Problem 12
Question
A \(1500 \Omega\) resistor is connected in series with a \(350 \mathrm{mH}\) inductor and an ac power supply. At what frequency will this combination have twice the impedance that it has at \(120 \mathrm{~Hz}\) ?
Step-by-Step Solution
Verified Answer
The new frequency at which the impedance is twice its value at 120 Hz is approximately 336 Hz.
1Step 1: Understanding Impedance in Series Circuits
In an R-L series circuit, impedance, denoted as \( Z \), is calculated using the formula \( Z = \sqrt{R^2 + (X_L)^2} \), where \( R \) is the resistance and \( X_L \) is the inductive reactance given by \( X_L = 2 \pi f L \).
2Step 2: Calculate Initial Impedance at 120 Hz
Given \( R = 1500 \ \Omega \), \( L = 350 \ \text{mH} = 0.35 \ \text{H} \), and \( f = 120 \ \text{Hz} \), first calculate \( X_L = 2 \pi \times 120 \times 0.35 \). Thus, \( X_L \approx 263.89 \ \Omega \). The impedance is then \( Z = \sqrt{1500^2 + 263.89^2} \approx 1523.54 \ \Omega \).
3Step 3: Impedance at New Frequency
We seek a frequency \( f' \) such that the impedance is twice the initial impedance \( 2 \times 1523.54 \ \approx 3047.08 \ \Omega \). Using the impedance formula, set up the equation: \( \sqrt{1500^2 + (2\pi f' \times 0.35)^2} = 3047.08 \).
4Step 4: Solving for the New Frequency
Square both sides of the equation from Step 3 and rearrange to solve for \( f' \). This gives: \( 1500^2 + (0.7 \pi f')^2 = 3047.08^2 \). Solve to find \( (0.7 \pi f')^2 = 3047.08^2 - 1500^2 \), leading to \( f' \approx 335.89 \ \text{Hz} \) after appropriate calculations.
Key Concepts
Impedance in an R-L Series CircuitUnderstanding Inductive ReactanceThe Concept of Resonant FrequencyExploring the Resistor-Inductor Circuit
Impedance in an R-L Series Circuit
Impedance is a measure of how much a circuit resists the flow of alternating current (AC) at a particular frequency. In an R-L series circuit, which consists of a resistor and an inductor connected in a line, the total impedance, denoted as \( Z \), combines the effects of the resistor and inductor.
To understand impedance, consider the formula:
To understand impedance, consider the formula:
- \( Z = \sqrt{R^2 + (X_L)^2} \)
- Where \( R \) is resistance and \( X_L \) is inductive reactance.
Understanding Inductive Reactance
Inductive Reactance is a specific property of inductors in AC circuits that determines how much they resist the change of current. The inductive reactance, denoted as \( X_L \), depends directly on the frequency of the AC flowing through the circuit.
Here’s the formula:
Here’s the formula:
- \( X_L = 2 \pi f L \)
- Where \( f \) is frequency and \( L \) is inductance.
The Concept of Resonant Frequency
In circuits with both inductors and capacitors, the resonant frequency is a special frequency where the impedance is minimized, resulting in a condition called resonance. Although this specific problem deals with an R-L circuit, understanding resonant frequency helps in broader circuit design.
The resonant frequency \( f_r \) is mostly calculated in RLC circuits, but the principle extends to understanding how frequency affects impedance:
The resonant frequency \( f_r \) is mostly calculated in RLC circuits, but the principle extends to understanding how frequency affects impedance:
- Resonance occurs when inductive and capacitive reactances are equal and cancel each other out.
- This leads to maximum current flow at resonance if there is little resistance.
Exploring the Resistor-Inductor Circuit
A Resistor-Inductor (R-L) circuit is a basic electrical circuit consisting of two components: resistors and inductors. These circuits are fundamental in electronics, serving purposes like filtering signals or controlling circuit responses to AC.
In an R-L series circuit:
In an R-L series circuit:
- The resistor, with constant resistance \( R \), opposes electrical current flow.
- The inductor, with inductance \( L \), opposes changes in current flow.
Other exercises in this chapter
Problem 10
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