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Chapter 27: Problem 24

The current in a series \(R L\) circuit increases to \(20 \%\) of its final value in \(3.1 \mu \mathrm{s} .\) If \(L=1.8 \mathrm{mH},\) what's the resistance?

Short Answer

Expert verified
The resistance in the circuit is approximately \(129.5 \, \Omega\).

Step by step solution

01

Understanding the Given Information

We have the following data: Current reaches 20% of its final value in \(3.1 \mu \mathrm{s}\), Inductance \(L=1.8 \mathrm{mH} = 1.8 \times 10^{-3} H\). We also know that the time needed for the current to reach 63.2% of its final value in a RL circuit equals one time constant \(\tau\). So, the given time duration doesn't represent one full time constant, instead, it is the time taken for the current to reach 20% of its final value.
02

Calculate Equivalent Time Constant

We know that the time constant \(\tau\) is the time required for the current to reach 63.2% of its final value. From a standard exponential growth chart (or mathematical property of exponentials), we know that 20% of the final value is reached at approximately 0.2231 of the time constant. So, we can say, \(0.2231 \tau = 3.1 \mu s \Rightarrow \tau = \frac{3.1 \mu s}{0.2231} = 13.9 \mu s\)
03

Calculate Resistance

Knowing the time constant, we can now calculate the resistance. The formula for the time constant for a RL circuit is \(\tau = \frac{L}{R}\). Rearranging this to solve for resistance gives \(R = \frac{L}{\tau} = \frac{1.8 \times 10^{-3} H}{13.9 \times 10^{-6} s} = 129.5 \, \Omega\).

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Key Concepts

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

Understanding Inductance
In any RL circuit, inductance is a crucial element. Inductance, represented by the letter "L," is a property of the coil or inductor that opposes changes in current flow.
It is measured in henries (H) and can be visualized as the coil's ability to store energy in its magnetic field. When the current through the coil changes, the inductance causes a delay, inhibiting rapid fluctuations.
This property is particularly beneficial in preventing excessive currents or shaping the waveform of AC signals. Inductors are commonly used in electrical circuits to:
  • Control the rise and fall rates of the current.
  • Eliminate noise or sudden surges of power.
  • Provide reactive power in alternating current (AC) applications.
Deciphering the Time Constant
The time constant, often symbolized as \(\tau\), plays a pivotal role in understanding RL circuits.
It is essentially the time it takes for the current to rise to approximately 63.2% of its maximum value in a series RL circuit.This value comes from the nature of exponential growth and decay processes.
For any RL circuit, the time constant can be determined using the formula:\[\tau = \frac{L}{R}\]where \(L\) is the inductance and \(R\) is the resistance.Understanding the time constant helps in:
  • Predicting how long it will take for circuits to respond to changes in voltage.
  • Identifying the damping effects of inductance on a circuit.
  • Designing circuits with precise timing control.
It is important to recognize that a shorter time constant means the circuit responds quickly to changes, while a longer time constant implies a slower response.
Resistance Calculation in RL Circuits
Calculating resistance in an RL circuit involves applying the relationship between the inductance, resistance, and time constant.
When you know the time constant \(\tau\), it provides a direct insight into the circuit's behavior and helps in determining resistance.The essential formula utilized here is:\[R = \frac{L}{\tau}\]In the context of this exercise, we utilized the computed time constant and the given inductance to find resistance.This calculation helps in:
  • Determining how much opposition the circuit offers to the flow of current.
  • Designing circuits that need precise current flow regulations.
  • Balancing the energy storage and dissipation dynamics within the circuit.
Having a clear understanding of resistance in RL circuits aids in optimizing circuit design for both efficiency and functionality.

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Most popular questions from this chapter

A magnetic field is given by \(\vec{B}=B_{0}\left(x / x_{0}\right)^{2} \hat{k},\) where \(B_{0}\) and \(x_{0}\) are constants. Find an expression for the magnetic flux through a square of side \(2 x_{0}\) that lies in the \(x\) -y plane with one corner at the origin and sides coinciding with the positive \(x\) - and \(y\) -axes.

During lab, you're given a circular wire loop of resistance \(R\) and radius \(a\) with its plane perpendicular to a uniform magnetic field. You're supposed to increase the field strength from \(B_{1}\) to \(B_{2}\) and measure the total charge that moves around the loop. Your lab partner claims that the details of how you vary the field will make a difference in the total charge; your hunch is that it won't. By integrating the loop current over time, determine who's right.

A conducting loop with area \(0.15 \mathrm{m}^{2}\) and resistance \(6.0 \Omega\) lies in the \(x-y\) plane. A spatially uniform magnetic field points in the z-direction. The field varies with time according to \(B_{z}=a t^{2}-b\) where \(a=2.0 \mathrm{T} / \mathrm{s}^{2}\) and \(b=8.0 \mathrm{T} .\) Find the loop current (a) at \(t=3.0 \mathrm{s}\) and \((\mathrm{b})\) when \(B_{z}=0.\)

You have a fixed length of wire to wind into an inductor. Will you get more inductance if you wind a short coil with large diameter, or a long coil with small diameter?

The induced electric field \(12 \mathrm{cm}\) from the axis of a 10 -cm-radius solenoid is \(45 \mathrm{V} / \mathrm{m} .\) Find the rate of change of the solenoid's magnetic field.
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