Question

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

Short answer

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$\frac{V\left(t=2\tau \right)}{E}=86.47\%$

Step-by-step solution

Concept Introduction

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.

Explanation

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,

$$\begin{gathered} V_C\left(T\right)=e\left(1-E^{-t/\tau }\right) \\ {V}_C\left(t=2\tau \right)=E\left(1-e^{-2\tau /\tau }\right) \\ {V}_C\left(t=2\tau \right)=E\left(0.8647\right) \end{gathered}$$

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

$$\begin{gathered} \frac{V\left(t=2\tau \right)}{E}=\left(0.8647\times 100\%\right) \\ =86.47\% \end{gathered}$$