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What percentage of the final current $I_o$ flows through an inductor $L$ in series with a resistor $R$, three time constants after the circuit is completed?
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OpenStax College Physics Solution, Chapter 23, Problem 75 (Problems & Exercises) (0:58)

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This is College Physics Answers with Shaun Dychko. When turning ON an <i>R</i> - <i>L</i> Circuit, the current will be the final peak current times one minus <i>e</i> to the power of negative time divided by time constant. And so we are told after three time constants, what will percentage of the final peak current [unclear] current be? So we take this formula and divide both sides by the peak current. So we're going to get the ratio of the current to the peak current and that's gonna equal 1 minus <i>e</i> to the negative <i>t</i> over <i>Tau</i>. So it's 1 minus <i>e</i> to the negative 3 <i>tau</i> which is the time, 3 time constants, divided by <i>Tau</i> so the <i>Tau</i>'s cancel and so in our calculator, we will plug 1 minus <i>e</i> to the negative 3 and we get 0.950 and as a percentage that's 95.0 percent. And so the current will be 95.0 percent of the peak by the time 3 time constants have passed.