Explanation Given information: Consider a series circuit consisting of a resistor of R ohms, an indicator of L, and a variable voltage source of V ( t ) volts (time t in seconds). The current through the circuit I ( t ) (in amperes) satisfies the differential equation d I d t + R L I = 1 L V ( t ) , And given units are L = 4 H , R = 500 Ω and V = 20 cos ( 80 t ) v o l t s . Formula used: The general solution of y ' + P ( x ) y = Q ( x ) is y = 1 α ( x ) ( ∫ α ( x ) Q ( x ) d x ) where α ( x ) is an integrating factor such that α ( x ) = e ∫ P ( x ) d x ∫ e a x cos ( b x ) d x = 1 a 2 + b 2 ( e a x ( a cos b x + b sin b x ) ) + C . Calculation: Consider the given differential equation. d I d t + R L I = 1 L V ( t ) L = 4 H , R = 500 Ω a n d V = 20 cos ( 80 t ) v o l t s After substituting the values, the differential equation becomes d I d t + 500 4 I = 1 4 × 20 cos ( 80 t ) d I d t +

ACHIEVE FOR CALCULUS 1 TERM >CSI CUSTOM<

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ISBN: 9781319411657

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 9.5, Problem 42E

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The solution for I(t), assuming that L=4 H,R=500 Ω and V=20cos(80t) volts.

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Chapter 9.5, Problem 41E

Chapter 9.5, Problem 43E

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ACHIEVE FOR CALCULUS 1 TERM >CSI CUSTOM<

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The solution for I ( t ) , assuming that L = 4 H , R = 500 Ω and V = 20 cos ( 80 t ) v o l t s .

Solution Summary: The author explains the solution for I(t), assuming that L=4H,R=500Omega andV=20mathrm

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The solution for I(t), assuming that L=4 H,R=500 Ω and V=20cos(80t) volts.

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Chapter 9 Solutions