Question

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 electromotive force attains its maximum absolute value \(E_{0}\). (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.

Short answer

c) Show that the maximum amplitude of the steady-state current occurs when \[\omega=\frac{1}{\sqrt{LC}}\] #tag_title# Impedance minimum #tag_content# The amplitude of the steady-state current is given by \(E_0/Z\). To maximize the amplitude, we need to minimize the impedance Z. Differentiating Z with respect to \(\omega\), we get: \[\frac{dZ}{d\omega}=\frac{d}{d\omega}\left(\sqrt{\left(L\omega-\frac{1}{C\omega}\right)^2 + R^2}\right)\] Setting \(\frac{dZ}{d\omega} = 0\) to find the minimum, and solving for \(\omega\), we obtain that \( \omega = \frac{1}{\sqrt{LC}} \). This value of \(\omega\) corresponds to the electrical resonance. d) Calculate the value of \(\omega\) for electrical resonance and determine the amplitude of the steady-state current #tag_title# Given values and electrical resonance #tag_content# We are given that R = 20, L = 1/4, C = 10^{-4}, and E_0 = 100. Using these values, we can find the value of \(\omega\) that gives rise to electrical resonance: \[\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{\frac{1}{4}\times 10^{-4}}}\] \[\omega = 200\mathrm{~rad/s}\] #tag_title# Steady-state current amplitude #tag_content# Using this value of \(\omega\), we can now determine the impedance Z and the amplitude of the steady-state current: \[Z = \sqrt{X^2 + R^2} = \sqrt{\left(\frac{1}{2} \cdot 200 - \frac{1}{200 \cdot 10^{-4}}\right)^2 + 20^2} = \sqrt{400}\] \[Z = 20\Omega\] The amplitude of the steady-state current at this frequency will be: \[I = \frac{E_0}{Z} = \frac{100V}{20\Omega} = 5A\] So, for R = 20, L = 1/4, C = 10^{-4}, and E_0 = 100, the value of \(\omega\) giving rise to electrical resonance is 200 rad/s, and the amplitude of the steady-state current at this frequency is 5A.

Step-by-step solution

Kirchhoff's second law

Kirchhoff's second law states that the algebraic sum of the emf and potential differences in any closed loop equals zero. In this case, let us consider the loop formed by the source, resistor, inductor and capacitor. Applying the second law to this loop, we get: \(E(t) - iR - L\frac{di(t)}{dt} - \frac{1}{C}\int i(t) dt = 0\)

Angular frequency

We're given that the electromotive force is given by: \(E(t) = E_0 \sin \omega t\). The goal is to find the expression for the steady-state current, \(i(t)\). To this end, we first differentiate both sides of the equation above with respect to time, \(t\), because taking a derivative will eliminate the integral term. \[\frac{dE(t)}{dt} - R\frac{di(t)}{dt} - L\frac{d^2i(t)}{dt^2} = 0\].

Characteristic equation

Differentiating the EMF and substituting the given form, we get: \(\frac{dE(t)}{dt} = E_0 \omega \cos(\omega t)\). Now, we can substitute this into the previous equation and solve for \(i(t)\). The equation will become: \[E_0 \omega \cos(\omega t) - R\frac{di(t)}{dt} - L\frac{d^2i(t)}{dt^2} = 0\]. This is a second-order linear differential equation, and its general solution can be written in the form: \[i(t) = A\sin(\omega t) + B\cos(\omega t)\], where A and B are constants to be determined by initial conditions.

