/*! 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 42 A coil of inductance \(L=300 \ma... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A coil of inductance \(L=300 \mathrm{mH}\) and resistance \(R=140 \mathrm{~m} \Omega\) is connected to a constant voltage source. Current in the coil will reach to \(50 \%\) of its steady value after time \(t\) equals to: (a) \(155 \mathrm{~s}\) (b) \(0.755 \mathrm{~s}\) (c) \(0.155 \mathrm{~s}\) (d) \(1.48 \mathrm{~s}\)

Short Answer

Expert verified
The current reaches 50% at \(1.48 \, \text{s}\), corresponding to option (d).

Step by step solution

01

Identify the Formula

To find out how long it takes for the current to reach 50% of its steady state value, we use the RL circuit formula for current growth: \[ I(t) = I_0 (1 - e^{-t/\tau}) \]where \(I_0\) is the steady state current, and \(\tau = \frac{L}{R}\) is the time constant of the circuit.
02

Calculate the Time Constant

First, compute the time constant \(\tau\) using the given values of inductance \(L = 300 \, \text{mH} = 0.3 \, \text{H}\) and resistance \(R = 140 \, \text{m}\Omega = 0.14 \, \Omega\):\[ \tau = \frac{L}{R} = \frac{0.3}{0.14} \approx 2.14 \, \text{s} \]
03

Determine the Time for 50% Current

We know that the current reaches 50% of its steady state value when:\[ 0.5I_0 = I_0 (1 - e^{-t/\tau}) \]This simplifies to:\[ 0.5 = 1 - e^{-t/\tau} \]Solving for \(e^{-t/\tau}\):\[ e^{-t/\tau} = 0.5 \]
04

Solve the Exponential Equation

Take the natural logarithm of both sides:\[ -\frac{t}{\tau} = \ln(0.5) \]Solving for \(t\), we find:\[ t = -\tau \ln(0.5) \]Substitute the values obtained in previous steps:\[ t = -2.14 \ln(0.5) \approx 1.48 \, \text{s} \]
05

Select the Correct Answer

The time when the current in the coil reaches 50% of its steady value is approximately \(1.48 \, \text{s}\), so the correct answer is (d).

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.

Inductance
Inductance is a fundamental property of an electrical circuit that opposes changes in current flow. Represented by the symbol \( L \) and measured in henrys (H), it is the ability of a coil to store energy in a magnetic field when electrical current passes through it. A coil designed with more turns or made from a material with higher magnetic permeability will typically have greater inductance. In the context of RL circuits (which include resistors and inductors), inductance plays a crucial role in determining how quickly current in the circuit can rise to its maximum or steady state value. It essentially acts as a temporary energy absorber that slows the rate at which current changes.
Time Constant
The time constant \( \tau \) of an RL circuit is an important parameter that indicates how quickly the circuit responds to changes in voltage. It is calculated as the ratio of the inductance \( L \) to the resistance \( R \) in the circuit, \( \tau = \frac{L}{R} \). Measured in seconds, the time constant gives us an idea of how long it will take for the current to reach a substantial portion of its final value. Specifically, after a time period of one time constant, the current in the circuit typically reaches about 63% of its final steady state value. The greater the time constant, the slower the process of reaching steady state, and vice versa. This is critical for understanding how quickly an RL circuit can adapt to sudden changes in voltage.
Exponential Growth
In the context of RL circuits, current growth follows an exponential pattern. Once a voltage is applied, the current through the circuit rises gradually until it approaches its steady state value. This growth can be described by the equation \( I(t) = I_0 (1 - e^{-t/\tau}) \), where \( I_0 \) is the steady state current, and \( \tau \) is the time constant of the circuit. The term \( e^{-t/\tau} \) represents the exponential decay factor, which slows down as \( t \) becomes large. Exponential growth is characterized by a rapid increase at the beginning, which flattens out as it gets closer to the maximum value, illustrating how RL circuits gradually reach stability over time.
Steady State Current
The steady state current, denoted \( I_0 \), is the maximum current a circuit can reach under a constant voltage once transient effects have dissipated. It represents the balance point where the rate of current growth through the inductor stops changing, resulting from the voltage overcoming the combined resistance and inductance. In RL circuits, achieving this steady state can take several time constants due to the inductor's initial opposition to changes in current. Once the steady state is achieved, the current remains constant as long as the voltage source remains unchanged. This concept is essential in designing circuits, ensuring stability and efficiency in their operation.

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

A conducting ring of radius \(r\) is rolling without slipping with a constant angular velocity \(\omega\). If the magnetic field strength is \(B\) and is directed into the page then emf induced across \(P Q\) is : (a) \(B \omega r^{2}\) (b) \(\frac{B \omega r^{2}}{2}\) (c) \(4 B \omega r^{2}\) (d) \(\frac{\pi^{2} r^{2} B \omega}{8}\)

Two straight super-conducting rails form an angle \(\theta\) where their ends are joined a conducting bar having \(R_{0}\) resistance per unit length in contact with the rails and forming an isosceles triangle with them. The bar starts at the vertex at time \(t=0\) and moves with constant velocity \(v\) to right. A magnetic field \(B\) is present into the region (shown in figure). Find the force exerted by external agent to maintain constant velocity to the rod: (a) \(\frac{2 B^{2} v^{2} t}{R_{0}} \tan \frac{\theta}{2}\) (b) \(\frac{B^{2} v^{2} t}{R_{0}} \tan \frac{\theta}{2}\) (c) \(\frac{B^{2} v^{2} t}{R_{0}}\) (d) none of these

An Indian ship with a vertical conducting mass navigates the Indian ocean in the latitude of magnetic equator. To induce the greatest emf in the mast, the ship should proceed : (a) northward (b) southward (c) eastward (d) none of these

The perfect formula used for calculating induced emf in a rod moving in a uniform magnetic field is: (a) \(e=\overrightarrow{\mathbf{B}} \cdot(\overrightarrow{\overrightarrow{1}} \times \overrightarrow{\mathbf{v}})\) (b) \(c=\overrightarrow{\mathrm{B}} \cdot(\overrightarrow{1} \cdot \overrightarrow{\mathrm{v}})\) (c) \(e-\overrightarrow{\mathrm{B}} \times(\overrightarrow{1} \cdot \overrightarrow{\mathrm{v}})\) (d) \(e=\overrightarrow{\mathrm{B}} \times(\overrightarrow{1} \times \overrightarrow{\mathrm{v}})\)

The inductance of a coil in which a current of \(0.1 \mathrm{~A}\) increasing at the rate of \(0.5 \mathrm{~A} / \mathrm{s}\) represents a power flow of \(\frac{1}{2}\) watt, is: (a) \(2 \mathrm{H}\) (b) \(8 \mathrm{H}\) (c) \(20 \mathrm{H}\) (d) \(10 \mathrm{H}\)

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.