91Ó°ÊÓ

A circuit has in series an electromotive force given by \(E(t)=E_{0} \sin \omega t \mathrm{~V}\), a resistor of \(R \Omega\), an inductor of \(L H\), and a capacitor of \(C\) farads. (a) Show that the steady-state current is $$ i=\frac{E_{0}}{Z}\left(\frac{R}{Z} \sin \omega t-\frac{X}{Z} \cos \omega t\right) $$ where \(X=L \omega-1 / C \omega\) and \(Z=\sqrt{X^{2}+R^{2}}\). The quantity \(X\) is called the reactance of the circuit and \(Z\) is called the impedance. (b) Using the result of part (a) show that the steady-state current may be written $$ i=\frac{E_{0}}{Z} \sin (\omega t-\phi) $$ where \(\phi\) is determined by the equations $$ \cos \phi=\frac{R}{Z}, \quad \sin \phi=\frac{X}{Z} $$ Thus show that the steady-state current attains its maximum absolute value \(E_{0} / Z\) at times \(t_{n}+\phi / \omega\), where $$ t_{n}=\frac{1}{\omega}\left[\frac{(2 n-1) \pi}{2}\right] \quad(n=1,2,3, \ldots) $$ are the times at which the clectromotive force attains its maximum absolute value \(E_{\mathrm{o}}\) - (c) Show that the amplitude of the steady-state current is a maximum when $$ \omega=\frac{1}{\sqrt{L C}} $$ For this value of \(\omega\) electrical resonance is said to occur. (d) If \(R=20, L=\frac{1}{4}, C=10^{-4}\), and \(E_{0}=100\), find the value of \(\omega\) that gives rise to electrical resonance and determine the amplitude of the steady-state current in this case.

Step by step solution

01

Write the Kirchhoff's voltage law equation

For the given circuit, we can write the equation using Kirchhoff's voltage law as follows: \( E(t) - iR - L \frac{di}{dt} - \frac{1}{C} \int_{0}^{t} i dt = 0\) #Step 2: Solve for steady-state current#

02

Take the time derivative of the Kirchhoff's voltage law equation

Let's take the time derivative of the equation obtained in Step 1: \(\frac{dE(t)}{dt} - R \frac{di}{dt} - L \frac{d^2 i}{dt^2} + \frac{1}{C} i = 0\) Now substitute \(E(t) = E_0 \sin(\omega t)\), then \(\omega E_0 \cos(\omega t) - R \frac{di}{dt} - L \frac{d^2 i}{dt^2} + \frac{1}{C} i = 0\) #Step 3: Solve the differential equation#

03

Solve the differential equation for steady-state current

We are looking for a solution in the form of: \(i = A \sin(\omega t) + B \cos(\omega t)\) Differentiating i once with respect to t, we get \(\frac{di}{dt} = \omega A\cos(\omega t) - \omega B \sin(\omega t)\) , and differentiating it twice, we obtain \(\frac{d^2 i}{dt^2} = -\omega^2 A \sin(\omega t) - \omega^2 B \cos(\omega t)\) . Substitute these expressions into the differential equation from Step 2, and we get: \(\omega E_0 \cos(\omega t) - R(\omega A \cos(\omega t) - \omega B \sin(\omega t)) - L(-\omega^2 A \sin(\omega t) - \omega^2 B \cos(\omega t)) + \frac{1}{C}(A \sin(\omega t) + B \cos(\omega t)) = 0\) #Step 4: Find coefficients A and B#

04

Find coefficients A and B using trigonometric identities

Using trigonometric identities, we can rewrite the equation as: \((-\omega^2 L A + \frac{R}{C} A + \frac{1}{C} B) \sin(\omega t) + (\frac{1}{C} A - \omega^2 L B + \frac{R}{C} B) \cos(\omega t) = \omega E_0 \cos(\omega t)\) To satisfy the equation, coefficients of the sine and cosine terms on both sides must be equal. Then, \( -\omega^2 L A + \frac{R}{C} A + \frac{1}{C} B = 0 \) and \(\frac{1}{C} A - \omega^2 L B + \frac{R}{C} B = \omega E_0 \). Solving this system of linear equations, we get: \(A = \frac{E_0 R}{R^2 + (L\omega - \frac{1}{C\omega})^2}\) and \(B = -\frac{E_0 (L\omega - \frac{1}{C\omega})}{R^2 + (L\omega - \frac{1}{C\omega})^2}\) #Step 5: Write steady-state current expression#

05

Combine coefficients A and B with the current expression

Now we can write the expression for the steady-state current: \(i = \frac{E_0}{Z}\left(\frac{R}{Z}\sin \omega t - \frac{X}{Z}\cos \omega t \right)\), where \(X = L\omega - \frac{1}{C\omega}\) and \(Z = \sqrt{X^2 + R^2}\). This is the expression required in part (a). #Step 6: Rewrite the steady-state current with phase difference#

