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

Expert verified

The proportion of the final voltage, emf, is on a capacitor C that was initially uncharged and is now charged via a resistance R isVt=2τE=86.47%

Step by step solution

01

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.

02

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,

VCT=e1-E-t/τVCt=2τ=E1-e-2τ/τVCt=2τ=E0.8647

As a result, the fraction of the original emf is transferred to the capacitor aftert=2τis,

Vt=2τE=0.8647×100%=86.47%

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Most popular questions from this chapter

Verify the second equation in Example 21.5 by substituting the values found for the currents l1andl2

(a) What is the internal resistance of a voltage source if its terminalvoltage drops by 2.00 V when the current supplied increases by 5.00 A?(b) Can the emf of the voltage source be found with the information supplied.

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(a) What is the resistance to ground of \(100\) of these insulators?

(b) Calculate the power dissipated by \(100\) of them.

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