An RLC circuit consists of a resistor R, an inductor L, and a capacitor C connected in series to a voltage source E(t). The charge q on the capacitor satisfies the following second-order ODE: Lď' + Rq' + ¿q = E(t) Suppose R = 0.02 N (ohms), L = 0.001 H (henrys), C = 2 F (farads), and E(t) = sin(100t) V (volts). a. Solve for q(t), using q(0) = 0 and d'(0) = 0 (initial charge and current are 0). dq b. Find I(t), the current in the circuit. Use I(t) = dt
An RLC circuit consists of a resistor R, an inductor L, and a capacitor C connected in series to a voltage source E(t). The charge q on the capacitor satisfies the following second-order ODE: Lď' + Rq' + ¿q = E(t) Suppose R = 0.02 N (ohms), L = 0.001 H (henrys), C = 2 F (farads), and E(t) = sin(100t) V (volts). a. Solve for q(t), using q(0) = 0 and d'(0) = 0 (initial charge and current are 0). dq b. Find I(t), the current in the circuit. Use I(t) = dt
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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![1. An RLC circuit consists of a resistor \( R \), an inductor \( L \), and a capacitor \( C \) connected in series to a voltage source \( E(t) \). The charge \( q \) on the capacitor satisfies the following second-order ordinary differential equation (ODE):
\[
Lq'' + Rq' + \frac{1}{C}q = E(t)
\]
Suppose \( R = 0.02 \, \Omega \) (ohms), \( L = 0.001 \, H \) (henrys), \( C = 2 \, F \) (farads), and \( E(t) = \sin(100t) \, V \) (volts).
a. Solve for \( q(t) \), using \( q(0) = 0 \) and \( q'(0) = 0 \) (initial charge and current are 0).
b. Find \( I(t) \), the current in the circuit. Use \( I(t) = \frac{dq}{dt} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7d7f8ba3-ac35-40be-9d69-f75baa192b25%2F19f09559-5c1d-460c-a6e0-476bb163a1d6%2Fnu9lpg_processed.png&w=3840&q=75)
Transcribed Image Text:1. An RLC circuit consists of a resistor \( R \), an inductor \( L \), and a capacitor \( C \) connected in series to a voltage source \( E(t) \). The charge \( q \) on the capacitor satisfies the following second-order ordinary differential equation (ODE):
\[
Lq'' + Rq' + \frac{1}{C}q = E(t)
\]
Suppose \( R = 0.02 \, \Omega \) (ohms), \( L = 0.001 \, H \) (henrys), \( C = 2 \, F \) (farads), and \( E(t) = \sin(100t) \, V \) (volts).
a. Solve for \( q(t) \), using \( q(0) = 0 \) and \( q'(0) = 0 \) (initial charge and current are 0).
b. Find \( I(t) \), the current in the circuit. Use \( I(t) = \frac{dq}{dt} \).
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