/*! 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 The current in an \(R L\) circui... [FREE SOLUTION] | 91Ó°ÊÓ

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The current in an \(R L\) circuit builds up to one-third of its steady-state value in \(3.00 \mathrm{~s}\). Find the inductive time constant.

Short Answer

Expert verified
The inductive time constant is approximately 7.4 seconds.

Step by step solution

01

Define the RL Circuit Model

In an RL circuit, the current builds up according to the formula \( I(t) = I_0(1 - e^{-t/\tau}) \), where \( I_0 \) is the steady-state current and \( \tau \) is the inductive time constant. Our task is to find \( \tau \).
02

Substitute Given Values

We know that the current reaches one-third of its maximum value in \(3 \text{ seconds}\). Thus, \( I(t) = \frac{I_0}{3} \). Substitute this into the formula: \( \frac{I_0}{3} = I_0(1 - e^{-3/\tau}) \).
03

Simplify the Equation

Cancel \( I_0 \) from both sides: \( \frac{1}{3} = 1 - e^{-3/\tau} \). Rearrange to find \( e^{-3/\tau} = 1 - \frac{1}{3} = \frac{2}{3} \).
04

Solve for the Time Constant

Take the natural logarithm of both sides: \( -\frac{3}{\tau} = \ln\left(\frac{2}{3}\right) \). Solve for \( \tau \) to get \( \tau = -\frac{3}{\ln(\frac{2}{3})} \).
05

Calculate the Inductive Time Constant

Compute \( \ln\left(\frac{2}{3}\right) \) which is approximately \(-0.4055\). Thus, \( \tau = -\frac{3}{-0.4055} \). This evaluates to \( \tau \approx 7.4 \text{ seconds} \).

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Key Concepts

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

Inductive Time Constant
The inductive time constant, often denoted as \( \tau \), is a fundamental concept in RL circuits, which consist of a resistor (R) and an inductor (L). The time constant \( \tau \) determines how quickly the current reaches its steady-state value.
In an RL circuit, when a voltage is applied, the current does not rise instantly to its maximum value due to the inductor resisting changes in current. Instead, it increases gradually following an exponential growth pattern, described by the formula:
  • \( I(t) = I_0(1 - e^{-t/\tau}) \)
where \( I_0 \) represents the maximum or steady-state current, and \( t \) is time.
This formula shows that the current will rise to about 63.2% of its maximum value in one time constant \( \tau \). The time constant is directly proportional to the inductor's inductance \( L \) and inversely proportional to the circuit's resistance \( R \):
  • \( \tau = \frac{L}{R} \)
Understanding \( \tau \) allows engineers and physicists to predict how quickly a circuit responds to changes in input.
Steady-State Current
Steady-state current is the current that flows through an RL circuit long after any transients have died away. This is the maximum current the circuit can sustain once the inductor no longer resists the flow due to changes.
In the case of the RL circuit, the steady-state current, \( I_0 \), is reached when the inductor is fully "charged," meaning the back EMF (electromotive force) is zero, and the entire voltage is dropped across the resistor. This implies that the circuit behaves like a simple resistive circuit in the steady state. The current can be expressed as:
  • \( I_0 = \frac{V}{R} \)
where \( V \) is the applied voltage and \( R \) is the resistance. In other words, once the current stabilizes, the circuit does not exhibit any inductive effects, and the current is stable unless the circuit parameters change. Understanding the steady-state current is crucial in analyzing and designing circuits for controlled and predictable operations.
Natural Logarithm
The natural logarithm, denoted \( \ln \), is a critical mathematical function frequently used in analyzing exponential growth or decay processes, such as those in RL circuits. The natural logarithm specifically refers to the logarithm with base \( e \), where \( e \) is approximately 2.718.
In the context of RL circuits, we use the natural logarithm to solve equations involving exponential functions like \( e^{-t/\tau} \). When solving for the inductive time constant, \( \tau \), the equation
  • \( -\frac{3}{\tau} = \ln\left(\frac{2}{3}\right) \)
illustrates the application of the natural logarithm to simplify the equation and isolate \( \tau \). Evaluating \( \ln(\frac{2}{3}) \), which is approximately \(-0.4055\), allows us to calculate \( \tau \) by simple arithmetic. Understanding natural logarithms is essential for solving RL circuit problems, as it allows us to interpret and manipulate the exponential time-dependent behavior of the circuit.

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Most popular questions from this chapter

A coil is connected in series with a \(23.0 \mathrm{k} \Omega\) resistor. An ideal \(50.0 \mathrm{~V}\) battery is applied across the two devices, and the current reaches a value of \(2.00 \mathrm{~mA}\) after \(5.00 \mathrm{~ms}\). (a) Find the inductance of the coil. (b) How much energy is stored in the coil at this same moment?

A battery is connected to a series \(R L\) circuit at time \(t=0\). At what multiple of \(\tau_{L}\) will the current be \(1.00 \%\) less than its equilibrium value?

One hundred turns of (insulated) copper wire are wrapped around a wooden cylindrical core of cross-sectional area \(1.90 \times 10^{-3} \mathrm{~m}^{2}\). The two ends of the wire are connected to a resistor. The total resistance in the circuit is \(9.50 \Omega\). If an externally applied uniform longitudinal magnetic field in the core changes from \(1.60 \mathrm{~T}\) in one direction to \(1.60 \mathrm{~T}\) in the opposite direction, how much charge flows through a point in the circuit during the change?

A solenoid having an inductance of \(9.70 \mu \mathrm{H}\) is connected in series with a \(1.20 \mathrm{k} \Omega\) resistor. (a) If a \(14.0 \mathrm{~V}\) battery is connected across the pair, how long will it take for the current through the resistor to reach \(40.0 \%\) of its final value? (b) What is the current through the resistor at time \(t=0.50 \tau_{L}\) ?

A long solenoid has a diameter of \(12.0 \mathrm{~cm}\). When a current \(i\) exists in its windings, a uniform magnetic field of magnitude \(B=30.0 \mathrm{mT}\) is produced in its interior. By decreasing \(i\), the field is caused to decrease at the rate of \(6.50 \mathrm{mT} / \mathrm{s}\). Calculate the magnitude of the induced electric field (a) \(4.20 \mathrm{~cm}\) and (b) \(10.3 \mathrm{~cm}\) from the axis of the solenoid.

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