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

Short Answer

Expert verified
(a) Approximate capacitive reactance is 1736.11 Ω. (b) Voltage across the resistor is approximately 1.10 cos(120t) V.

Step by step solution

01

Identify Given Values for Capacitive Reactance

We are given the capacitance, \( C = 4.80 \, \mu\text{F} = 4.80 \times 10^{-6} \, \text{F} \), and the angular frequency of the voltage, \( \omega = 120 \, \text{rad/s} \). We need these values to calculate the capacitive reactance.
02

Calculate Capacitive Reactance

The formula for capacitive reactance is \( X_C = \frac{1}{\omega C} \). Substituting the known values, we get:\[X_C = \frac{1}{120 \times 4.80 \times 10^{-6}} = \frac{1}{5.76 \times 10^{-4}} \approx 1736.11 \, \Omega\]Thus, the capacitive reactance is approximately \(1736.11 \, \Omega\).
03

Write Expression for Current

The current through the series circuit can be found by differentiating the voltage across the capacitor:\( v_C = (7.60 \, \text{V}) \sin (120 \, \text{rad/s} \, t) \).The current \( i(t) \) is:\[ i(t) = C \frac{d}{dt} v_C = C \left( 7.60 \times 120 \right) \cos(120t) \= 4.80 \times 10^{-6} \times 912 \cos(120t) \\approx 4.80 \times 10^{-6} \time 912 \cos(120t) = 4.38 \times 10^{-3} \cos(120t) \, \text{A}\]
04

Calculate Voltage Across Resistor

Ohm’s Law for the resistor gives \( v_R(t) = i(t) \times R \), where \( R = 250 \, \Omega \). Substituting the expression for the current:\[v_R(t) = 4.38 \times 10^{-3} \times 250 \cos(120t) \= 1.095 \cos(120t) \v_R(t) \approx 1.10 \cos(120t) \, \text{V}\]This results in the expression for the voltage across the resistor.

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Key Concepts

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

Angular Frequency
Angular frequency is a critical concept when working with alternating current (AC) circuits, such as those involving capacitors and resistors in series, like the one in the original exercise. It is a measure of how quickly an AC voltage waveform oscillates in radians per second. The formula for angular frequency is given by \( \omega = 2\pi f \), where
  • \( \omega \) represents the angular frequency in radians per second.
  • \( f \) is the frequency of the waveform in hertz (Hz).
For our problem, we were provided directly with \( \omega = 120 \, \text{rad/s} \). Angular frequency helps to determine how the capacitive reactance, \( X_C \), can vary with the frequency of the applied voltage.
You can think of it as a rotational equivalent of the regular frequency, crucial for deriving other AC circuit parameters.
Series Circuit
A series circuit is one where electrical components are connected end-to-end, so the same current flows through each component in succession. In the exercise, the resistor and capacitor are connected in series:
  • The current flowing through each element is the same.
  • The total voltage across the series circuit is the sum of voltages across each element.
  • This circuit is subject to Ohm's Law, which we'll discuss in detail.
For a series RC circuit like this one, while the current remains constant, the capacitive reactance dictates how the voltage is shared between the capacitor and resistor.
This setup is particularly interesting because it combines resistive effects with reactive effects from capacitors, allowing the analysis of phase shifts between voltage and current.
Ohm's Law
Ohm's Law is foundational in electrical engineering, particularly in analyzing circuits. It states that the voltage (\( V \)) across a resistor is directly proportional to the current (\( I \)) flowing through it, expressed with:\[ V = I \times R \]where
  • \( V \) is the voltage across the resistor in volts (V).
  • \( I \) is the current in amperes (A).
  • \( R \) is the resistance in ohms (\( \Omega \)).
In the step-by-step solution, Ohm's Law was used to find the voltage across the resistor by multiplying the current by the resistance: \( v_R(t) = 1.10 \cos(120t) \, \text{V} \). This principle helps analyze how AC signals interact with resistors, establishing the basis for calculating voltage across other components like our capacitor in a circuit.
Differential Calculus
Differential calculus is an essential mathematical tool used to find rates of change, and it's crucial in analyzing alternating signals in physics and engineering. In the context of a capacitive circuit, differential calculus helps us determine the behavior of current when we know how voltage changes over time.
In the exercise, we differentiated the voltage equation to find the expression for current: \( i(t) = C \frac{d}{dt} v_C \).
  • This represents the proportionality between current and the rate of change of voltage across a capacitor.
  • The derivative of the sinusoidal voltage function, \( v_C(t) = 7.60 \sin(120t) \), led to the cosine function expressing current: \( 4.38 \times 10^{-3} \cos(120t) \, \text{A} \).
Differential calculus thus gives us insight into the dynamic behaviors of circuits over time, crucial in circuits involving capacitors and alternating currents.

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Most popular questions from this chapter

31\. 46. A circuit consists of a resistor and a capacitor in series with an ac source that supplies an rms voltage of 240 \(\mathrm{V}\) . At the frequency of the source the reactance of the capacitor is 50.0\(\Omega\) . The rms current in the circuit is 3.00 A. What is the average power supplied by the source?

31.67. You want to double the resonance angular frequency of a series \(R-L-C\) circuit by changing only the pertinent circuit elements all by the same factor. (a) Which ones should you change? (b) By what factor should you change them?

31.73. In an \(L-R-C\) series circuit the current is given by \(i=I \cos \omega t .\) The voltage amplitudes for the resistor, inductor, and capacitor are \(V_{R}, V_{L},\) and \(V_{C}\) (a) Show that the instantaneous power into the resistor is \(p_{R}=V_{R} I \cos ^{2} \omega t=\frac{1}{2} V_{R} I(1+\cos 2 \omega t) .\) What does this expression give for the average power into the resistor? (b) Show that the instantaneous power into the inductor is \(p_{L}=\) \(-V_{L}\) Isin \(\omega t \cos \omega t=-\frac{1}{2} V_{L} I \sin 2 \omega t .\) What does this expression give for the average power into the inductor?(c) Show that the instantaneous power into the capacitor is \(p_{C}=V_{C}\) Isin\omegat cos \(\omega t=\) \(\frac{1}{2} V_{C}\) Isin \(2 \omega t .\) What does this expression give for the average power into the capacitor? (d) The instantancous power delivered by the source is shown in Section 31.4 to be \(p=V I \cos \omega t\) (cos \(\phi \cos \omega t-\) sin \(\phi \sin \omega t ) .\) Show that \(p_{R}+p_{L}+p_{C}\) equals \(p\) at each instant of time.

31.10. A Radio Inductor. You want the current amplitude through a \(0.450-\mathrm{mH}\) inductor (part of the circuitry for a radio receiver) to be 2.60 \(\mathrm{mA}\) when a sinusoidal voltage with amplitude 12.0 \(\mathrm{V}\) is applied across the inductor. What frequency is required?

31.35. A series circuit consists of an ac source of variable frequency, a \(115-\Omega\) resistor, a \(1.25-\mu F\) capacitor, and a \(4.50-\mathrm{mH}\) inductor. Find the impedance of this circuit when the angular frequency of the ac source is adjusted to (a) the resonance angular frequency; (b) twice the resonance angular frequency; (c) half the resonance angular frequency.

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