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Chapter 10: Problem 13

A \(35.0-\mathrm{V}\) battery with negligible internal resistance, a \(50.0-\Omega\) resistor, and a \(1.25-\mathrm{mH}\) 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) Approximately \(1.73 \times 10^{-5}\) s. (b) Approximately \(3.72 \times 10^{-5}\) s.

Step by step solution

01

Define the given quantities

Identify the given values from the problem: - Battery voltage \( V = 35.0 \text{ V} \).- Resistance \( R = 50.0 \Omega \).- Inductance \( L = 1.25 \text{ mH} = 1.25 \times 10^{-3} \text{ H} \).
02

Write the formula for current in an RL circuit

After the switch is closed, the current in the circuit increases following the equation: \[ i(t) = I_{\text{max}} (1 - e^{-t/\tau}) \]where \( I_{\text{max}} = \frac{V}{R} \) is the maximum current, and \( \tau = \frac{L}{R} \) is the time constant of the circuit.
03

Calculate the time constant

Calculate the time constant \( \tau \) of the circuit: \[ \tau = \frac{L}{R} = \frac{1.25 \times 10^{-3} \text{ H}}{50.0 \Omega} = 2.5 \times 10^{-5} \text{ s} \].
04

Find time for half maximum current

Set \( i(t) = \frac{1}{2} I_{\text{max}} \) and solve for \( t \):\[ \frac{1}{2} I_{\text{max}} = I_{\text{max}} (1 - e^{-t/\tau}) \]which simplifies to\[ \frac{1}{2} = 1 - e^{-t/\tau} \]\[ e^{-t/\tau} = \frac{1}{2} \]Taking the natural log:\[ -\frac{t}{\tau} = \ln\left(\frac{1}{2}\right) \]\[ t = -\tau \ln\left(\frac{1}{2}\right) \]\[ t = 2.5 \times 10^{-5} \text{ s} \ln(2) \approx 1.73 \times 10^{-5} \text{ s} \].
05

Relate energy to current

The energy stored in the inductor is given by: \[ U(t) = \frac{1}{2} L i(t)^2 \].When \( U(t) = \frac{1}{2} U_{\text{max}} \), \( i(t) \) is 0.707 times \( I_{\text{max}} \) due to the square relationship (from \( U \propto i^2 \)).
06

Find time for half maximum energy

Set \( i(t) = 0.707 I_{\text{max}} \) and use the RL current formula:\[ 0.707 I_{\text{max}} = I_{\text{max}} (1 - e^{-t/\tau}) \]which simplifies to\[ 0.707 = 1 - e^{-t/\tau} \]\[ e^{-t/\tau} = 0.293 \]Taking the natural log:\[ -\frac{t}{\tau} = \ln(0.293) \]\[ t = -\tau \ln(0.293) \]\[ t = 2.5 \times 10^{-5} \text{ s} (-\ln(0.293)) \approx 3.72 \times 10^{-5} \text{ s} \].

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Key Concepts

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

Time Constant
The time constant in an RL circuit plays a vital role in determining how quickly the current reaches its maximum value after a switch is closed. It is denoted by \( \tau \) and is calculated using the formula \( \tau = \frac{L}{R} \), where \( L \) is the inductance in henrys and \( R \) is the resistance in ohms.
For this exercise, with \( L = 1.25 \times 10^{-3} \) H and \( R = 50.0 \) \( \Omega \), the time constant is \( \tau = 2.5 \times 10^{-5} \) seconds.

The time constant \( \tau \) is crucial because it tells us how fast the circuit responds to changes. Specifically, in an RL circuit, after one time constant \( \tau \), the current reaches about 63.2% of the total possible current, or \( I_{\text{max}} \).
Knowing this value helps predict how the current behaves over time.
Inductor Energy
Energy in an RL circuit is stored in the inductor. This stored energy is based on the current flowing through the inductor and is given by the formula \( U(t) = \frac{1}{2} L i(t)^2 \).

