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In a certain \(\mathrm{RLC}\) circuit, a \(20.0-\Omega\) resistor, a 10.0 - \(\mathrm{mH}\) inductor, and a \(5.00-\mu \mathrm{F}\) capacitor are connected in series with an AC power source for which \(V_{\mathrm{rms}}=10.0 \mathrm{~V}\) and \(f=100 . \mathrm{Hz} .\) Calculate a) the amplitude of the current, b) the phase between the current and the voltage, and c) the maximum voltage across each component.

Short Answer

Expert verified
Answer: The amplitude of the current is \(0.047 A\), the phase difference between the current and the voltage is \(-1.539\) radians, and the maximum voltage across the resistor, inductor, and capacitor are \(0.94 V\), \(0.295 V\), and \(14.96 V\), respectively.

Step by step solution

01

Find the circuit impedance

We need to find the impedance of the circuit, which is given by: $$Z = \sqrt{R^2 + (X_L - X_C)^2}$$ where `\(R\)' is the resistance, `\(X_L\)' is the inductive reactance, and `\(X_C\)' is the capacitive reactance. First, calculate the reactances using the following formulas: $$X_L = 2\pi fL$$ $$X_C = \frac{1}{2\pi fC}$$ where `\(f\)' is the frequency, `\(L\)' is the inductor value, and `\(C\)' is the capacitor value.
02

Calculate the reactances values

Now, we can plug in the given values to calculate the reactances: $$X_L = 2\pi (100 Hz)(0.010 H) = 6.283 Hz$$ $$X_C = \frac{1}{2\pi (100 Hz)(5\times10^{-6} F)} = 318.31 Hz$$
03

Calculate the circuit impedance

With the reactances calculated, find the impedance of the RLC circuit: $$Z = \sqrt{(20.0 \Omega)^2 + (6.283 \Omega - 318.31 \Omega)^2} = 299.22 \Omega$$
04

Find the current amplitude

Use Ohm's Law to calculate the amplitude of the current: $$I_0 = \frac{V_0}{Z} = \frac{\sqrt{2} V_{rms}}{Z} = \frac{\sqrt{2}(10.0 V)}{299.22 \Omega} = 0.047 A$$
05

Calculate the phase difference

The phase difference between the current and the voltage can be found using the relation: $$\tan \phi = \frac{X_L - X_C}{R}$$ $$\phi = \arctan \left(\frac{6.283 \Omega - 318.31 \Omega}{20.0 \Omega}\right) = -1.539 \, \mathrm{rad}$$
06

Calculate the maximum voltage across each component

Finally, determine the maximum voltage across each component using the following formulas: $$V_R = I_0R$$ $$V_L = I_0X_L$$ $$V_C = I_0X_C$$ Now, plug in the values: $$V_R = (0.047 A)(20.0 \Omega) = 0.94 V$$ $$V_L = (0.047 A)(6.283 \Omega) = 0.295 \, V$$ $$V_C = (0.047 A)(318.31 \Omega) = 14.96 V$$ So, the amplitude of the current is \(0.047 A\), the phase difference between the current and the voltage is \(-1.539\) radians, and the maximum voltage across the resistor, inductor, and capacitor are \(0.94 V\), \(0.295 V\), and \(14.96 V\), respectively.

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Key Concepts

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

Impedance
Impedance is a fundamental concept in analyzing RLC circuits. It measures the opposition that a circuit presents to the flow of alternating current (AC). Unlike resistance, which applies to direct current (DC) circuits and is purely real, impedance is a complex quantity that includes both resistance and reactance. It is represented as a vector sum in rectangular form or as a magnitude and phase angle in polar form. In this exercise, the impedance of the circuit is calculated using the formula:
\[Z = \sqrt{R^2 + (X_L - X_C)^2}\]
Here, \(R\) is the resistance, \(X_L\) is the inductive reactance, and \(X_C\) is the capacitive reactance. Impedance is important because it determines how the current and voltage relate in a circuit with multiple reactive and resistive elements. A higher impedance means the circuit resists the current flow more effectively.
  • Resistance (R) is purely real and opposes current uniformly.
  • Reactance (X) contributes to changes in phase and amplitude.
  • Impedance is key to determining the AC circuit's behavior.
Reactance
Reactance quantifies the opposition to AC that changes with frequency and is different for capacitors and inductors. It comes in two types:
  • Inductive Reactance \(X_L\): It increases with frequency and is given by the formula: \(X_L = 2\pi fL\), where \(L\) is inductance and \(f\) is frequency. In our exercise, \(X_L = 6.283\) Ohm.
  • Capacitive Reactance \(X_C\): It decreases with frequency and is calculated by \(X_C = \frac{1}{2\pi fC}\), where \(C\) is capacitance. For this circuit, \(X_C = 318.31\) Ohm.
Reactive components do not dissipate energy but store it temporarily, affecting the current and voltage phase angle.

