/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 88 A \(1.15-k \Omega\) resistor and... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(1.15-k \Omega\) resistor and a \(505-\mathrm{mH}\) inductor are connected in series to a \(1250-\mathrm{Hz}\) generator with an rms voltage of \(14.2 \mathrm{V}\). (a) What is the rms current in the circuit? (b) What capacitance must be inserted in series with the resistor and inductor to reduce the rms current to half the value found in part (a)?

Short Answer

Expert verified
(a) 3.44 mA, (b) 0.127 µF.

Step by step solution

01

Calculate Inductive Reactance

The inductive reactance \( X_L \) is calculated using the formula \( X_L = 2 \pi f L \), where \( f \) is the frequency and \( L \) is the inductance. Given \( f = 1250 \) Hz and \( L = 505 \) mH, we convert \( L \) to henrys by dividing by 1000, resulting in 0.505 H. Then we calculate \( X_L = 2 \pi \times 1250 \times 0.505 = 3974.43 \) ohms.
02

Calculate Total Impedance

The total impedance \( Z \) for this RL circuit is given by \( Z = \sqrt{R^2 + X_L^2} \), where \( R \) is the resistance. Substituting \( R = 1150 \) ohms and \( X_L = 3974.43 \) ohms, we find \( Z = \sqrt{1150^2 + 3974.43^2} = 4123.77 \) ohms.
03

Calculate RMS Current

The rms current \( I_{rms} \) is calculated using Ohm's law: \( I_{rms} = \frac{V_{rms}}{Z} \), where \( V_{rms} = 14.2 \) V. Therefore, \( I_{rms} = \frac{14.2}{4123.77} = 0.00344 \) A or 3.44 mA.
04

Calculate Desired RMS Current

The desired rms current is half of what we found in Step 3. Thus, \( I_{rms, desired} = \frac{3.44}{2} = 1.72 \) mA.
05

Calculate New Impedance

To find the capacitance, we first calculate the new impedance needed. Using \( I_{rms, desired} = \frac{14.2}{Z_{new}} \), we find \( Z_{new} = \frac{14.2}{0.00172} = 8254.65 \) ohms.
06

Calculate Capacitive Reactance

The capacitive reactance \( X_C \) is needed such that it increases the impedance to \( Z_{new} \). The relationship in an RLC circuit is \( Z_{new} = \sqrt{R^2 + (X_L - X_C)^2} \). Rearrange for \( X_C \) and solve: \( X_C = X_L - \sqrt{Z_{new}^2 - R^2} \). Substitute values to get \( X_C = 3974.43 - \sqrt{8254.65^2 - 1150^2} = 1007.25 \) ohms.
07

