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A \(250 \Omega\) resistor is connected in series with a \(4.80 \mu \mathrm{F}\) capacitor and an ac source. The voltage across the capacitor is \(v_{C}=(7.60 \mathrm{~V}) \sin [(120 \mathrm{rad} / \mathrm{s}) t] .\) (a) Determine the capacitive reac- tance of the capacitor. (b) Derive an expression for the voltage \(v_{R}\) across the resistor.

Short Answer

Expert verified
The capacitive reactance is \(X_C = 1 / (2 \pi fC)\), where the frequency \(f = \omega / 2\pi \) with \(\omega = 120 rad/s\) and the capacitance \(C = 4.80 \mu F\). The expression for the voltage across the resistor is given by \(v_R = V_m sin(\omega t) X_C / R\).

Step by step solution

01

Determine the Capacitive Reactance

This can be done using the formula for capacitive reactance \(X_C = 1/(2\pi fC)\), where \(f\) is the frequency of the ac source and \(C\) is the capacitance. The frequency can be found from the given source voltage, which has a form that suggests a simple harmonic motion \(v_C = V_m*sin(\omega t)\). Equating \(v_C\) with the given expression, we can obtain the value of \(\omega = 120 rad/s\) and the frequency \(f = \omega / 2\pi \). Substituting the given capacitor value \(C = 4.80 \mu F\) and the calculated frequency into the equation for capacitive reactance gives \(X_C\).
02

Derive Voltage Across Resistor

This involves applying Ohm's law (\(v_R = I R\)) in combination with \(I = V / X_C\), where \(V\) is the voltage across the capacitor. Substituting \(I\) in the Ohm's law gives \(v_R = V R / X_C\), which can be rewritten as \(v_R = V XC / R\ = \(V_m sin(\omega t) X_C / R\) by substituting the expression of capacitive voltage. Finally, substituting the computed value of \(X_C\) in \(v_R\) gives the expression for voltage across the resistor.
03

Final Simplification

Simplify the expression of \(v_R\) to have a form similar to that of simple harmonic oscillation \(v(t) = V_m' sin(\omega t + \phi')\).

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Key Concepts

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

Understanding AC Circuits
Alternating current (AC) circuits are fundamental to the distribution and utilization of electrical energy. Unlike direct current (DC) circuits, where the current flows in a constant direction, AC circuits carry current that changes direction periodically. This current variation is sinusoidal, reflecting a simple harmonic motion typical of AC systems. The sine wave expression for voltage across a component in an AC circuit is given by v(t) = V_m * sin(Ó¬t), where V_m is the maximum voltage, Ó¬ is the angular frequency, and t is time.

When working with AC circuits, it's important to understand the concept of reactance, which is the opposition to current flow caused by capacitors and inductors. In the given exercise, a capacitor adds capacitive reactance to the circuit, which is calculated using the frequency of the AC source and the capacitance value. The lower the reactance, the easier it is for the alternating current to pass through the capacitor. This dynamic behavior greatly influences how we analyze and design such circuits for various applications, from power transmission to electronic devices.

Capacitive reactance also affects other parameters of the circuit, such as phase angle and impedance. These factors collectively determine the overall current and voltage distribution within the circuit. For students, understanding the interplay between these parameters reveals how AC circuits can be tuned for desired performance, whether it's for achieving resonance in radio frequencies or for ensuring efficient power supply in household appliances.
Applying Ohm's Law in AC Circuits
At the core of electrical circuit analysis lies Ohm's Law, an indispensable tool for students and engineers alike. In its simplest form, Ohm's Law states that the current (I) through a conductor between two points is directly proportional to the voltage (V) across the two points and inversely proportional to the resistance (R) of the conductor. This gives us the fundamental formula V = IR. However, AC circuits add a layer of complexity to this basic law.

In an AC circuit containing reactive components like capacitors or inductors, Ohm's Law must take into account the reactance. For capacitors, the capacitive reactance (X_C) replaces the resistance component in Ohm's Law. Therefore, when calculating the voltage across a resistor in an AC circuit, as in our exercise, the effective resistance becomes the reactance of the capacitor when there's no other resistance in the path of the current.

