BuyFindarrow_forward

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

Solutions

Chapter
Section
BuyFindarrow_forward

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

A series circuit contains a resistor with R = 40 Ω, an inductor with L = 2 H, a capacitor with C = 0.0025 F, and a 12-V battery. The initial charge is Q = 0.01 C and the initial current is 0. Find the charge at time t.

To determine

To find: The charge at time t.

Explanation

Given data:

The values of, the resistor R is 40Ω , an inductor L is 2 H, a capacitor C is 0.0025 F, and battery E is 12 V.

Write the expression for the differential equation,

LQ+RQ+1CQ=E (1)

Substitute 2 H for L, 40Ω for R, 0.0025 F for C and 12 V for E in equation (1),

2Q+40Q+10.0025Q=12

2Q+40Q+400Q=12 (2)

Consider the auxiliary equation,

2r2+40r+400=0

r2+20r+200=0 (3)

Roots of equation (3) are,

r=20±(20)24(1)(200)2(1){r=b±b24ac2afortheequationofar2+br+c=0}=20±i202=10±i10

Write the expression for the complementary solution of two complex roots r=α±iβ ,

Qc(t)=eαt(c1cosβt+c2sinβt) (4)

Substitute 10 for α and 10 for β in equation (4),

Qc(t)=e10t(c1cos10t+c2sin10t) (5)

The Right hand side (RHS) of a differential equation contains only a numerical value, and then the particular solution for this case can be expressed as follows

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Chapter 17 Solutions

Show all chapter solutions add
Sect-17.1 P-11ESect-17.1 P-12ESect-17.1 P-13ESect-17.1 P-14ESect-17.1 P-15ESect-17.1 P-16ESect-17.1 P-17ESect-17.1 P-18ESect-17.1 P-19ESect-17.1 P-20ESect-17.1 P-21ESect-17.1 P-22ESect-17.1 P-23ESect-17.1 P-24ESect-17.1 P-25ESect-17.1 P-26ESect-17.1 P-27ESect-17.1 P-28ESect-17.1 P-29ESect-17.1 P-30ESect-17.1 P-31ESect-17.1 P-32ESect-17.1 P-33ESect-17.1 P-34ESect-17.2 P-1ESect-17.2 P-2ESect-17.2 P-3ESect-17.2 P-4ESect-17.2 P-5ESect-17.2 P-6ESect-17.2 P-7ESect-17.2 P-8ESect-17.2 P-9ESect-17.2 P-10ESect-17.2 P-11ESect-17.2 P-12ESect-17.2 P-13ESect-17.2 P-14ESect-17.2 P-15ESect-17.2 P-16ESect-17.2 P-17ESect-17.2 P-18ESect-17.2 P-19ESect-17.2 P-20ESect-17.2 P-21ESect-17.2 P-22ESect-17.2 P-23ESect-17.2 P-24ESect-17.2 P-25ESect-17.2 P-26ESect-17.2 P-27ESect-17.2 P-28ESect-17.3 P-1ESect-17.3 P-2ESect-17.3 P-3ESect-17.3 P-4ESect-17.3 P-5ESect-17.3 P-6ESect-17.3 P-7ESect-17.3 P-8ESect-17.3 P-9ESect-17.3 P-10ESect-17.3 P-11ESect-17.3 P-12ESect-17.3 P-13ESect-17.3 P-14ESect-17.3 P-15ESect-17.3 P-16ESect-17.3 P-17ESect-17.3 P-18ESect-17.4 P-1ESect-17.4 P-2ESect-17.4 P-3ESect-17.4 P-4ESect-17.4 P-5ESect-17.4 P-6ESect-17.4 P-7ESect-17.4 P-8ESect-17.4 P-9ESect-17.4 P-10ESect-17.4 P-11ESect-17.4 P-12ECh-17 P-1RCCCh-17 P-2RCCCh-17 P-3RCCCh-17 P-4RCCCh-17 P-5RCCCh-17 P-1RQCh-17 P-2RQCh-17 P-3RQCh-17 P-4RQCh-17 P-1RECh-17 P-2RECh-17 P-3RECh-17 P-4RECh-17 P-5RECh-17 P-6RECh-17 P-7RECh-17 P-8RECh-17 P-9RECh-17 P-10RECh-17 P-11RECh-17 P-12RECh-17 P-13RECh-17 P-14RECh-17 P-15RECh-17 P-16RECh-17 P-17RECh-17 P-18RECh-17 P-19RECh-17 P-20RECh-17 P-21RE

Additional Math Solutions

Find more solutions based on key concepts

Show solutions add

(a) By comparing areas, show that ln 2 1 ln 3. (b) Deduce that 2 e 3.

Single Variable Calculus: Early Transcendentals, Volume I

(t+6)(60)(60t+180)(t+6)2

Applied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach

Simplify: 99

Elementary Technical Mathematics

By the methods of trigonometric integrals, sin3 x cos2 x dx should be rewritten as: a) sin3 x (1 sin2 x)dx b...

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th