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A 35.0-V battery with negligible internal resistance, a 50.0-\(\Omega\) resistor, and a 1.25-mH inductor with negligible resistance are all connected in series with an open switch. The switch is suddenly closed. (a) How long after closing the switch will the current through the inductor reach one-half of its maximum value? (b) How long after closing the switch will the energy stored in the inductor reach one-half of its maximum value?

Short Answer

Expert verified
(a) 1.73 µs; (b) 4.15 µs.

Step by step solution

01

Identify the maximum current

In an RL circuit, the maximum current, denoted as \( I_{max} \), is determined by the formula: \[ I_{max} = \frac{V}{R} \]where \( V = 35.0 \text{ V} \) is the voltage and \( R = 50.0 \Omega \) is the resistance. Therefore, \[ I_{max} = \frac{35.0}{50.0} = 0.7 \text{ A} \].
02

Use the RL time constant

The RL time constant, represented by \( \tau \), is given by the formula:\[ \tau = \frac{L}{R} \]where \( L = 1.25 \text{ mH} = 1.25 \times 10^{-3} \text{ H} \) and \( R = 50.0 \Omega \). So, the time constant \( \tau \) is:\[ \tau = \frac{1.25 \times 10^{-3}}{50.0} = 2.5 \times 10^{-5} \text{ s} \].
03

Calculate time for current to reach half of its maximum value

The current in the RL circuit at time \( t \) is given by:\[ I(t) = I_{max} \left(1 - e^{-\frac{t}{\tau}}\right) \]We are solving for \( t \) when \( I(t) = \frac{I_{max}}{2} \). Substituting the values:\[ \frac{0.7}{2} = 0.7 \left(1 - e^{-\frac{t}{\tau}}\right) \]Simplifying:\[ 0.5 = 1 - e^{-\frac{t}{\tau}} \]\[ e^{-\frac{t}{\tau}} = 0.5 \] Taking natural log on both sides:\[ -\frac{t}{\tau} = \ln(0.5) \]\[ \frac{t}{\tau} = -\ln(0.5) \]Thus, \[ t = -\tau \ln(0.5) \approx 1.73 \times 10^{-5} \text{ s} \].
04

Calculate time for energy to reach half of its maximum value

The energy stored in the inductor is determined by:\[ E(t) = \frac{1}{2} L I(t)^2 \].The maximum energy, \( E_{max} \), is when \( I = I_{max} \):\[ E_{max} = \frac{1}{2} L I_{max}^2 \].We need the value of \( t \) when \( E(t) = \frac{1}{2} E_{max} \). This equates to \( I(t) = \frac{I_{max}}{\sqrt{2}} \).Setting \( I(t) = 0.7 \left(1 - e^{-\frac{t}{\tau}} \right) = \frac{0.7}{\sqrt{2}} \):\[ 1 - e^{-\frac{t}{\tau}} = \frac{1}{\sqrt{2}} \]\[ e^{-\frac{t}{\tau}} = 1 - \frac{1}{\sqrt{2}} \]Taking natural logs, we solve:\[ -\frac{t}{\tau} = \ln\left(1 - \frac{1}{\sqrt{2}}\right) \]\[ t = -\tau \ln\left(1 - \frac{1}{\sqrt{2}}\right) \approx 4.15 \times 10^{-5} \text{ s} \].

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Key Concepts

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

RL time constant
The RL time constant, often denoted as \( \tau \), is a crucial concept in analyzing RL (Resistor-Inductor) circuits. It is a measure of the time it takes for the current to either build up to its maximum value when a circuit is closed, or decay when the circuit is opened. Mathematically, the RL time constant is given by the formula: \[ \tau = \frac{L}{R} \] where \( L \) is the inductance of the inductor measured in henrys (H), and \( R \) is the resistance in ohms (\( \Omega \)). This time constant helps in understanding how quickly an RL circuit responds to changes in voltage.
  • A smaller \( \tau \) indicates a rapid response with a quick rise or fall of current.
  • A larger \( \tau \) suggests a slower response with a gradual change in current.
The concept is pivotal in applications like filters and timing circuits because it governs the transient behavior when the circuit is switching states.
maximum current in RL circuit
In an RL circuit, the maximum current, also called the steady-state current, can be found once the inductor has fully charged, and the circuit reaches equilibrium. We calculate the maximum current using Ohm's Law given as: \[ I_{max} = \frac{V}{R} \] where \( V \) is the voltage across the whole circuit and \( R \) is the resistance. An inductor initially opposes changes in current but allows the current to achieve this maximum value after time elapses, defined by the voltage and resistance values.

