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After two time constants, what percentage of the final voltage, emf, is on an initially uncharged capacitor $C$ , charged through a resistance $R$?
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$86.5 \%$

Solution Video

OpenStax College Physics Solution, Chapter 21, Problem 67 (Problems & Exercises) (1:16)

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Video Transcript

This is College Physics Answers with Shaun Dychko. The question we’re answering here is to what percentage of the final voltage will the capacitor be charged after two times constants. So assuming that charging started completely discharged capacitor. We have the voltage will be the final EMF times one minus <i>e</i> to the power of negative the note of time passed divided by the time constant <i>RC</i>. And so the voltage when you have passed the time of two time constant, that’s two times <i>RC</i> will be EMF times one times <i>e</i> to the power of negative two <i>RC</i> which are substituting for the time divided by <i>RC</i>. And the <i>RC</i> is cancel. And this is going to make negative two for the exponent now. So when you divide that for the final maximum possible voltage which is the EMF, the EMF is cancel giving us one minus <i>e</i> to the negative two. And one minus <i>e</i> to the negative two is 86.5 percent.