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Chapter 33: Problem 73

Sinusoidal Voltage , \(R=15.0 \Omega, C=4.70 \quad \mu \mathrm{F}\), and \(L=\) \(25.0 \mathrm{mH}\). The generator provides a sinusoidal voltage of \(75.0 \mathrm{~V}(\mathrm{rms})\) and frequency \(f=550 \mathrm{~Hz}\). (a) Calculate the rms current. (b) Find the rms voltages \(\Delta V_{a b}, \Delta V_{b c}\), \(\Delta V_{c d}, \Delta V_{b d}, \Delta V_{a d} \cdot\) (c) At what average rate is energy dissipated by each of the three circuit elements?

Short Answer

Expert verified
RMS current is 2.16 A. RMS voltages: \(\Delta V_{ab} = 32.4 \, V\), \(\Delta V_{bc} = 186.59 \, V\), \(\Delta V_{cd} = 133.17 \, V\), \(\Delta V_{bd} = 319.76 \, V\), \(\Delta V_{ad} = 35.09 \, V\). Power dissipated: 69.888 W (resistor only).

Step by step solution

01

- Calculate Reactance of Inductor and Capacitor

The inductive reactance, \(X_L\), is given by \(X_L = 2\pi f L\). For the capacitor, the capacitive reactance, \(X_C\), is given by \(X_C = 1/(2\pi f C)\).
02

- Calculations for Inductive and Capacitive Reactance

Calculate \(X_L\): \[X_L = 2\pi \times 550 \times 25 \times 10^{-3} = 86.3938\,\Omega.\]Calculate \(X_C\): \[X_C = \frac{1}{2\pi \times 550 \times 4.7 \times 10^{-6}} = 61.6566\, \Omega.\]
03

- Calculate Impedance

The impedance \(Z\) of an RLC series circuit is given by \[Z = \sqrt{R^2 + (X_L - X_C)^2}. \]Substitute the known values to find impedance: \[Z = \sqrt{15^2 + (86.3938 - 61.6566)^2} = 34.72\, \Omega.\]
04

- Calculate the RMS Current

The rms current \(I_{rms}\) is found by the equation: \[I_{rms} = \frac{V_{rms}}{Z}\].Given \(V_{rms} = 75.0\, V\), calculate \(I_{rms}\): \[I_{rms} = \frac{75.0}{34.72} = 2.16\, A.\]
05

- Find RMS Voltages across Each Element

Use the relations \((\Delta V_{a b} = I_{rms} R\), \((\Delta V_{b c} = I_{rms} X_L\), and \((\Delta V_{c d} = I_{rms} X_C\). Calculate \((\Delta V_{b d}\) and \((\Delta V_{a d}\) using Pythagoras theorem on impedance: \[\Delta V_{a b} = 2.16 \times 15.0 = 32.4\, V.\]\[\Delta V_{b c} = 2.16 \times 86.3938 = 186.59\, V.\]\[\Delta V_{c d} = 2.16 \times 61.6566 = 133.1743\, V.\]Using Kirchoff's Voltage Law, find \((\Delta V_{a d}, \Delta V_{b d}\) using impedance: \[\Delta V_{a d} = \sqrt{\Delta V_{a b}^2 + (\Delta V_{b c}-\Delta V_{c d})^2} = \sqrt{32.4^2 + (186.59-133.1743)^2} 35.09\, V.\]\[\Delta V_{b d} = \Delta V_{b c}+\Delta V_{c d}= 133.1743+186.59 \approx 319.7643 \]
06

- Find Average Energy Dissipated

The average power dissipated in each component:For resistor: \[P_R = I_{rms}^2 R = 2.16^2 \times 15 = 69.888\, W.\]For inductor and capacitor (reactive components do not dissipate real power): \[P_L = P_C = 0\, W.\]Power dissipation is only in the resistor.

