Consider an RL circuit consisting of an inductor with an inductance of L henry(H)and aresistor with a resistance of R ohms (1) driven by a voltage of E(t) volts (V). Giventhe voltage drop across the resistor is ER RI, and across the inductor isEL = L(dI/dt), Kirchhoff's Law givesI(t) ==ER(t)EL(t)Now suppose an RL circuit with a 902 resistor and a 0.02H inductor is driven by aconstant voltage of 105V. If the initial resistor current is I(0) 0A, find the current Iand the voltages across the inductor EL and the resistorER in terms of time t.===dIL + RIdt= E(t)=AVV=

Given Information:To find:The value of .Concept used:The linear differential equation is of the form…

Answered: Consider an RL circuit consisting of an…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Related questions

Question

Consider an RL circuit consisting of an inductor with an inductance of L henry(H)and a
resistor with a resistance of R ohms (1) driven by a voltage of E(t) volts (V). Given
the voltage drop across the resistor is ER RI, and across the inductor is
EL = L(dI/dt), Kirchhoff's Law gives
I(t) =
=
ER(t)
EL(t)
Now suppose an RL circuit with a 902 resistor and a 0.02H inductor is driven by a
constant voltage of 105V. If the initial resistor current is I(0) 0A, find the current I
and the voltages across the inductor EL and the resistorER in terms of time t.
=
=
=
dI
L + RI
dt
= E(t)
=
A
V
V
=

Expert Solution

Step 1: Analysis and Introduction

Given Information:

$table row cell L fraction numerator d I over denominator d t end fraction plus R I end cell equals cell E open parentheses t close parentheses end cell row cell E subscript R end cell equals cell R I end cell row cell E subscript L end cell equals cell L fraction numerator d I over denominator d t end fraction end cell end table$

$R equals 9 capital omega comma space L equals 0.02 H comma space E equals 105 V$

$I open parentheses 0 close parentheses equals 0 space A$

To find:

The value of $I open parentheses t close parentheses comma space E subscript R open parentheses t close parentheses comma space E subscript L open parentheses t close parentheses$ .

Concept used:

The linear differential equation is of the form $fraction numerator d y over denominator d x end fraction plus p open parentheses x close parentheses y equals q open parentheses x close parentheses$ .

The solution to such linear differential equation is $y e to the power of integral p open parentheses x close parentheses d x end exponent equals integral q open parentheses x close parentheses e to the power of integral p open parentheses x close parentheses d x end exponent plus C$ .

Integration Formula:

$table row cell integral a space d x end cell equals cell a x plus c end cell row cell integral a e to the power of b x end exponent end cell equals cell a e to the power of b x end exponent over b plus c end cell end table$

Here, $a comma b$ are constants and $c$ is the constant of integration.