Here's why understanding maximum current is essential:
  • It determines the upper limit of current that can flow through the circuit, which is critical for preventing circuit damage.
  • When designing circuits, knowing the maximum current helps select suitable components to ensure reliability and safety.
energy stored in inductor
Inductors have the fascinating ability to store energy in their magnetic fields when current flows through them. The amount of energy stored in an inductor is pivotal for your grasp of RL circuits.
The energy stored is expressed by the formula: \[ E = \frac{1}{2} L I^2 \] where \( E \) is the energy in joules, \( L \) is the inductance, and \( I \) is the current at any given time. This formula represents the fact that energy storage in magnetic fields is dependent on both the inductance of the inductor and the square of the current.

Some key points about energy in inductors:
  • When a circuit is opened, an inductor can release its stored energy, which can be desirable or damaging.
  • Stored energy allows inductors to act as a stabilizing element in circuits by providing continuous current flow during changes.
Understanding energy storage in an inductor not only aids in appreciating their role in circuits but helps to avoid scenarios of energy discharge surprises.
natural logarithm in circuits
Natural logarithms frequently appear in the analysis of RL circuits, particularly when calculating how a variable like current or energy evolves over time. This is because the current \( I(t) \) and energy \( E(t) \) in circuits often exhibit exponential behavior:
The formula for the current in an RL circuit over time is: \[ I(t) = I_{max} \left(1 - e^{-\frac{t}{\tau}}\right) \] where \( e \) is the base of natural logarithms, approximately equal to 2.718.
The natural logarithm is the inverse operation of the exponential function. So when solving for time-related expressions such as those where \( I(t) \) or \( E(t) \) are known, taking the natural logarithm helps us isolate the variable \( t \).
  • Using natural logs, you can solve for variables in exponential decay or growth situations.
  • They can simplify calculations involving time constants and current or energy change over time.
Mastery over the use of natural logarithms in RL circuits is not just a mathematical exercise—it is essential for designing and troubleshooting circuits effectively.

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Most popular questions from this chapter

An air-filled toroidal solenoid has 300 turns of wire, a mean radius of 12.0 cm, and a cross-sectional area of 4.00 cm\(^2\). If the current is 5.00 A, calculate: (a) the magnetic field in the solenoid; (b) the self inductance of the solenoid; (c) the energy stored in the magnetic field; (d) the energy density in the magnetic field. (e) Check your answer for part (d) by dividing your answer to part (c) by the volume of the solenoid.

It has been proposed to use large inductors as energy storage devices. (a) How much electrical energy is converted to light and thermal energy by a 150-W light bulb in one day? (b) If the amount of energy calculated in part (a) is stored in an inductor in which the current is 80.0 A, what is the inductance?

In an \(L\)-\(C\) circuit, \(L = 85.0\) mH and \(C = 3.20 \, \mu \mathrm{F}\). During the oscillations the maximum current in the inductor is 0.850 mA. (a) What is the maximum charge on the capacitor? (b) What is the magnitude of the charge on the capacitor at an instant when the current in the inductor has magnitude 0.500 mA?

(a) A long, straight solenoid has \(N\) turns, uniform cross sectional area \(A\), and length \(l\). Show that the inductance of this solenoid is given by the equation \(L = \mu_0 AN^2/l\). Assume that the magnetic field is uniform inside the solenoid and zero outside. (Your answer is approximate because \(B\) is actually smaller at the ends than at the center. For this reason, your answer is actually an upper limit on the inductance.) (b) A metallic laboratory spring is typically 5.00 cm long and 0.150 cm in diameter and has 50 coils. If you connect such a spring in an electric circuit, how much self-inductance must you include for it if you model it as an ideal solenoid?

A long, straight solenoid has 800 turns. When the current in the solenoid is 2.90 A, the average flux through each turn of the solenoid is \(3.25 \times 10^{-3}\) Wb. What must be the magnitude of the rate of change of the current in order for the self-induced emf to equal 6.20 mV?

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