Problem 1 Solve the one-loop \(L-R-C\) cir... [FREE SOLUTION] (2024)

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Chapter 7: Problem 1

Solve the one-loop \(L-R-C\) circuit with \(L=\) \(3 \mathrm{H}, R=10 \Omega, C=1 /10 \mathrm{~F}\), and (a) \(E(t)=0 \mathrm{~V}\); (b) \(E(t)=e^{-t} \mathrm{~V} .(Q(0)=0 \mathrm{C}, I(0)=1 \mathrm{~A}\).)

Short Answer

Expert verified

(a) \(Q(t) = \frac{3}{\sqrt{5}} e^{-\frac{5}{3}t} \sin \left( \frac{\sqrt{5}}{3}t \right).\) (b) \(Q(t) = e^{-\frac{5}{3} t} \left( -\frac{1}{7}\cos \left( \frac{\sqrt{5}}{3} \right) + \frac{16}{7\sqrt{5}} \sin \left( \frac{\sqrt{5}}{3}t \right)\right) + \frac{1}{7}e^{-t}.\)

Step by step solution

01

Identify the given parameters

The given parameters for the circuit are inductance: \(L = 3 \text{ H}\), resistance: \(R = 10 \text{ }\text{Ω}\), and capacitance: \(C = \frac{1}{10} \text{ F}\).

02

(a) Derive the differential equation for the circuit with \[E(t) = 0\]

For the L-R-C circuit with zero input voltage \(E(t) = 0\), the differential equation is derived using Kirchhoff's voltage law: \[L\frac{dI}{dt} + RI + \frac{Q}{C} = E(t).\] Given \( E(t) = 0\), it simplifies to: \[3\frac{dI}{dt} + 10I + 10Q = 0.\]

03

Express current I in terms of charge Q

The current \(I\) can be expressed in terms of charge \(Q\) as \( I = \frac{dQ}{dt} \). Substituting this into the differential equation: \[3\frac{d^2Q}{dt^2} + 10\frac{dQ}{dt} + 10Q = 0.\]

04

Solve the hom*ogeneous differential equation

This is a second-order hom*ogeneous linear differential equation. The characteristic equation is: \[3r^2 + 10r + 10 = 0.\] Solving this using the quadratic formula: \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-10 \pm \sqrt{100 - 120}}{6} = \frac{-10 \pm \sqrt{-20}}{6} = \frac{-10 \pm 2i\sqrt{5}}{6} = -\frac{5}{3} \pm \frac{i\sqrt{5}}{3}.\]

05

Find the general solution for Q

With the characteristic roots \(r_1 = -\frac{5}{3} + \frac{i\sqrt{5}}{3}\) and \(r_2 = -\frac{5}{3} - \frac{i\sqrt{5}}{3}\), the general solution for \(Q\) is: \[Q(t) = e^{-\frac{5}{3}t} \left( A\cos \left( \frac{\sqrt{5}}{3}t \right) + B\sin \left( \frac{\sqrt{5}}{3}t \right) \right).\]

06

Determine constants using initial conditions for part (a)

Using \( Q(0) = 0 \): \[0 = A \implies A = 0.\] For initial current \( I(0) = 1 = \frac{dQ}{dt}\bigg|_{t=0} \): \[1 = B \cdot \frac{\sqrt{5}}{3} e^{0} \implies B = \frac{3}{\sqrt{5}}.\] Hence, solution becomes: \[Q(t) = \frac{3}{\sqrt{5}} e^{-\frac{5}{3}t} \sin \left( \frac{\sqrt{5}}{3}t \right).\]

07

(b) Derive the differential equation for the circuit with \[E(t) = e^{-t}\]

For \[E(t) = e^{-1}\], using Kirchhoff's voltage law: \[3\frac{dI}{dt} + 10I + 10Q = e^{-t}.\]

08

Solve the non-hom*ogeneous differential equation

The differential equation can be solved using the method of undetermined coefficients. First, solve the hom*ogeneous equation as found in (a). The particular solution can be assumed of the form: \[Q_p(t) = Ae^{-t}.\] Substituting \(Q_p\) into the non-hom*ogeneous equation: \[3Ae^{-t} - 3Ae^{-t} + 10A = e^{-t}\] Matching coefficients, obtain \[A = \frac{1}{7}.\] Thus the particular solution: \[Q_p(t) = \frac{1}{7} e^{-t}.\]

09

Combine hom*ogeneous and particular solutions

The general solution is the sum of the hom*ogeneous and particular solutions: \[Q(t) = e^{-\frac{5}{3}t} \left( A\cos \left( \frac{\sqrt{5}}{3}t \right) + B\sin \left( \frac{\sqrt{5}}{3}t \right) \right) + \frac{1}{7} e^{-t}.\]

