/*! 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 29 (II) A \(12 - \mathrm { V }\) ba... [FREE SOLUTION] | 91Ó°ÊÓ

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(II) A \(12 - \mathrm { V }\) battery has been connected to an \(L R\) circuit for sufficient time so that a steady current flows through the resistor \(R = 2.2 \mathrm { k } \Omega\) and inductor \(L = 18 \mathrm { mH } .\) At \(t = 0\) the battery is removed from the circuit and the current decays exponentially through \(R\) . Determine the emf \(\mathscr { Y }\) across the inductor as time \(t\) increases. At what time is 8 greatest and what is this maximum value (V)?

Short Answer

Expert verified
The EMF is greatest at \( t = 0 \) and equals 12 V.

Step by step solution

01

Understanding the Problem

We have an LR circuit where the battery is removed at time \( t = 0 \), allowing the current to decay through the resistor and inductor. We need to find the EMF across the inductor over time and determine when it is maximized and its maximum value.
02

Initial Current Calculation

Before the battery is removed, a steady current \( I_0 \) flows through the circuit. This can be calculated using Ohm's Law: \( I_0 = \frac{V}{R} = \frac{12 \text{ V}}{2200 \text{ } \Omega} = 0.00545 \text{ A (or 5.45 mA)} \).
03

Current Decay Formula

After the battery is removed, the current \( I(t) \) decays exponentially: \( I(t) = I_0 e^{-\frac{t}{\tau}} \), where \( \tau = \frac{L}{R} \) is the time constant. Substitute the values: \( \tau = \frac{18 \times 10^{-3}}{2200} = 8.18 \times 10^{-6} \text{ s} \).
04

EMF Across the Inductor Formula

The EMF \( \mathscr{E}(t) \) across the inductor can be described by the formula: \( \mathscr{E}(t) = -L\frac{dI}{dt} \). Substitute \( I(t) = I_0 e^{-\frac{t}{\tau}} \) to find \( \frac{dI}{dt} = -\frac{I_0}{\tau} e^{-\frac{t}{\tau}} \). Thus, \( \mathscr{E}(t) = L \times \frac{I_0}{\tau} e^{-\frac{t}{\tau}} \).
05

Maximize EMF Calculation

The EMF \( \mathscr{E}(t) \) is maximized at \( t = 0 \) since it is the product of a decaying exponential, \( I(t) \), and a constant. Hence, \( \mathscr{E}_{\text{max}} = L \times \frac{I_0}{\tau} = 18 \times 10^{-3} \times \frac{0.00545}{8.18 \times 10^{-6}} \approx 12 \text{ V} \).

