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(I) A \(10.0 - \mathrm { k } \Omega\) resistor is in series with a \(26.0 - \mathrm { mH }\) inductor and an ac source. Calculate the impedance of the circuit if the source frequency is \(( a ) 55.0 \mathrm { Hz } ; ( b ) 55,000 \mathrm { Hz }\) .

Short Answer

Expert verified
The impedance is approximately 10.0 kΩ for 55 Hz and 13,448 Ω for 55,000 Hz.

Step by step solution

01

Understand Circuit Components

The given circuit consists of a resistor with resistance \( R = 10.0 \text{ k}\Omega \) and an inductor with inductance \( L = 26.0 \text{ mH} \). These are connected in series in an AC circuit.
02

Calculate Reactance of Inductor

For an inductor, the reactance \( X_L \) can be calculated using the formula \( X_L = 2 \pi f L \), where \( f \) is the frequency.
03

Calculate Inductive Reactance for (a) 55.0 Hz

Using the formula from Step 2, calculate \( X_L \) for \( f = 55.0 \) Hz: \[ X_L = 2 \pi \times 55.0 \times 26.0 \times 10^{-3} = 8.987 \text{ } \Omega \]
04

Calculate Total Impedance for (a) 55.0 Hz

The impedance \( Z \) of a series resistor-inductor circuit is given by: \[ Z = \sqrt{R^2 + X_L^2} \]. Using \( R = 10,000 \) \( \Omega \) and \( X_L = 8.987 \) \( \Omega \), we calculate: \[ Z = \sqrt{(10,000)^2 + (8.987)^2} = 10,000 \text{ } \Omega \approx 10.0 \text{ } \text{k}\Omega \]
05

Calculate Inductive Reactance for (b) 55,000 Hz

Calculate \( X_L \) for \( f = 55,000 \) Hz: \[ X_L = 2 \pi \times 55,000 \times 26.0 \times 10^{-3} = 8,987 \text{ } \Omega \]
06

Calculate Total Impedance for (b) 55,000 Hz

The impedance \( Z \) for the case of \( 55,000 \) Hz is given by: \[ Z = \sqrt{(10,000)^2 + (8,987)^2} = \sqrt{100,000,000 + 80,667,769} = \sqrt{180,667,769} \approx 13,448 \text{ } \Omega \]

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Key Concepts

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

Impedance Calculation
Impedance is a key concept when dealing with AC circuits. It is the total resistance that a circuit presents to the flow of alternating current (AC). In an AC circuit, impedance is not just about resistors as in DC circuits, but it also includes reactance from inductors and capacitors.

Impedance is calculated using both the resistance (R) and the reactance (X) in the circuit. In mathematical terms, the impedance \( Z \) is given by the formula:
  • \( Z = \sqrt{R^2 + X_L^2} \)
The unit of impedance is the ohm (\( \Omega \)), and it is represented as a complex number because it includes both real and imaginary components.

Understanding how to calculate impedance in AC circuits is essential, as it affects how the circuit responds to different frequencies.
Reactance
Reactance is the component of impedance that accounts for the opposition a circuit element provides to AC flows due to inductance and capacitance. There are two types of reactances, inductive and capacitive, but we'll focus on inductive reactance since the original exercise involves an inductor.

Inductive reactance (\( X_L \)) measures how much an inductor opposes the change in current flowing through it and is determined by the frequency of the AC source and the inductance of the inductor:
  • \( X_L = 2 \pi f L \)
Where \( f \) is the frequency in hertz (Hz) and \( L \) is the inductance measured in henrys (H).

The higher the frequency or the inductance, the greater the reactance. This is why the reactance appears larger when the frequency increases, as shown in the exercise solution for 55 Hz compared to 55,000 Hz.
Inductor in AC Circuit
In an AC circuit, an inductor plays a crucial role in affecting current and voltage due to its property of inductance. It resists changes in current due to the magnetic field generated around the coil with current flow.

An inductor's behavior in an AC circuit can be summarized as:
  • It generates an opposing voltage when the current changes, known as back emf.
  • The reactance of the inductor increases with higher frequencies.
When an inductor is connected in series with a resistor in an AC circuit, it forms an RL-series circuit. The current in the RL circuit is determined by both the resistance and the inductive reactance as components of the impedance.

At low frequencies, the inductor has a smaller impact, behaving almost like a short for very low frequencies. However, at high frequencies, the inductive reactance increases significantly, altering the circuit's impedance greatly, as seen with the jump in calculated impedance between 55 Hz and 55,000 Hz in the example given.

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Most popular questions from this chapter

A \(3.5-\mathrm{k} \Omega\) resistor in series with a \(440-\mathrm{mH}\) inductor is driven by an ac power supply. At what frequency is the impedance double that of the impedance at \(60 \mathrm{~Hz} ?\)

Ignoring any mutual inductance, what is the equivalent inductance of two inductors connected \((a)\) in series, \((b)\) in parallel?

An inductor \(L\) in series with a resistor \(R,\) driven by a sinusoidal voltage source, responds as described by the following differential equation: $$V_{0} \sin \omega t=L \frac{d I}{d t}+R I$$ Show that a current of the form \(I=I_{0} \sin (\omega t-\phi)\) flows through the circuit by direct substitution into the differential equation. Determine the amplitude of the current \(\left(I_{0}\right)\) and the phase difference \(\phi\) between the current and the voltage source.

At \(t = 0 ,\) the current through a 60.0 -mH inductor is 50.0\(\mathrm { mA }\) and is increasing at the rate of 78.0\(\mathrm { mA } / \mathrm { s } .\) What is the initial energy stored in the inductor, and how long does it take for the energy to increase by a factor of 5.0 from the initial value?

(II) In some experiments, short distances are measured by using capacitance. Consider forming an \(L C\) circuit using a parallel-plate capacitor with plate area \(A ,\) and a known inductance \(L . ( a )\) If charge is found to oscillate in this circuit at frequency \(f = \omega / 2 \pi\) when the capacitor plates are separated by distance \(x ,\) show that \(x = 4 \pi ^ { 2 } A \epsilon _ { 0 } f ^ { 2 } L\) . (b) When the plate separation is changed by \(\Delta x ,\) the circuit's oscillation frequency will change by \(\Delta f .\) Show that \(\Delta x / x \approx 2 ( \Delta f / f ) . ( c )\) If \(f\) is on the order of 1\(\mathrm { MHz }\) and can be measured to a precision of \(\Delta f = 1 \mathrm { Hz } ,\) with what percent accuracy can \(x\) be determined? Assume fringing effects at the capacitor's edges can be neglected.

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