Problem 46 - University Physics with Modern Physics

Question

A \(0.250-\mathrm{H}\) inductor carries a time-varying current given by the expression \(i=(124 \mathrm{mA}) \cos [(240 \pi / \mathrm{s}) t] .\) (a) Find an expression for the induced emf as a function of time. Graph the current and induced emf as functions of time for \(t=0\) to \(t=\frac{1}{60} \mathrm{s}\) (b) What is the maximum emf? What is the current when the induced emf is a maximum? (c) What is the maximum current? What is the induced emf when the current is a maximum?

Step-by-Step Solution

Answer

(a) \( \varepsilon(t) = 23.365 \sin(240\pi t) \); graph current and emf from \( t=0 \) to \( t=\frac{1}{60} \text{s} \). (b) Max emf: 23.365 V, current is 0 A when emf is max. (c) Max current: 0.124 A; emf is 0 V when current is max.

Step 1: Find the expression for the induced emf

The induced emf (electromotive force) in an inductor can be found using the formula \( \varepsilon = -L \frac{di}{dt} \), where \( L = 0.250 \text{ H} \) is the inductance. The current is given by \( i(t) = 124 \text{ mA} \cdot \cos[(240\pi/\text{s})t] \), which can be rewritten as \( i(t) = 0.124 \text{ A} \cdot \cos[(240\pi)t] \). Let us find \( \frac{di}{dt} \):\[ \frac{di}{dt} = \frac{d}{dt}(0.124 \cdot \cos(240\pi t)) = -0.124 \cdot 240\pi \cdot \sin(240\pi t) = -29.760\pi \cdot \sin(240\pi t) \] Substituting \( \frac{di}{dt} \) into the formula for \( \varepsilon \), we get:\[ \varepsilon(t) = -0.250 \cdot (-29.760\pi \cdot \sin(240\pi t)) = 7.440\pi \cdot \sin(240\pi t) \] Thus, the induced emf as a function of time is \( \varepsilon(t) = 23.365 \cdot \sin(240\pi t) \) V.

Step 2: Determine the maximum emf

The maximum emf occurs when the sine function reaches its maximum value of 1. Substituting into the expression for \( \varepsilon(t) \):\[ \varepsilon_{\text{max}} = 23.365 \times 1 = 23.365 \text{ volts} \]Thus, the maximum emf is 23.365 V.

Step 3: Evaluate the current when emf is maximum

The emf is maximum when the derivative of the cosine function (sine) is maximum, which corresponds to when the cosine function itself is zero (due to the sine being at its peak, the cosine is at zero crossing). Thus, at the peak of emf, \( i(t) = 0 \) A. Hence, the current is zero when the induced emf is at its maximum.

Step 4: Determine the maximum current

The maximum current occurs when the cosine function takes on its maximum value, i.e., \( \cos(240\pi t) = 1 \). From the expression \( i(t) = 0.124 \cos(240\pi t) \):\[ i_{\text{max}} = 0.124 \times 1 = 0.124 \text{ A} \]Thus, the maximum current is 0.124 A (which corresponds to 124 mA).

Step 5: Evaluate the emf when the current is maximum

When the current is maximum, the cosine function is at its maximum and the derivative \( \sin(240\pi t) = 0 \) since \( \sin \) is at zero for cosine's peak. Hence, from the expression \( \varepsilon(t) = 23.365 \sin(240\pi t) \), the emf is:\[ \varepsilon(t) = 23.365 \times 0 = 0 \text{ V} \]Thus, the induced emf is zero when the current is maximum.

Step 6: Graph current and emf as functions of time

To graph the current and emf as functions of time for \( t = 0 \) to \( t = \frac{1}{60} \text{ s} \), use the functions derived:- Current: \( i(t) = 0.124 \cos(240\pi t) \)- Emf: \( \varepsilon (t) = 23.365 \sin(240\pi t) \)For one period, plot both curves. The current function will have a cosine shape, starting at the maximum 0.124 A at \( t=0 \) and reaching zero at one quarter of the period. The emf function will start at zero and reach its maximum \( 23.365 \text{ V} \) one quarter period into the cycle. Use graphing software or graph paper for a precise sketch.

Key Concepts

Electromotive Force (emf)

Electromotive force, or emf, is a fundamental concept in inductor circuit analysis. It represents the energy per unit charge provided by a source such as a battery or in our case, an inductor. Inductors, when experiencing a change in current, generate an emf to counteract that change due to Lenz's law. This property is expressed in the formula:

\[\varepsilon = -L \frac{di}{dt}\]
where \( L \) is the inductance and \( \frac{di}{dt} \) is the rate of change of current with respect to time. The negative sign signifies the direction of the induced emf, which opposes the change in current. In our exercise, the inductor's emf fluctuates over time in response to the given time-varying current, peaking when the current changes most rapidly.

Time-varying Current

In circuits, the current does not always remain constant. When we talk about time-varying currents, it means the amplitude of the current changes with time. In this exercise, the current is given as a cosine function:
\[i(t) = 0.124 \text{ A} \cdot \cos(240\pi t)\]
This indicates a sinusoidal variation, which often occurs in alternating current (AC) circuits. The frequency of this variation depends on the term \( 240\pi \), dictating how fast the current oscillates per second. Time-varying currents are crucial because they directly influence the behavior of the inductor, affecting both the amplitude and phase of the induced emf.

Maxima and Minima in Circuits

Identifying maxima and minima in circuits is crucial for understanding the performance and limitations of electronic components. In our context, these refer to the maximum and minimum values of current and emf. For our current expression
\[i(t) = 0.124 \cos(240\pi t)\]
we observe peaks at values of \( 0.124 \text{ A} \) and troughs at \( -0.124 \text{ A} \). The induced emf peaks at 23.365 V as calculated earlier when the sinusoidal function of sine reaches its maximum. Being able to calculate these extremes helps in predicting how circuits behave under different conditions and ensuring they operate within safe and efficient limits.

Derivative in Calculus

In the analysis of circuits, derivatives play a major role, especially with components like inductors where the change in current over time matters. The derivative \( \frac{di}{dt} \) represents how quickly the current changes at any given point in time. By taking the derivative of the time-varying current
\[i(t) = 0.124 \cos(240\pi t)\]
we find
\[\frac{di}{dt} = -29.760\pi \sin(240\pi t)\]
This derivative is negative because the cosine function is decreasing as it passes through zero. Understanding this concept is key in many areas of physics and engineering where rates of change affect system behavior. Calculus provides the tools to not only determine these rates but also predict future behavior based on current conditions.