An electric circuit with a resistor, a capacitor, an inductor, and a voltage source can be...

 

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An electric circuit with a resistor, a capacitor, an inductor, and a voltage source can be described by the ODE $$ \begin {equation) L (dl\over dt} + RI+ (Q\over C} = E(t). \tag (76) \end{equation) $$ where \( LdI/dt) is the voltage drop across the inductor, \(I) is the current (measured in amperes, A), (L) is the inductance (measured in henrys, H). (R) is the resistance (measured in ohms, \(\Omega \)), \(Q) is the charge on the capacitor (measured in coulombs, C). (C) is the capacitance (measured in farads, F), (E(t)\) is the time-variable voltage source (measured in volts, V), and (t) is time (measured in seconds, s). There is a relation between \(I) and (Q): SS \begin{equation) {dQ\over dt} = I\tp \tag (77}\end{equation) SS Equations (76)-(77) is a system two ODES. Solve these for \(L=1 \) H₂ \( E(t)=2\sin omega t \) V, \(\omega^2=3.5\hbox { s}^{-2}\). \(C=0.25 ) C₂ (R=0.2\ \Omega \), \(I(0)=1 \) A, and \(Q(0)=1 C). Use the Forward Euler scheme with \( \Delta t = 2\pi/(60\omega) 1). The solution will, after some time, oscillate with the same period as \(E(t)), a period of ( 2\pi/omega \). Simulate 10 periods. Filename: electric_circuit. Remarks It turns out that the Forward Euler scheme overestimates the amplitudes of the oscillations. The more accurate 4th-order Runge-Kutta method is much better for this type of differential equation model. An electric circuit with a resistor, a capacitor, an inductor, and a voltage source can be described by the ODE $$ \begin {equation) L (dl\over dt} + RI+ (Q\over C} = E(t). \tag (76) \end{equation) $$ where \( LdI/dt) is the voltage drop across the inductor, \(I) is the current (measured in amperes, A), (L) is the inductance (measured in henrys, H). (R) is the resistance (measured in ohms, \(\Omega \)), \(Q) is the charge on the capacitor (measured in coulombs, C). (C) is the capacitance (measured in farads, F), (E(t)\) is the time-variable voltage source (measured in volts, V), and (t) is time (measured in seconds, s). There is a relation between \(I) and (Q): SS \begin{equation) {dQ\over dt} = I\tp \tag (77}\end{equation) SS Equations (76)-(77) is a system two ODES. Solve these for \(L=1 \) H₂ \( E(t)=2\sin omega t \) V, \(\omega^2=3.5\hbox { s}^{-2}\). \(C=0.25 ) C₂ (R=0.2\ \Omega \), \(I(0)=1 \) A, and \(Q(0)=1 C). Use the Forward Euler scheme with \( \Delta t = 2\pi/(60\omega) 1). The solution will, after some time, oscillate with the same period as \(E(t)), a period of ( 2\pi/omega \). Simulate 10 periods. Filename: electric_circuit. Remarks It turns out that the Forward Euler scheme overestimates the amplitudes of the oscillations. The more accurate 4th-order Runge-Kutta method is much better for this type of differential equation model.

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