Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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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
=
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Transcribed Image Text: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
Check Mark
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.

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