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Q67PE

Expert-verifiedFound in: Page 778

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

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

The proportion of the final voltage, emf, is on a capacitor C that was initially uncharged and is now charged via a resistance R is $\frac{V\left(t=2\tau \right)}{E}=86.47\%$

**The opposition to current flow in an electrical circuit is measured by resistance (also known as ohmic resistance or electrical resistance). The Greek letter omega (Ω) is used to represent resistance in ohms. **

We determine the voltage across a capacitor in an RC circuit after two-time constants in this issue. When a capacitor is charging, its voltage is,

${V}_{C}\left(T\right)=e\left(1-{E}^{-t/\tau}\right)\phantom{\rule{0ex}{0ex}}{V}_{C}\left(t=2\tau \right)=E\left(1-{e}^{-2\tau /\tau}\right)\phantom{\rule{0ex}{0ex}}{V}_{C}\left(t=2\tau \right)=E\left(0.8647\right)$

As a result, the fraction of the original emf is transferred to the capacitor after $t=2\tau is,$

$\frac{V\left(t=2\tau \right)}{E}=\left(0.8647\times 100\%\right)\phantom{\rule{0ex}{0ex}}=86.47\%$

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