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An inductor with an inductance of 2.50 \(\mathrm{H}\) and a resistance of 8.00\(\Omega\) is connected to the terminals of a battery with an emf of 6.00 \(\mathrm{V}\) and negligible internal resistance. Find (a) the initial rate of increase of current in the circuit; (b) the rate of increase of current at the instant when the current; is \(0.500 \mathrm{A} ;\) (c) the current 0.250 \(\mathrm{s}\) after the circuit is closed; (d) the final steady-state current.

Short Answer

Expert verified
(a) 2.4 A/s; (b) 0.8 A/s; (c) 0.145 A; (d) 0.750 A.

Step by step solution

01

Understanding the circuit

This is an RL circuit with an inductor (L = 2.50 H) and a resistor (R = 8.00 \(\Omega\)) connected to a battery with emf of 6.00 V.
02

Calculate the initial rate of increase of current

At the initial moment when the circuit is turned on, the current is zero. According to the formula \(\frac{di}{dt} = \frac{\varepsilon}{L}\), we have: \(\frac{di}{dt} = \frac{6.00}{2.50} = 2.4 \mathrm{A/s}\).
03

Rate of increase of current when current is 0.5 A

Using the formula \(\frac{di}{dt} = \frac{\varepsilon - iR}{L}\), substitute \(\varepsilon = 6.00\), \(i = 0.500\), \(R = 8.00\), and \(L = 2.50\). \(\frac{di}{dt} = \frac{6.00 - 8.00 \cdot 0.500}{2.50} = 0.8 \mathrm{A/s}\).
04

Calculate the current after 0.250 s

Using the formula \(i(t) = \frac{\varepsilon}{R}(1 - e^{-\frac{R}{L}t})\), substitute \(\varepsilon = 6.00\), \(R = 8.00\), \(L = 2.50\), and \(t = 0.250\). Calculate \(i(0.250) = \frac{6.00}{8.00}(1 - e^{-\frac{8.00}{2.50} \cdot 0.250}) = 0.145 \mathrm{A}\).
05

Calculate the final steady-state current

At steady state, the current is given by Ohm's Law \(i = \frac{\varepsilon}{R}\). So, the steady-state current is \(i = \frac{6.00}{8.00} = 0.750 \mathrm{A}\).