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Key Concepts

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

Ohm's Law
In the realm of electric circuits, Ohm's Law is a fundamental principle that describes the relationship between voltage, current, and resistance. The law is mathematically expressed as \( V = I \times R \), where \( V \) is the voltage across the circuit, \( I \) is the current flowing through it, and \( R \) is the resistance present in the circuit.
One helpful way to understand Ohm's Law is to think of it as a bridge.
It connects the potential difference to the flow of electric charge, giving us a way to calculate the current if the voltage and resistance are known.
In our problem, before the battery is removed, the circuit is in a steady state with a constant current. Using Ohm’s Law, we calculate this initial current \( I_0 \) by dividing the battery voltage \( 12\, \text{V} \) by the circuit's total resistance \( R = 2200\, \Omega \).
- **Key Insight**: Ohm’s Law is crucial for determining how much current is flowing based on known voltage and resistance values. It provides the initial condition for analyzing further circuit dynamics.
Inductor EMF
Inductors play a critical role in circuits, storing energy in a magnetic field when current flows through them. When the current changes, an Electromotive Force (EMF) is induced in the inductor according to Lenz's Law.
This induced EMF opposes the change in current, described by the equation \( \mathscr{E} = -L \frac{dI}{dt} \), where \( L \) is the inductance and \( \frac{dI}{dt} \) represents the rate of change of current.
As the battery is removed from the circuit, the current begins to decrease, and consequently, the inductor produces an EMF to oppose this change. In our solution, \( \mathscr{E}(t) \) is directly derived from the decaying current \( I(t) \). By substituting the current decay expression into the EMF formula, we calculate how this EMF changes over time.
- **Key Insight**: Understanding how EMF works in inductors helps us see how inductors resist changes in current and how energy is dissipated when a battery is removed.
Exponential Current Decay
When a battery is disconnected from an LR circuit, the current doesn't stop immediately.
Instead, it decays exponentially over time, a pattern described by \( I(t) = I_0 e^{-\frac{t}{\tau}} \).
Here, \( I_0 \) is the initial current and \( \tau \) represents the time constant.
This exponential behavior is due to the energy stored in the inductor's magnetic field that continues to drive current even after the power source is removed.
The rate at which current decreases is determined by the circuit’s time constant \( \tau \).
The solution emphasizes how current follows this mathematical model, which is exceptionally common in circuits with inductors or capacitors.
Understanding this pattern is key to predicting how circuit quantities will behave over time, especially in circuits experiencing sudden component changes such as a power source disconnection.
- **Key Insight**: Recognizing exponential decay in current allows us to predict circuit behavior accurately after changes, using foundational mathematical relationships.
Time Constant in Circuits
The time constant, \( \tau \), is an essential parameter in the analysis of LR circuits. It is calculated as \( \tau = \frac{L}{R} \), where \( L \) is the inductance and \( R \) is the resistance.
The time constant defines the speed at which the current decreases in the circuit. In simpler terms, it represents the time it takes for the current to fall to around 37% of its initial value \( I_0 \).
In our given problem, the calculated time constant is \( 8.18\, \mu s \).
This explains how quickly the circuit loses current after the battery is removed.
Knowing \( \tau \) aids in understanding how fast responses occur and how long one can expect energy storage in inductors.
This insight is crucial for designing circuits with specific timing requirements.
- **Key Insight**: The time constant is a direct measure of the pace at which reactive elements (like inductors) respond to changes.
Grasping \( \tau \) is vital for predicting how quickly circuit dynamics settle after disturbances.

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Most popular questions from this chapter

To detect vehicles at traffic lights, wire loops with dimensions on the order of 2\(\mathrm { m }\) are often buried horizontally under roadways. Assume the self-inductance of such a loop is \(L = 5.0 \mathrm { mH }\) and that it is part of an \(L R C\) circuit as shown in Fig. 37 with \(C = 0.10 \mu \mathrm { F }\) and \(R = 45 \Omega\) . The ac voltage has frequency \(f\) and rms voltage \(V _ { \mathrm { rms } }\) . (a) The frequency \(f\) is chosen to match the resonant frequency \(f _ { 0 }\) of the circuit. Find \(f _ { 0 }\) and determine what the rms voltage \(\left( V _ { R } \right) _ { \text { Tms across the resistor will be when } }\) \(f = f _ { 0 } . ( b )\) Assume that \(f , C ,\) and \(R\) never change, but that, when a car is located above the buried loop, the loop's self-inductance decreases by 10\(\%\) (due to induced eddy currents in the car's metal parts). Determine by what factor the voltage \(\left( V _ { R } \right) _ { \text { rms decreases in this situation in } }\) comparison to no car above the loop. [Monitoring \(\left( V _ { R } \right) _ { \mathrm { rms } }\) detects the presence of a car.

(II) At \(t = 0 ,\) let \(Q = Q _ { 0 } ,\) and \(I = 0\) in an \(L C\) circuit. (a) At the first moment when the energy is shared equally by the inductor and the capacitor, what is the charge on the capacitor? (b) How much time has elapsed (in terms of the period \(T ) ?\)

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A \(35-\mathrm{mH}\) inductor with \(2.0-\Omega\) resistance is connected in series to a \(26-\mu \mathrm{F}\) capacitor and a \(60-\mathrm{Hz}, 45-\mathrm{V}(\mathrm{rms})\) source. Calculate \((a)\) the rms current, \((b)\) the phase angle, and \((c)\) the power dissipated in this circuit.

(I) If the current in a \(280-\mathrm{mH}\) coil changes steadily from \(25.0 \mathrm{~A}\) to \(10.0 \mathrm{~A}\) in \(360 \mathrm{~ms}\), what is the magnitude of the induced emf?

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