This modified form of Ohm's Law allows us to understand and predict the behavior of electrical components in response to AC signals. It's a powerful extension of the principle that aids in the analysis of more complex circuits involving both resistors and capacitors, broadening our grasp on electronic design and functionality in AC systems.
Simple Harmonic Motion in Electrical Circuits
The concept of simple harmonic motion (SHM) is not just confined to mechanical systems but is also prevalent in electrical circuits, particularly in AC circuits. SHM refers to a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the opposite direction. In the context of electrical circuits, SHM describes the sinusoidal waveforms of alternating current and voltage.

The relationship between SHM and AC circuits becomes clear when looking at the voltage across a capacitor, which varies sinusoidally with time as v_C(t) = V_m * sin(Ó¬t). This equation reflects the defining characteristics of SHM, with V_m representing the maximum voltage (akin to amplitude in mechanical SHM), and Ó¬ denoting the angular frequency of the oscillation.

When analyzing circuits, students should recognize that the sinusoidal behavior of voltage and current implicates SHM principles. These principles underlie the time-dependent variations in the circuit and are essential for understanding the phase relationships and resonance conditions in AC systems. Grasping the SHM aspects of electrical behavior paves the way to a deeper insight into wave phenomena, signal processing, and the oscillatory nature of electromagnetic fields in electronics and physics at large.

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Most popular questions from this chapter

\(\begin{array}{lllll} \text {The } & \text { L- } R \text { - } C & \text { Parallel } & \text { Circuit. A resistor, an induc- }\end{array}\) tor, and a capacitor are connected in parallel to an ac source with voltage amplitude \(V\) and angular frequency \(\omega .\) Let the source voltage be given by \(v=V \cos \omega t\). (a) Show that each of the instantaneous voltages \(v_{R}, v_{L},\) and \(v_{C}\) at any instant is equal to \(v\) and that \(i=i_{R}+i_{L}+i_{C},\) where \(i\) is the current through the source and \(i_{R}, i_{L}\) and \(i_{C}\) are the currents through the resistor, inductor, and capacitor, respectively. (b) What are the phases of \(i_{R}, i_{L},\) and \(i_{C}\) with respect to \(v ?\) Use current phasors to represent \(i, i_{R}, i_{L},\) and \(i_{C} .\) In a phasor diagram, show the phases of these four currents with respect to \(v\). (c) Use the phasor diagram of part (b) to show that the current amplitude \(I\) for the current \(i\) through the source is \(I=\sqrt{I_{R}^{2}+\left(I_{C}-I_{L}\right)^{2}}\) (d) Show that the result of part (c) can be written as \(I=V / Z,\) with \(1 / Z=\sqrt{\left(1 / R^{2}\right)+[\omega C-(1 / \omega L)]^{2}}\)

\(\mathrm{A}\) series ac circuit contains a \(250 \Omega\) resistor, a \(15 \mathrm{mH}\) inductor, a \(3.5 \mu \mathrm{F}\) capacitor, and an ac power source of voltage amplitude \(45 \mathrm{~V}\) operating at an angular frequency of \(360 \mathrm{rad} / \mathrm{s}\). (a) What is the power factor of this circuit? (b) Find the average power delivered to the entire circuit. (c) What is the average power delivered to the resistor, to the capacitor, and to the inductor?

You plan to take your hair dryer to Europe, where the electrical outlets put out \(240 \mathrm{~V}\) instead of the \(120 \mathrm{~V}\) seen in the United States. The dryer puts out \(1600 \mathrm{~W}\) at \(120 \mathrm{~V}\). (a) What could you do to operate your dryer via the \(240 \mathrm{~V}\) line in Europe? (b) What current will your dryer draw from a European outlet? (c) What resistance will your dryer appear to have when operated at \(240 \mathrm{~V}\) ?

You have a special light bulb with a very delicate wire filament. The wire will break if the current in it ever exceeds \(1.50 \mathrm{~A}\), even for an instant. What is the largest root-mean-square current you can run through this bulb?

A transformer connected to a \(120 \mathrm{~V}\) (rms) ac line is to supply \(13,000 \mathrm{~V}\) (ms) for a neon sign. To reduce shock hazard, a fuse is to be inserted in the primary circuit; the fuse is to blow when the rons current in the secondary circuit exceeds \(8.50 \mathrm{~mA}\). (a) What is the ratio of secondary to primary turns of the transformer? (b) What power must be supplied to the transformer when the \(\mathrm{rms}\) secondary current is \(8.50 \mathrm{~mA} ?\) (c) What current rating should the fuse in the primary circuit have?

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