   Chapter 17.3, Problem 14E

Chapter
Section
Textbook Problem

A series circuit contains a resistor with R = 24 Ω, an inductor with L = 2 H, a capacitor with C = 0.005 F. and a 12-V battery. The initial charge is Q = 0.001 C and the initial current is 0.(a) Find the charge and current at time t.(b) Graph the charge and current functions.

(a)

To determine

To find: The charge of series RLC circuit at time t and the current of series RLC circuit at time t .

Explanation

Given data:

Series RLC circuit with following values.

R=24Ω , L=2H , C=0.005F , E(t)=12V , Q(0)=0.001 and Q(0)=I(0)=0 .

Formula used:

Write the expression for differential equation of electric circuit.

Ld2Qdt2+RdQdt+1CQ=E(t)

LQ+RQ+1CQ=E(t) (1)

Write the expression for general solution with complex roots.

Q(t)=eαt[c1cos(βt)+c2sin(βt)] (2)

Write the expression for r .

r=α+βi (3)

Write the expression for auxiliary equation.

ar2+br+c=0 (4)

Write the expression for differential equation.

ay+by+cy=0 (5)

Find the differential equation for electric circuit using equation (1).

Substitute 2 for L, 24 for R, 0.005 for C and 12 for E(t) in equation (1),

2Q+24Q+10.005Q=12

2Q+24Q+200Q=12 (6)

Modify equation (5) as follows.

aQ+bQ+cQ=0 (7)

Compare equation (6) and (7).

a=2b=24c=200

Substitute 2 for a, 24 for b and 200 for c in equation (4),

2r2+24r+200=0r2+12r+100=0

Find the value of r .

r=12±(12)24(1)(100)2(1)=12±1444002(1)=12±2562=12±16i2

Simplify r as follows.

r=6±8i (8)

Compare equations (3) and (8).

α=6β=8

Substitute 6 for α and 8 for β in equation (2),

Q(t)=e6t[c1cos(8t)+c2sin(8t)] (9)

Modify equation (9) for complementary equation as follows.

Qc(t)=e6t[c1cos(8t)+c2sin(8t)]

Consider the value of Qp(t) as follows.

Qp(t)=A (10)

Differentiate equation (10) with respect to t .

Qp(t)=0

Differentiate equation with respect to t .

Qp(t)=0

Modify equation (6) as follows.

2Qp(t)+24Qp(t)+200Qp(t)=12

Substitute 0 for Qp(t) , 0 for Qp(t) and A for Qp(t) ,

2(0)+24(0)+200(A)=12200(A)=12A=12200A=350

Substitute 350 for A in equation (10),

Qp(t)=350

Write the expression for general solution.

Q(t)=Qc(t)+Qp(t)

Substitute e6t[c1cos(8t)+c2sin(8t)] for Qc(t) and 350 for Qp(t) ,

Q(t)=e6t[c1cos(8t)+c2sin(8t)]+350 (11)

Substitute 0 for t ,

Q(0)=e6(0)[c1cos(8×0)+c2sin(8×0)]+350=c1+350

Substitute 0

(b)

To determine

To graph: The charge functions and the current functions.

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