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Expert-verified Found in: Page 778 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # Figure 21.55 shows how a bleeder resistor is used to discharge a capacitor after an electronic device is shut off, allowing a person to work on the electronics with less risk of shock. (a) What is the time constant? (b) How long will it take to reduce the voltage on the capacitor to ${\mathbf{0}}{\mathbf{.}}{\mathbf{250}}{\mathbf{%}}$(5% of 5%) of its full value once discharge begins? (c) If the capacitor is charged to a voltage V0 through a 100-Ω resistance, calculate the time it takes to rise to 0.865V0 (This is about two-time constants.) (a) The time constant is$\tau =20s$

(b) Time to reduce the voltage on the capacitor to $0.250%$is: $\mathrm{t}=119.8\mathrm{s}$.

(c) The time it takes to rise to 0.865V0 is:$\mathrm{t}=0.016\mathrm{s}$.

See the step by step solution

## Step 1: Concept Introduction

A capacitor is an electrical energy storage device that operates in an electric field. It's a two-terminal passive electrical component. Capacitance is the term used to describe the effect of a capacitor.

## Step 2: Time Constant

(a)

The following formula may be used to compute the time constant,

$\mathrm{\tau }=\mathrm{RC}\phantom{\rule{0ex}{0ex}}=\left(2.50×{10}^{5}\mathrm{\Omega }\right)×\left(8.0×{10}^{5}\mathrm{F}\right)\phantom{\rule{0ex}{0ex}}=20\mathrm{s}$

## Step 3: Voltage Reduction

(b)

We will utilize anequation to determine how long it will take to drop the voltage on the capacitor to $0.250%$.

$\mathrm{V}={\mathrm{V}}_{0}{\mathrm{e}}^{\left(\frac{\mathrm{\tau }}{\mathrm{RC}}\right)}\phantom{\rule{0ex}{0ex}}\mathrm{t}=-\left[\mathrm{RC}\left(\mathrm{ln}\frac{\mathrm{V}}{{\mathrm{V}}_{0}}\right)\right]\phantom{\rule{0ex}{0ex}}\mathrm{t}=\left[\left(2.50×{10}^{5}\mathrm{\Omega }\right)×\left(8.0×{10}^{-5}\mathrm{F}\right)\mathrm{ln}\left(0.0025\right)\right]\phantom{\rule{0ex}{0ex}}\mathrm{t}=119.8\mathrm{s}$

Hence, the time to reduce the voltage on the capacitor to $0.250%$is$119.8\mathrm{s}$.

## Step 4: Rising Time of Capacitor

(c)

For charging, we'll use the following equation,

$\mathrm{V}={\mathrm{V}}_{0}\left(1-{\mathrm{e}}^{-\left(\frac{\mathrm{t}}{\mathrm{RC}}\right)}\right)\phantom{\rule{0ex}{0ex}}\mathrm{t}=-\left[\mathrm{RC}\left(\mathrm{ln}\left(\frac{{\mathrm{V}}_{0}-\mathrm{V}}{{\mathrm{V}}_{0}}\right)\right)\right]\phantom{\rule{0ex}{0ex}}\mathrm{t}=\left[\left(100\mathrm{\Omega }\right)×\left(8.0×{10}^{-5}\mathrm{F}\right)\mathrm{ln}\left(1-0.865\right)\right]\phantom{\rule{0ex}{0ex}}\mathrm{t}=0.016\mathrm{s}$

Therefore, the time it takes to rise to 0.865V0 is $0.016s$. ### Want to see more solutions like these? 