10

Determine constants using initial conditions for part (b)

Using \( Q(0) = 0 \): \[0 = A + \frac{1}{7} \implies A = -\frac{1}{7}.\] For \(1 = \left. \frac{dQ}{dtA} \right|_{t=0} \), obtain: \[ B = \frac{16}{7 \sqrt{5}}.\] Thus final solution: \[Q(t) = e^{-\frac{5}{3} t} \left( -\frac{1}{7}\cos \left( \frac{\sqrt{5}}{3} \right) + \frac{16}{7\sqrt{5}} \sin \left( \frac{\sqrt{5}}{3}t \right)\right) + \frac{1}{7}e^{-t} \]

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Second-order Differential Equations

Second-order differential equations involve derivatives up to the second order. In the context of an L-R-C circuit, this means that the equation governs how the charge or current changes over time. Specifically, for an L-R-C circuit, we derive a second-order differential equation using Kirchhoff's voltage law. The general form of a second-order differential equation is: \[ a\frac{d^2Q}{dt^2} + b\frac{dQ}{dt} + cQ = 0 \] where \( a \), \( b \), and \( c \) are constants. In our exercise involving a resistor, inductor, and capacitor, the coefficients depend on the circuit components, leading to an equation that reflects the interplay between resistance, inductance, and capacitance.

hom*ogeneous Differential Equations

hom*ogeneous differential equations have terms that can all be set to zero without any external input, such as external voltage in an L-R-C circuit. For our particular case with \( E(t) = 0 \), we derived the equation: \[ 3\frac{d^2Q}{dt^2} + 10\frac{dQ}{dt} + 10Q = 0 \] The characteristic equation, \[3r^2 + 10r + 10 = 0 \], helps us find roots that describe the system's natural response. The roots can be real or complex, determining whether the system's response will decay exponentially, oscillate, or both. Solving this differential equation gives us the general solution for the charge, \(Q(t)\), as a combination of exponential and trigonometric functions.

Kirchhoff's Voltage Law

Kirchhoff's voltage law (KVL) is a fundamental principle in circuit analysis, stating that the sum of all electrical potential differences around a loop is zero. For an L-R-C circuit, this law helps us set up the differential equation governing the circuit's behavior by considering the potential drops across the inductor (L), resistor (R), and capacitor (C): \[ L\frac{dI}{dt} + RI + \frac{Q}{C} = 0 \] This relationship accounts for the inductive reactance, resistive voltage drop, and capacitive reactance. By substituting known values for \( L \), \( R \), and \( C \) and simplifying, we obtain the specific second-order differential equation for the circuit.

Method of Undetermined Coefficients

The method of undetermined coefficients is used to find particular solutions to non-hom*ogeneous differential equations. When our L-R-C circuit has an external voltage \(E(t)\), the governing equation modifies to include this term, making it non-hom*ogeneous: \[ 3\frac{d^2Q}{dt^2} + 10\frac{dQ}{dt} + 10Q = e^{-t} \] To solve this, we first solve the hom*ogeneous part to find the complementary solution. Next, we assume a particular form for \(Q_p(t)\), such as \(Q_p(t) = Ae^{-t}\), and determine the coefficient \(A\) by substituting into the original equation. The general solution is then the sum of the complementary and particular solutions, providing a complete description of the circuit's response.

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Problem 1 Solve the one-loop \(L-R-C\) cir... [FREE SOLUTION] (3)

Most popular questions from this chapter

Solve the one-loop \(L-R-C\) circuit with \(L=\) \(1 \mathrm{H}, R=40 \Omega, C=1 /250 \mathrm{~F}\), and \(E(t)=\) \(120 \sin t \mathrm{~V} .(Q(0)=0 \mathrm{C},I(0)=0 \mathrm{~A}\). \()\)\(x^{\prime \prime}+10 x^{\prime}+50 x=0\)Suppose that a satellite is in flight on the line between a planet of mass\(M_{1}\) and its moon of mass \(M_{2}\), which are a constant distance \(R\) apart.The distance \(x\) between the satellite and the planet satisfies the nonlinearsecond order equation \(x^{\prime \prime}=-g M_{1} x^{-2}+\) \(g M_{2}(R-x)^{-2}\)where \(g\) is the gravitational constant. Transform this equation into a systemof first order equations. Find and classify the equilibrium point of thelinearized system.
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Problem 1 Solve the one-loop \(L-R-C\) cir... [FREE SOLUTION] (2024)

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