Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

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 $\frac{V (t = 2 τ)}{E} = 86.47 %$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: 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.

Step 2: 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,

$V_{C} (T) = e (1 - E^{- t / τ}) V_{C} (t = 2 τ) = E (1 - e^{- 2 τ / τ}) V_{C} (t = 2 τ) = E (0.8647)$

As a result, the fraction of the original emf is transferred to the capacitor after $t = 2 τ i s,$

$\frac{V (t = 2 τ)}{E} = (0.8647 \times 100 %) = 86.47 %$

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Short Answer

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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Step 1: Concept Introduction

Step 2: Explanation

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College Physics (Urone)

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Short Answer

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

Step by Step Solution

Step 1: Concept Introduction

Step 2: Explanation

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