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Chapter 3: Problem 44

At \(t=0\), the current flowing in a \(0.5-\mathrm{H}\) inductance is \(2 \mathrm{~A}\). What constant voltage must be applied to increase the current to \(4 \mathrm{~A}\) at \(t=0.4 \mathrm{~s}\) ?

Short Answer

Expert verified
A voltage of 2.5 V is needed.

Step by step solution

01

Understanding the Inductor's Voltage-Current Relationship

The voltage across an inductor is related to the rate of change of current through it. This is given by the formula: \[ V = L \frac{di}{dt} \] where \( V \) is the voltage across the inductor, \( L \) is the inductance, and \( \frac{di}{dt} \) is the rate of change of current.
02

Determine the Rate of Change of Current

The current is supposed to increase from \( 2 \, \text{A} \) to \( 4 \, \text{A} \) in \( 0.4 \, \text{s} \). Thus, the change in current \( \Delta i \) is \( 4 - 2 = 2 \, \text{A} \). Since the time period over which this change occurs is \( 0.4 \, \text{s} \), the rate of change of current \( \frac{di}{dt} \) can be calculated as follows: \[ \frac{di}{dt} = \frac{\Delta i}{\Delta t} = \frac{2}{0.4} = 5 \, \text{A/s} \].
03

Calculate the Required Voltage

Using the formula \( V = L \frac{di}{dt} \), where \( L = 0.5 \, \text{H} \) and \( \frac{di}{dt} = 5 \, \text{A/s} \), we can find the required voltage. Substitute the known values into the formula: \[ V = 0.5 \times 5 = 2.5 \, \text{V} \]. This is the constant voltage required to increase the current.

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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 electrical circuits. It measures an inductor's ability to store energy in a magnetic field when electrical current passes through it. If you imagine an inductor, think of it as a coil of wire. The more the coil's turns, the higher the inductance.

Inductance is symbolized by the letter 'L' and is measured in Henrys (H). One Henry of inductance is quite large; it's the inductance of a circuit in which a change of current of one ampere per second induces a voltage of one volt. So, when you have a smaller inductance like 0.5 H, the system is relatively smaller and responds faster to changes.

Inductors are essential for controlling current and voltage in many electronic devices. They work well with resistors and capacitors to form filters or oscillators, and they help smooth out variations in supply currents.
Rate of Change of Current
The rate of change of current essentially tells us how quickly the current is increasing or decreasing in a circuit. It is a crucial part of understanding how an inductor behaves.

When the current changes over time, the inductor generates a voltage to oppose this change. This opposition adheres to Lenz's Law, which states the induced electromotive force will always oppose the change in current.

To calculate the rate of change of current, \(\frac{di}{dt}\), you simply take the difference in current and divide it by the time over which the change occurs. In our original exercise, the current went from 2 A to 4 A in 0.4 seconds, so the calculation was:
  • \(\Delta i = 4 \, \mathrm{A} - 2 \, \mathrm{A} = 2 \, \mathrm{A}\)
  • \(\Delta t = 0.4 \, \mathrm{s}\)
  • \(\frac{di}{dt} = \frac{2}{0.4} = 5 \, \mathrm{A/s}\)
This tells us the current is increasing at a rate of 5 amperes per second.
Voltage Calculation
Calculating the voltage across an inductor can seem challenging, but it follows a straightforward formula. This derived formula from the principles of inductance is:
  • \( V = L \frac{di}{dt} \)
Here, \( V \) is the voltage across the inductor, \( L \) is the value of inductance, and \( \frac{di}{dt} \) is the rate at which the current is changing.

In the scenario from our original problem, you were given an inductance \( L \) of 0.5 H and a rate of change of current \( \frac{di}{dt} \) as 5 A/s. Thus, you can plug these values into the formula to get: \( V = 0.5 \times 5 = 2.5 \, \mathrm{V} \).

This means a constant voltage of 2.5 volts needs to be applied across the inductor to increase the current from 2 A to 4 A over 0.4 seconds. Understanding this concept is crucial for designing circuits and ensuring they function correctly under varying electrical loads.

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Most popular questions from this chapter

The current through a 200-mH inductance is given by \(i_{L}(t)=\exp (-2 t) \sin (4 \pi t)\) A in which the angle is in radians. Using your knowledge of calculus, find an expression for the voltage across the inductance. Then, use MATLAB to verify your answer for the voltage and to plot both the current and the voltage for \(0 \leq\) \(t \leq 2 \mathrm{~s}\).

At \(t=t_{0}\) the voltage across a certain capacitance \(C\) is zero. A pulse of current flows through the capacitance between \(t_{0}\) and \(t_{0}+\) \(\Delta t\), and the voltage across the capacitance increases to \(V_{f}\). What can you say about the peak amplitude \(I_{m}\) and area under the pulse waveform (i.e., current versus time)? What are the units and physical significance of the area under the pulse? What must happen to the peak amplitude and area under the pulse as \(\Delta t\) approaches zero, assuming that \(V_{f}\) remains the same?

Determine the capacitance of a parallel-plate capacitor having plates \(100 \mathrm{~mm}\) by \(0.3 \mathrm{~m}\) separated by \(0.04 \mathrm{~mm}\). The dielectric constant \(\epsilon=100 \mathrm{pF} / \mathrm{m}\).

Prior to \(t=0, \mathrm{a} 100-\mu \mathrm{F}\) capacitance is uncharged. Starting at \(t=0\), the voltage across the capacitor is increased linearly with time to \(100 \mathrm{~V}\) in \(2 \mathrm{~s}\). Then, the voltage remains constant at \(100 \mathrm{~V}\). Sketch the voltage, current, power, and stored energy to scale versus time.

We have capacitor \(C_{1}=200 \mu \mathrm{F}\), which is charged to an initial voltage of \(50 \mathrm{~V}\), and capacitor \(C_{2}=200 \mu \mathrm{F}\), which is charged to \(100 \mathrm{~V}\). If they are placed in series with the positive terminal of the first connected to the negative terminal of the second, determine the equivalent capacitance and its initial voltage. Now, compute the total energy stored in the two capacitors. Compute the energy stored in the equivalent capacitance. Why is it less than the total energy stored in the original capacitors?
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