a})$ the time constant $\tau$ . (b) What is the emf across the inductor at $t=0.100$ s? (c) What is the emf across the inductor in the instant after the switch is closed at $t=0 ?$

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Cornell University

Numerade Educator

University of Washington

McMaster University

All right. So for part a, uh, we're going to use the fact that, um it reaches half its maximum current in 0.1 seconds so I over my max will be half. This ratio is actually absolute absolute of our PMF of resistance times one minus this exponential factor, the time constant in it over the max, which is just enough over are So those things cancel like you're left with. Um, so you have 1/2 is equal to to one minus exporter of dynasty. Roberto. So, uh, you can take, um Won't have to the right hand side expert Exponential term to the left hand side. What you get is exponents physical to 1/2. Take the log of that. And so TV will be negative. Tao o r. Um heard me, uh, Tao, uh, will be if negative TV over log of quite five and negative t is they just 0.1 seconds over larger point and that is divided by a lot of 0.5. Gives you a time constant of point 144 seconds. All right. Uh, part B, um, in part B, we're just using homes law. So enough across the conductor is I r current times resistance and appoint 15 seconds. I is half of too. So that's one NPR and Resistance is four and five homes. This gives you 4.5 volts. Ah, and in part, see same homes a lot. But this time met after the switch of clothes. So this is two amps. That's its maximum current times. Uh, the same resistance. So this gives you nine volts.