   # Electrical Circuits In Exercises 29 and 30, use the electrical circuit differentialequation d 2 q d t 2 + ( R L ) d q d t + ( 1 L C ) q = ( 1 L ) E ( t ) where R is theresistance (in ohms), C is the capacitance(in farads), L is theinductance (in henrys), E ( t )is the electromotiveforce (in volts), and q is the charge on thecapacitor (in coulombs).Find the charge q asa function of time t for the electrical circuit described. Assume that q ( 0 ) = 0 and q ' ( 0 ) = 0 . R = 20 , C = 0.02 , L = 2 , E ( t ) = 12 sin 5 t ### Multivariable Calculus

11th Edition
Ron Larson + 1 other
Publisher: Cengage Learning
ISBN: 9781337275378

#### Solutions

Chapter
Section ### Multivariable Calculus

11th Edition
Ron Larson + 1 other
Publisher: Cengage Learning
ISBN: 9781337275378
Chapter 16.3, Problem 29E
Textbook Problem
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## Electrical CircuitsIn Exercises 29 and 30, use the electrical circuit differentialequation d 2 q d t 2 + ( R L ) d q d t + ( 1 L C ) q = ( 1 L ) E ( t ) where R is theresistance (in ohms),C is the capacitance(in farads), L is theinductance (in henrys),E(t)is the electromotiveforce (in volts), and q is the charge on thecapacitor (in coulombs).Find the charge q asa function of time t for the electrical circuit described. Assume that q ( 0 ) = 0 and q ' ( 0 ) = 0 . R = 20 , C = 0.02 , L = 2 , E ( t ) = 12 sin 5 t

To determine

To calculate: The charge q as a function of time t for the electrical circuit describedfor the differential equation is d2qdt2+(RL)dqdt+(1LC)q=(1L)E(t) where, R=20,C=0.02,L=2,E(t)=12sin5t.

### Explanation of Solution

Given information:

The given differential equation is d2qdt2+(RL)dqdt+(1LC)q=(1L)E(t) where R=20,C=0.02,L=2,E(t)=12sin5t and assume that q(0)=0and q'(0)=0.

Concept Used:

If yp is the particular solution of the differential equation y''+ay'+by=F(x) and yh is the general solution, then y=yh+yp is the general solution of the non homogeneous equation.

Calculation:

Consider the given differential equation is d2qdt2+(RL)dqdt+(1LC)q=(1LC)E(t).

The coefficients of the differential terms are,

RL=202=101LC=12×0.02=251L=12=0.5

Now putting these in the differential equation

d2qdt2+10dqdt+25q=0.5×12sin5td2qdt2+10dqdt+25q=6sin5t

The characteristics equation is m2+10m+25=0.

First, we have to find the solution of the characteristics equation m2+10m+25=0.

m2+10m+25=0(m+5)2=0m=5

Therefore,

qh=(C1+tC2)e5t

Let qp be generalized form of 6sin5t

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