Understanding reactance helps predict how an RLC circuit will react to changes in frequency and component values. It's a dynamic interplay between inductors and capacitors that modify the impedance of a circuit and influence its response.
Ohm's Law
Ohm's Law is a foundational principle of electrical circuits that relates voltage \((V)\), current \((I)\), and resistance \((R)\) through the formula:
\[V = IR\]
For AC circuits, the formula becomes \(V = IZ\), where \(Z\) is the impedance. This formula allows us to calculate current amplitude in RLC circuits by rearranging it to:
\[I_0 = \frac{V_0}{Z} = \frac{\sqrt{2} V_{rms}}{Z}\]
Here, \(V_{rms}\) is the root-mean-square voltage, and \(I_0\) is the peak current. In our specific example, Ohm's Law is used to find the current: \(I_0 = 0.047 A\). This relationship is crucial because it provides a simple way to understand how voltage and current interact in any circuit, predicting the circuit behavior effectively.
Phase Difference
Phase difference describes the timing difference between the peaks of the voltage and current waveforms in an AC circuit. This aspect is crucial in RLC circuit analysis. It indicates how the voltage and current are out of sync, affecting power delivery and efficiency.

The phase difference \(\phi\) is determined using:
\[\phi = \arctan \left(\frac{X_L - X_C}{R}\right)\]
In our case, \(\phi = -1.539 \) radians, meaning the current lags behind the voltage. This lag is typically due to the inductive elements since inductors delay current changes.

Phase difference is significant in AC power systems, where it affects power factor and can lead to less efficient energy transfer. Understanding it helps control and improve how energy flows through an electrical system.
  • Positive phase angle: Current leads voltage.
  • Negative phase angle: Current lags voltage.
  • Zero phase angle: Current and voltage peaks coincide.

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Most popular questions from this chapter

The quality factor, \(Q\), of a circuit can be defined by \(Q=\omega_{0}\left(U_{E}+U_{B}\right) / P .\) Express the quality factor of a series RLC circuit in terms of its resistance \(R\), inductance \(L\), and capacitance \(C .\)

An AC power source with \(V_{\mathrm{m}}=220 \mathrm{~V}\) and \(f=60.0 \mathrm{~Hz}\) is connected in a series RLC circuit. The resistance, \(R\), inductance, \(L\), and capacitance, \(C\), of this circuit are, respectively, \(50.0 \Omega, 0.200 \mathrm{H},\) and \(0.040 \mathrm{mF}\). Find each of the following quantities: a) the inductive reactance b) the capacitive reactance c) the impedance of the circuit d) the maximum current through the circuit e) the maximum potential difference across each circuit element

A label on a hair dryer reads "110V \(1250 \mathrm{~W}\)." What is the peak current in the hair dryer, assuming that it behaves like a resistor?

The figure shows a simple FM antenna circuit in which \(L=8.22 \mu \mathrm{H}\) and \(C\) is variable (the capacitor can be tuned to receive a specific station). The radio signal from your favorite FM station produces a sinusoidal time-varying emf with an amplitude of \(12.9 \mu \mathrm{V}\) and a frequency of \(88.7 \mathrm{MHz}\) in the antenna. a) To what value, \(C_{0}\), should you tune the capacitor in order to best receive this station? b) Another radio station's signal produces a sinusoidal time-varying emf with the same amplitude, \(12.9 \mu \mathrm{V}\), but with a frequency of \(88.5 \mathrm{MHz}\) in the antenna. With the circuit tuned to optimize reception at \(88.7 \mathrm{MHz}\), what should the value, \(R_{0}\), of the resistance be in order to reduce by a factor of 2 (compared to the current if the circuit were optimized for \(88.5 \mathrm{MHz}\) ) the current produced by the signal from this station?

a) A loop of wire \(5.00 \mathrm{~cm}\) in diameter is carrying a current of \(2.00 \mathrm{~A}\). What is the energy density of the magnetic field at its center? b) What current has to flow in a straight wire to produce the same energy density at a point \(4.00 \mathrm{~cm}\) from the wire?

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