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Consider an RLC circuit with resistance 2 ohms, inductance 0.05 henrys, capacitance 0.006 farads, and an impressed voltage, provided by an alternating current generator, of E(t) = 5 sin(25t). At t = 0 the charge is 1 coulomb and the current is 1 amps. a. Complete the ODE below in terms of the charge in the circuit Q(t) Q" ] Q² + [ Q' Q = b. Solve the ODE and complete the equation for the charge Q(t) =

Consider an RLC circuit with resistance 2 ohms, inductance 0.05 henrys, capacitance 0.006 farads, and an impressed voltage, provided by an alternating current generator, of E(t) = 5 sin(25t). At t = 0 the charge is 1 coulomb and the current is 1 amps. a. Complete the ODE below in terms of the charge in the circuit Q(t) Q" ] Q² + [ Q' Q = b. Solve the ODE and complete the equation for the charge Q(t) =

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### Solving an RLC Circuit with Given Parameters Consider an RLC circuit with: - Resistance: \( 2 \) ohms - Inductance: \( 0.05 \) henrys - Capacitance: \( 0.006 \) farads This circuit is driven by an alternating current generator with an impressed voltage defined by \( E(t) = 5 \sin(25t) \). At \( t = 0 \), the initial conditions are: - Charge: \( 1 \) coulomb - Current: \( 1 \) amp #### a. Completing the ODE Given the information: \[ \boxed{0.05}Q'' + \boxed{2}Q' + \boxed{\frac{1}{0.006}}Q = 5 \sin(25t) \] Substituting the provided values: \[ 0.05Q'' + 2Q' + \frac{1}{0.006}Q = 5 \sin(25t) \] Simplifying the coefficient for \(Q\): \[ \frac{1}{0.006} = 166.\overline{6} \] Thus, the completed ordinary differential equation (ODE) in terms of the charge \( Q(t) \) is: \[ 0.05Q'' + 2Q' + 166.67Q = 5 \sin(25t) \] #### b. Solving the ODE To solve for the charge \( Q(t) \), one must follow the procedures for solving second-order linear differential equations, potentially involving methods such as undetermined coefficients or variation of parameters. You'd generally: 1. Solve the corresponding homogeneous equation: \[ 0.05Q'' + 2Q' + 166.67Q = 0 \] 2. Find particular solutions to the non-homogeneous equation: \[ 0.05Q'' + 2Q' + 166.67Q = 5 \sin(25t) \] Combining the solutions will yield the general form of \( Q(t) \). For detailed, step-by-step solutions, please refer to resources on solving second-order linear differential equations with non-homogeneous terms. #### Solution: Given \( Q(t) \): \[ \boxed{Q(t) = \text{(The detailed solved equation in terms of } t)} \]
Transcribed Image Text:### Solving an RLC Circuit with Given Parameters Consider an RLC circuit with: - Resistance: \( 2 \) ohms - Inductance: \( 0.05 \) henrys - Capacitance: \( 0.006 \) farads This circuit is driven by an alternating current generator with an impressed voltage defined by \( E(t) = 5 \sin(25t) \). At \( t = 0 \), the initial conditions are: - Charge: \( 1 \) coulomb - Current: \( 1 \) amp #### a. Completing the ODE Given the information: \[ \boxed{0.05}Q'' + \boxed{2}Q' + \boxed{\frac{1}{0.006}}Q = 5 \sin(25t) \] Substituting the provided values: \[ 0.05Q'' + 2Q' + \frac{1}{0.006}Q = 5 \sin(25t) \] Simplifying the coefficient for \(Q\): \[ \frac{1}{0.006} = 166.\overline{6} \] Thus, the completed ordinary differential equation (ODE) in terms of the charge \( Q(t) \) is: \[ 0.05Q'' + 2Q' + 166.67Q = 5 \sin(25t) \] #### b. Solving the ODE To solve for the charge \( Q(t) \), one must follow the procedures for solving second-order linear differential equations, potentially involving methods such as undetermined coefficients or variation of parameters. You'd generally: 1. Solve the corresponding homogeneous equation: \[ 0.05Q'' + 2Q' + 166.67Q = 0 \] 2. Find particular solutions to the non-homogeneous equation: \[ 0.05Q'' + 2Q' + 166.67Q = 5 \sin(25t) \] Combining the solutions will yield the general form of \( Q(t) \). For detailed, step-by-step solutions, please refer to resources on solving second-order linear differential equations with non-homogeneous terms. #### Solution: Given \( Q(t) \): \[ \boxed{Q(t) = \text{(The detailed solved equation in terms of } t)} \]
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