06

Express the steady-state current with phase difference

Following the expression from step 5, we may rewrite the current equation as: \(i = \frac{E_0}{Z}(\cos\phi \sin\omega t - \sin\phi \cos\omega t)\), where \(\cos\phi = \frac{R}{Z}\) and \(\sin\phi = \frac{X}{Z}\). Using trigonometric identity, the current equation becomes: \(i = \frac{E_0}{Z}\sin(\omega t - \phi)\). #Step 7: Find times for maximum absolute value of current#

07

Find times for maximum current value by taking derivative with respect to time

When the sine term reaches the maximum value \(1\), we have \(\omega t - \phi = \frac{-\phi + (2n-1)\frac{\pi}{2}}{\omega}\), where n is an integer. Thus, the times at which the steady-state current attains its maximum absolute value are: \(t_n = \frac{(2n-1)\pi/2 - \phi}{\omega}\), for \(n = 1, 2, 3, ...\). #Step 8: Find the condition for electrical resonance#

08

Differentiate i with respect to omega and find the resonance condition

To find the electrical resonance condition, we need to maximize the amplitude of the steady-state current: \(\frac{\partial}{\partial\omega}\left(\frac{E_0}{Z}\right) = 0\) Differentiating the impedance Z with respect to \(\omega\) and setting it to zero, we get \(\omega = \frac{1}{\sqrt{LC}}\), which is the condition for electrical resonance. #Step 9: Calculate the resonance frequency and amplitude of steady-state current#

09

Substitute given values and find the resonance frequency and amplitude

For given values \(R = 20, L = \frac{1}{4}, C = 10^{-4},\) and \(E_0 = 100\), first, find the resonance frequency \(\omega\): \(\omega = \frac{1}{\sqrt{L C}} = \frac{1}{\sqrt{(\frac{1}{4})(10^{-4})}} = 20 \mathrm{~rad/s}\). Now find the impedance Z at resonance: \(Z = \sqrt{X^2 + R^2} = \sqrt{(\frac{1}{\omega C} - L\omega)^2 + R^2} = \sqrt{(\frac{1}{20(10^{-4})} - \frac{1}{4}(20))^2 + (20)^2} = 20 \Omega\). Finally, find the amplitude of the steady-state current at resonance: \(I_0 = \frac{E_0}{Z} = \frac{100}{20} = 5 \mathrm{~A}\).

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Key Concepts

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

Electrical Resonance

Electrical resonance occurs in a circuit when the inductive reactance and capacitive reactance are equal in magnitude but opposite in phase, causing them to cancel each other out. This leads to a condition where the impedance of the circuit is minimized and is equal to the resistance. At resonance, the circuit can easily transfer energy between the inductor and the capacitor without much loss, resulting in a maximized current amplitude. The frequency at which this phenomenon happens is known as the resonant frequency, given by the formula \( \omega = \frac{1}{\sqrt{LC}} \) for a simple LC circuit.

In the context of the exercise provided, the circuit reaches electrical resonance when \( \omega \) is set to \( \frac{1}{\sqrt{LC}} \) which is achieved through solving the provided differential equations. This principle is particularly critical in radio and telecommunications, where it is used for tuning circuits to select desired frequencies and reject others.

Impedance and Reactance in Circuits

In an electrical circuit, impedance and reactance play pivotal roles. Impedance (Z) is the total opposition a circuit offers to the flow of alternating current (AC), and it comprises both resistance (R) and reactance (X). Reactance itself can be inductive (XL) or capacitive (XC), with the former increasing with frequency and the latter decreasing.

Inductive reactance is given by \( XL = L\omega \) while capacitive reactance is \( XC = \frac{1}{C\omega} \) where \( L \) is inductance, \( C \) is capacitance, and \( \omega \) is the angular frequency of the AC source. The net reactance (X) is the difference between inductive and capacitive reactances, shown as \( X = L\omega - \frac{1}{C\omega} \) in the provided exercise. Impedance is then calculated using \( Z = \sqrt{X^2 + R^2} \), a vital step in determining the behavior of the circuit under various conditions.

Solving Differential Equations

Solving differential equations is a fundamental part of understanding dynamic systems such as electrical circuits. These equations describe how physical quantities change with respect to one another and over time. In the case of an electrical circuit with AC sources, the current and voltage are often described by differential equations that require solving to understand the circuit's behavior.

The exercise presented involves solving a second-order linear differential equation with constant coefficients. This method includes proposing a particular solution form that contains sines and cosines, and then finding the coefficients that satisfy the equation, a crucial step in predicting the circuit response. The understanding and application of solving such differential equations allow for the prediction of current and voltage at any given time, providing essential insights into the functioning of the circuit.

Kirchhoff's Voltage Law

Kirchhoff's Voltage Law (KVL) is a principle in electrical engineering stating that the sum of the electrical potential differences (voltage) around any closed network is zero. This law is based on the conservation of energy where the total amount of energy gained in a circuit must equal the amount of energy lost.

In application, KVL is used for writing equations that represent the voltage sources and drops across various components in a circuit. For AC circuits, it's important to consider reactances in addition to resistances. The provided exercise begins by applying KVL to write an equation that equates the source electromotive force to the sum of voltage drops across the resistor, inductor, and capacitor. This first step is critical for setting up the differential equation that ultimately helps in determining the circuit’s behavior over time.