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Key Concepts

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

sinusoidal voltage
A sinusoidal voltage is a type of alternating current (AC) where the voltage varies sinusoidally with time. In the given exercise, the generator provides a sinusoidal voltage of 75.0 V (rms) with a frequency of 550 Hz. Sinusoidal voltages are common in AC circuits, providing a smooth and consistent wave that makes calculations predictable.
inductive reactance
Inductive reactance (\(X_L\)) is the resistance offered by an inductor to the change in current. It depends on the frequency of the voltage and the inductance of the coil. The formula to calculate inductive reactance is:
\[X_L = 2\pi f L\]
For this problem, \(f = 550 \text{ Hz}\) and \(L = 25.0 \text{ mH}\). Plugging in the values, we get:
\[X_L = 2\pi \times 550 \times 25 \times 10^{-3} = 86.3938 \Omega\].
This means the inductor impedes the flow of current with a resistance of 86.3938 鈩�.
capacitive reactance
Capacitive reactance \((X_C)\) is the resistance offered by a capacitor to the change in voltage across its plates. It also depends on the frequency of the applied voltage and the capacitance. The formula for capacitive reactance is:
\[X_C = \frac{1}{2\pi f C}\]
For this exercise, \(f = 550 \text{ Hz} \text{ and } C = 4.7 \text{ 碌F}\). Substituting the values, we find:
\[X_C = \frac{1}{2\pi \times 550 \times 4.7 \times 10^{-6}} = 61.6566 \Omega\].
This results in the capacitor resisting the current flow with a reactance of 61.6566 鈩�.
impedance calculation
Impedance (\(Z)\) is the total resistance to the flow of current in an RLC circuit, combining resistive and reactive components. It can be calculated for a series RLC circuit using the formula:
\[Z = \sqrt{R^2 + (X_L - X_C)^2}\]
Given \(R = 15.0 \Omega, X_L = 86.3938 \Omega,\) and \(X_C = 61.6566 \Omega,\) we find:
\[Z = \sqrt{15^2 + (86.3938 - 61.6566)^2} = 34.72 \Omega\].
Impedance represents the total opposition to current flow.
RMS current
The RMS (root mean square) current is a measure of the effective value of the alternating current. It can be found using the formula:
\[I_{rms} = \frac{V_{rms}}{Z}\]
Given \(V_{rms} = 75.0 \text{ V}\) and \(Z = 34.72 \Omega,\) the RMS current is:
\[I_{rms} = \frac{75.0}{34.72} = 2.16 \text{ A}\].
This means that the effective current flowing through the circuit is 2.16 A.
RMS voltage
RMS voltage is the effective voltage of an AC source. In the context of this exercise, we need to determine the RMS voltages across individual components. Using the formula where RMS voltage across each component can be calculated by multiplying the RMS current with corresponding reactance or resistance:
\( \Delta V_{a b} = I_{rms} \times R = 2.16 \times 15 = 32.4 \text{ V}\)
\( \Delta V_{b c} = I_{rms} \times X_L = 2.16 \times 86.3938 = 186.59 \text{ V}\)
\( \Delta V_{c d} = I_{rms} \times X_C = 2.16 \times 61.6566 = 133.1743 \text{ V}\)
Additional voltages can be found using Kirchoff's law and impedance relationships.
energy dissipation
Energy dissipation in an RLC circuit primarily occurs across the resistor. Power dissipated by the resistor is found using:\[P_R = I_{rms}^2 R\]Given \(I_{rms}= 2.16 \text{ A}\) and \( R= 15 \text{ 惟}\), we get: \[P_R = (2.16)^2 \times 15 = 69.888 \text{ W}\].
Inductors and capacitors only store and release energy; they do not dissipate it as heat. Therefore, power dissipated by the inductor and capacitor is zero: \[P_L = P_C = 0 \text{ W}\].The total energy dissipation in the circuit is solely due to the resistor.

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Most popular questions from this chapter

Initially a Maximum In an oscillating \(L C\) circuit with \(L=50 \mathrm{mH}\) and \(C=4.0 \mu \mathrm{F}\), the current is initially a maximum. How long will it take before the capacitor is fully charged for the first time?

Oscillating \(L \boldsymbol{C}\) Circuit In an oscillating \(L C\) circuit, \(L=3.00 \mathrm{mH}\) and \(C=2.70 \mu \mathrm{F}\). At \(t=0 \mathrm{~s}\) the charge on the capacitor is zero and the current is \(2.00\) A. (a) What is the maximum charge that will appear on the capacitor? (b) In terms of the period \(T\) of oscillation, how much time will elapse after \(t=0\) until the energy stored in the capacitor will be increasing at its greatest rate? (c) What is this greatest rate at which energy is transferred to the capacitor?

Current as Function of Time In an oscillating \(L C\) circuit with \(C=64.0 \mu \mathrm{F}\), the current as a function of time is given by \(i=(1.60 \mathrm{~A})\) \(\sin \left[\begin{array}{llll}(2500 & \mathrm{rad} / \mathrm{s}) & t+0.680 & \mathrm{rad}]\end{array}\right.\) where \(t\) is in seconds. (a) How soon after \(t=0 \mathrm{~s}\) will the current reach its maximum value? What are (b) the inductance \(L\) and \((\mathrm{c})\) the total energy?

SHM A \(0.50 \mathrm{~kg}\) body oscillates in simple harmonic motion on a spring that, when extended \(2.0 \mathrm{~mm}\) from its equilibrium, has an \(8.0 \mathrm{~N}\) restoring force. (a) What is the angular frequency of oscillation? (b) What is the period of oscillation? (c) What is the capacitance of an \(L C\) circuit with the same period if \(L\) is chosen to be \(5.0 \mathrm{H}\) ?

Fraction of Energy Lost In an oscillating series \(R L C\) circuit, show that the fraction of the energy lost per cycle of oscillation, \(\Delta U / U\), is given to a close approximation by \(2 \pi R / \omega L\). The quantity \(\omega L / R\) is often called the \(Q\) of the circuit (for quality). A high- \(Q\) circuit has low resistance and a low fractional energy loss \((=2 \pi / Q)\) per cycle.
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