Calculate Capacitance

Finally, calculate the capacitance \( C \) using \( X_C = \frac{1}{2 \pi f C} \). Solve for \( C \): \( C = \frac{1}{2 \pi \times 1250 \times 1007.25} = 0.127 \) \( \mu F \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Inductive Reactance
Inductive reactance is the property of an inductor that opposes changes in the current flowing through a circuit. It is denoted by the symbol \(X_L\) and is measured in ohms. The inductive reactance can be calculated using the formula:
  • \(X_L = 2 \pi f L \)
where \(f\) is the frequency in hertz and \(L\) is the inductance in henrys.
Inductive reactance increases with frequency, meaning that higher frequencies result in greater opposition to current change. In our original exercise, we calculated the inductive reactance of a 505 mH inductor at 1250 Hz. After converting the inductance to henrys by dividing by 1000, we used the formula to find that \(X_L\) equaled 3974.43 ohms.
This information is essential when analyzing RLC circuits, as it helps determine how the circuit behaves at different frequencies.
Capacitive Reactance
Capacitive reactance is the tendency of a capacitor to resist the change in voltage across its plates. It's represented by \(X_C\) and is also measured in ohms. The calculation involves:
  • \(X_C = \frac{1}{2 \pi f C} \)
where \(f\) is the frequency and \(C\) is the capacitance.
Capacitive reactance, unlike inductive reactance, decreases with increasing frequency, making it an essential component in balancing RLC circuits. In the exercise, a capacitor was added to the circuit to alter the impedance, thereby changing the rms current.
The goal was to find a capacitance that made the current half of its original value, leading to the calculation of \(X_C = 1007.25\) ohms, subsequently determining the capacitance needed as 0.127 \(\mu F\). Understanding this helps modify circuits according to specific electrical requirements.
Ohm's Law
Ohm's Law is a fundamental principle in the study of electrical circuits, describing the relationship between voltage, current, and resistance. It is typically stated as:
  • \(V = I \, R\)
where \(V\) is the voltage, \(I\) is the current, and \(R\) is the resistance. In the context of AC circuits, we modify this equation for impedance (\(Z\)) rather than just resistance, using:
  • \(V_{rms} = I_{rms} \, Z\)
Ohm's Law allows for calculating the rms current in an AC circuit when the rms voltage and impedance are known. In the original exercise, given a voltage of 14.2 V and calculated impedance of 4123.77 ohms, we used Ohm’s Law to find the rms current to be 3.44 mA.
Understanding and applying Ohm's Law is critical for circuit analysis and design, ensuring that circuits function as intended under various loads.
Impedance
Impedance extends the concept of resistance to AC circuits, accounting for the frequency-dependent aspects of capacitors and inductors. It combines resistance (R), inductive reactance (\(X_L\)), and capacitive reactance (\(X_C\)) into a single quantity, providing a complete description of a circuit’s opposition to current flow. Impedance is represented by \(Z\) and calculated as:
  • \(Z = \sqrt{R^2 + (X_L - X_C)^2}\)
In RLC circuits, impedance can vary significantly with frequency changes, impacting the current and voltage characteristics. In the exercise, we used the impedance concept to determine the rms current and the necessary capacitance to adjust the circuit's performance.
The challenge was to manipulate the impedance by adjusting the capacitive reactance. Successfully doing this achieved the desired reduction in rms current, demonstrating the power of impedance analysis in electrical engineering and circuit design.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

-An ac generator with a frequency of \(30.0 \mathrm{Hz}\) and an \(\mathrm{rms}\) voltage of \(12.0 \mathrm{V}\) is connected to a \(45.5-\mu \mathrm{F}\) capacitor. (a) What is the maximum current in this circuit? (b) What is the current in the circuit when the voltage across the capacitor is \(5.25 \mathrm{V}\) and increasing? (c) What is the current in the circuit when the voltage across the capacitor is \(5.25 \mathrm{V}\) and decreasing?

CE BIO Persistence of Vision Although an incandescent lightbulb appears to shine with constant intensity, this is an artifact of the eye's persistence of vision. In fact, the intensity of a bulb's light rises and falls with time due to the alternating current used in household circuits. If you could perceive these oscillations, would you see the light attain maximum brightness 60 or 120 times per second? Explain.

A 3.33-k \(\Omega\) resistor is connected to a generator with a maximum voltage of \(141 \mathrm{V}\). Find \((\mathrm{a})\) the average and (b) the maximum power delivered to this circuit.

IP Tuning a Radio A radio tuning circuit contains an RLC circuit with \(R=5.0 \Omega\) and \(L=2.8 \mu \mathrm{H.}\) (a) What capacitance is needed to produce a resonance frequency of \(85 \mathrm{MHz}\) ? (b) If the capacitance is increased above the value found in part (a), will the impedance increase, decrease, or stay the same? Explain. (c) Find the impedance of the circuit at reso- nance. (d) Find the impedance of the circuit when the capacitance is \(1 \%\) higher than the value found in part (a).

IP (a) Determine the power factor for an \(R C\) circuit with \(R=4.0 \mathrm{k} \Omega\) and \(C=0.35 \mu \mathrm{F}\) that is connected to an ac generator with an rms voltage of \(24 \mathrm{V}\) and a frequency of \(150 \mathrm{Hz}\) (b) Will the power factor for this circuit increase, decrease, or stay the same if the frequency of the generator is increased? Explain.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.