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Key Concepts

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

Inductance
Inductance is a fundamental concept in the study of RL circuits and plays a key role in understanding how currents change over time. Essentially, inductance is the property of an inductor, which is a coil of wire within a circuit, to resist changes in the current flowing through it. This resistance occurs because a change in current generates a change in magnetic flux, thereby inducing a voltage across the inductor that counteracts the current change. This is described by the formula:
\[ V = L \frac{di}{dt} \]
where \( V \) is the voltage induced, \( L \) is the inductance, and \( \frac{di}{dt} \) is the rate of change of current.
- Inductance allows us to predict how quickly a current can rise or fall in an RL circuit. - The greater the inductance, the slower the rate of change of current, making inductors crucial in controlling circuits where soft start or voltage surge protection is desired.
In our exercise, we have an inductance of 2.50 H. This means the inductor will moderate how swiftly the circuit reaches its steady-state current.
Rate of Current Change
The rate of current change is a vital parameter when analyzing the behavior of RL circuits, especially during switches like turning the circuit on or off. It tells us how fast the current increases or decreases within the circuit over time. The expression for the initial rate of increase in an RL circuit, when just connected to a voltage source, is described by
\[ \frac{di}{dt} = \frac{\varepsilon}{L} \]
At the start when the switch is closed, there is no current, and this rate signifies how fast this current begins to flow.
In the discussed exercise:
  • The initial rate of change was determined to be 2.4 A/s, meaning the current increases by this rate immediately upon closing the switch.
  • When the current reaches 0.5 A, the rate of change diminishes to 0.8 A/s, as the voltage across the inductor breaks the uniformity of current flow according to \( \frac{di}{dt} = \frac{\varepsilon - iR}{L} \).
Understanding this helps in predicting how circuits--especially those starting with zero current--behave over time.
Steady-State Current
In an RL circuit, the steady-state current is the current that eventually flows once all transient effects have settled. This happens after the time constant, which is denoted by \( \tau = \frac{L}{R} \), has passed and the circuit reaches equilibrium. During this state, the effect of the inductor diminishes as it no longer opposes the current change.
- The steady-state current can be found using Ohm's Law: \( i = \frac{\varepsilon}{R} \).- It is typically the maximum current that will flow in the circuit when the inductor is fully magnetized and no longer affecting the current.
In our exercise, this current was calculated to be 0.750 A. This indicates that once the circuit stabilizes, this is the continuous current that will be present.
Ohm's Law
Ohm's Law is one of the fundamental principles used to analyze electrical circuits, providing a simple relationship between voltage, current, and resistance. In the context of our RL circuit, Ohm's Law is essential for calculating both the instantaneous and steady-state currents. The law states that
\[ V = IR \]
where \( V \) is the voltage across the circuit, \( I \) is the current, and \( R \) is the resistance. In scenarios where an emf, akin to a battery, is applied:
  • The voltage supplied by the battery sets the pace at which current flows through the resistor, once the transient period is over.
  • Ohm's Law directly gives us the steady-state current: \( i = \frac{\varepsilon}{R} \) as mentioned earlier.
This law provides a foundation for circuit analysis by simplifying the understanding of how each component in an electrical circuit affects the flow of electric current. In our specific exercise, it helped in determining that the final steady-state current would reach 0.750 A.

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Most popular questions from this chapter

An \(L \cdot R-C\) series circuit has \(L=0.450 \mathrm{H}, C=2.50 \times \)10^{-5} \mathrm{F},\( and resistance\)R . (a) What is the angular frequency of the circuit when \(R=0 ?\) (b) What value must \(R\) have to give a 5.0\(\%\) decrease in angular frequency compared to the value calculated in part (a)?

Two coils have mutual inductance \(M=3.25 \times 10^{-4} \mathrm{H}\) The current \(i_{1}\) in the first coil increases at a uniform rate of 830 \(\mathrm{A} / \mathrm{s} .\) (a) What is the magnitude of the induced emf in the second coil? Is it constant? (b) Suppose that the current described is in the second coil rather than the first. What is the magnitude of the induced emf in the first coil?

Inductance of a Solenoid. (a) A long, straight sole- noid has \(N\) turns, uniform crossectional area \(A,\) and length \(l .\) Show that the inductance of this solenoid is given by the equation \(L=\mu_{0} A N^{2} / l .\) 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 \(\mathrm{cm}\) long and 0.150 \(\mathrm{cm}\) 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 capacitor with capacitance \(6.00 \times\) \(10^{-5} \mathrm{F}\) is charged by connecting it to a \(12.0-\mathrm{V}\) battery. The capacitor is disconnected from the battery and connected across an inductor with \(L=1.50 \mathrm{H}\) (a) What are the angular frequency \(\omega\) of the electrical oscillations and the period of these oscillations (the time for one oscillation)? (b) What is the initial charge on the capacitor? (c) How much energy is initially stored in the capacitor? (d) What is the charge on the capacitor 0.0230 s after the connection to the inductor is made? Interpret the sign of your answer.(e) At the time given in part (d), what is the current in the inductor' Interpret the sign of your answer. (f) At the time given in par (d), how much electrical energy is stored in the capacitor and how much is stored in the inductor?

In an \(L-C\) circuit, \(L=85.0 \mathrm{mH}\) and \(C=3.20 \mu \mathrm{F} .\) During the oscillations the maximum current in the inductor is 0.850 \(\mathrm{mA}\) . (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 \(\mathrm{mA}\) ?

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