As the current builds, the energy stored also grows until it reaches a maximum value when the current is at its peak. When we examine the energy at half of its maximum, it correlates to having a current that is about 70.7% of the maximum current \( I_{\text{max}} \).
This is because energy and current have a square relationship鈥攇reater currents lead to a rapid increase in the stored energy.

To calculate this time, set \( i(t) = 0.707 I_{\text{max}} \), and solve for the time using the RL circuit equation for current.
Exponential Growth in Circuits
Exponential growth is a fundamental concept in RL circuits, portraying how quickly the current ramps up over time. After the switch closes, the current changes following an exponential growth equation:
  • \( i(t) = I_{\text{max}} (1 - e^{-t/\tau}) \)
This equation shows the gradual increase of current until it reaches \( I_{\text{max}} \), regulated by the time constant \( \tau \).

At the core of this, when \( t = \tau \), the current achieves about 63.2% of its maximum value. This slow start and rapid increase are pivotal traits of exponential growth in circuits.
Understanding this growth pattern is essential because it helps us model and predict how the currents and energies respond to time in real circuits.

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Most popular questions from this chapter

A \(12-\mu \mathrm{F}\) capacitor is placed across a \(25 \mathrm{~V}\) battery for several seconds and is then connected across a \(1.2 \mathrm{mH}\) inductor that has no appreciable resistance. (a) After the capacitor and inductor are connected together, find the maximum current in the circuit. When the current is a maximum, what is the charge on the capacitor? (b) How long after the capacitor and inductor are connected together does it take for the capacitor to be completely discharged for the first time? For the second time? (c) Sketch graphs of the charge on the capacitor plates and the current through the inductor as functions of time.

A toroidal solenoid has 500 turns, cross-sectional area \(6 \mathrm{~cm}^{2}\), and mean radius \(4.00 \mathrm{~cm}\). (a) Calculate the coil's selfinductance. (b) If the current decreases uniformly from \(5.00 \mathrm{~A}\) to \(2.00 \mathrm{~A}\) in \(3.00 \mathrm{~ms}\), calculate the self-induced emf in the coil. (c) The current is directed from terminal \(a\) of the coil to terminal \(b\). Is the direction of the induced emf from \(a\) to \(b\) or from \(b\) to \(a\) ?

A toroidal solenoid with mean radius \(r\) and cross-sectional area \(A\) is wound uniformly with \(N_{1}\) turns. A second toroidal solenoid with \(N_{2}\) turns is wound uniformly on top of the first, so that the two solenoids have the same crosssectional area and mean radius. (a) What is the mutual inductance of the two solenoids? Assume that the magnetic field of the first solenoid is uniform across the cross section of the two solenoids. (b) If \(N_{1}=500\) turns, \(N_{2}=300\) turns, \(r=10.0 \mathrm{~cm}\), and \(A=0.800 \mathrm{~cm}^{2}\), what is the value of the mutual inductance?

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 200-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 \(80.0 \mathrm{~A}\), what is the inductance?

A small solid conductor with radius \(a\) is supported by insulating, nonmagnetic disks on the axis of a thin-walled tube with inner radius \(b\). The inner and outer conductors carry equal currents \(i\) in opposite directions. (a) Use Ampere's law to find the magnetic field at any point in the volume between the conductors. (b) Write the expression for the flux \(d \Phi_{B}\) through a narrow strip of length \(l\) parallel to the axis, of width \(d r\), at a distance \(r\) from the axis of the cable and lying in a plane containing the axis. (c) Integrate your expression from part (b) over the volume between the two conductors to find the total flux produced by a current \(i\) in the central conductor. (d) Show that the inductance of a length \(l\) of the cable is \(L=l \frac{\mu_{0}}{2 \pi} \ln \left(\frac{b}{a}\right)\) (e) Use Eq. (10.9) to calculate the energy stored in the magnetic field for a length \(l\) of the cable.
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