Explanation Given: Voltage, E ( t ) = 10 sin 60 t Capacitance, C = 0.01 F Resistance is, R = 2 Ω Initial charge, Q ( 0 ) = 0 Derivative rule: Product rule d ( f ⋅ g ) d x = f ⋅ d ( g ) d x + g ⋅ d ( f ) d x The First order linear differential equation is of the form d y d x + P ( x ) y = Q ( x ) Kirchhoff’s law The voltage drop across the capacitor is Q C ,(where Q is the charge) so, R I + Q C = E ( t ) Calculation: Consider the given equation R I + Q C = E ( t ) Substitute I = d Q d t R d Q d t + Q C = E ( t ) (1) Substitute the given values in equation (1) 2 d Q d t + 1 0.01 Q = 10 sin 60 t 2 d Q d t + 1 0.01 ⋅ 100 100 Q = 10 sin 60 t 2 d Q d t + 100 Q = 10 sin 60 t Divide both sides by 2 d Q d t + 50 Q = 5 sin 60 t (2) Compare the above differential equation to the first order linear differential equation. P ( x ) = 50 Integrating factor is I ( x ) = e ∫ P ( x ) d x Substitute P ( x ) = 50 in integrating factor I ( t ) = e ∫ 50 d t = e 50 ∫ d t = e 50 t Multiply both sides by I ( x ) in differential equation (2) e 50 t d Q d t + e 50 t 50 Q = e 50 t 5 sin 60 t e 50 t d Q d t + d ( e 50 t ) d t Q = 5 e 50 t sin 60 t Apply the product rule in above equation ( e 50 t Q ) ′ = 5 e 50 t sin 60 t Integrate both sides with respect to t

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ISBN: 9781285741550

Author: James Stewart

Publisher: Cengage Learning

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To find: The charge and current at time t

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The charge and current at time t

Solution Summary: The author explains that the voltage drop across the capacitor is QC, where Q is the charge.

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