Impedance and Reactance

Now, in an RLC circuit, we define the reactance as \(X = L\omega - \frac{1}{C\omega}\) and impedance as \(Z = \sqrt{X^2 + R^2}\). Substituting the general solution for \(i(t)\) back into the first-order differential equation, we obtain: \[E_0 \sin(\omega t) = iR + L\frac{di(t)}{dt} = R\left(A\sin(\omega t) + B\cos(\omega t)\right) + \omega L\left(A\cos(\omega t) - B\sin(\omega t)\right)\] Comparing the coefficients of \(\sin(\omega t)\) and \(\cos(\omega t)\), we get two equations: \(A = \frac{E_0}{Z}\frac{R}{Z}\) and \(B = \frac{E_0}{Z}\frac{-X}{Z}]}\). The steady-state current in the circuit can now be expressed as: \[i(t)=\frac{E_{0}}{Z}\left(\frac{R}{Z} \sin \omega t-\frac{X}{Z} \cos \omega t\right)\] b) Express the steady-state current in an alternate form

Define phase angle

Let the phase angle \(\phi\) be determined by the equations: \[\cos \phi=\frac{R}{Z}\] and \[\sin \phi=\frac{X}{Z}\] Using the trigonometric identity, \(\sin(x+y) = \sin x \cos y + \cos x \sin y\), we can now rewrite the steady-state current as: \[i(t) = \frac{E_0}{Z} \sin(\omega t - \phi)\]

Key concepts

Kirchhoff's Second Law

In the context of RLC circuits, Kirchhoff's Second Law provides a fundamental rule: the sum of the electromotive force (emf) and the electrical potential differences around any closed loop is zero. This principle helps describe how voltage distributes across circuit elements like resistors, inductors, and capacitors. In our exercise, the loop involves:

• an electromotive force, \( E(t) \),
• a resistor, with voltage drop \( iR \),
• an inductor, modeled by \( L\frac{di(t)}{dt} \),
• and a capacitor, represented by \( \frac{1}{C}\int i(t) dt \).

Applying Kirchhoff's Second Law results in the equation:\[ E(t) = iR + L\frac{di(t)}{dt} + \frac{1}{C}\int i(t) dt \].This effectively underpins our understanding of how different elements impact the current and voltage relations in the circuit.

Steady-State Current

Steady-state current is the current that remains constant over time in a circuit despite transient components dying out. In our RLC circuit, after initial fluctuations, the current oscillates in a steady pattern dictated by the input voltage frequency. The formula for steady-state current is given by:\[ i(t) = \frac{E_{0}}{Z}\left(\frac{R}{Z} \sin \omega t - \frac{X}{Z} \cos \omega t\right) \]where:

• \( E_{0} \) is the amplitude of the emf,
• \( Z \) is the impedance of the circuit,
• \( X \) is the reactance.

Understanding steady-state current is key to analyzing RLC circuits, especially when considering features like phase difference between voltage and current, as it describes the sustained behavior of the system.

Electrical Resonance

Electrical resonance occurs in an RLC circuit when the inductive reactance equals the capacitive reactance, resulting in maximum amplitude of the steady-state current. This condition is crucial because it minimizes the total effective impedance in the circuit. At resonance, the angular frequency \( \omega \) satisfies the condition:\[ \omega = \frac{1}{\sqrt{LC}} \]where:

• \( L \) is the inductance,
• and \( C \) is the capacitance.

Reaching electrical resonance makes circuits highly efficient, as it allows for maximum energy transfer. Real-world applications exploit this principle in devices such as radios and TVs, where precise frequency tuning is necessary.

Impedance and Reactance

Impedance and reactance are fundamental concepts in AC circuits that affect how circuits respond to alternating currents. **Impedance (Z)**: This is the overall opposition a circuit offers to the flow of alternating current. It combines both resistance and reactance in a circuit, and is defined as:\[ Z = \sqrt{X^2 + R^2} \]where:

• \( R \) is the resistance,
• and \( X \) is the reactance.

The impedance sums up all forms of opposition in the circuit, adapting the resistance value to accommodate the reactive elements.**Reactance (X)**:This refers to the opposition to change in current caused by the presence of capacitors and inductors. It can be expressed as:\[ X = L\omega - \frac{1}{C\omega} \]where:

• \( L \omega \) is the inductive reactance,
• and \( \frac{1}{C\omega} \) is the capacitive reactance.

Reactance impacts the phase difference between the voltage and the current in the circuit, shifting energy between the electric and magnetic fields, rather than dissipating it as heat.