Problem 33 - Fundamentals of Physics

Question

An ac generator has emf \(\mathscr{E}=\mathscr{E}_{m} \sin \left(\omega_{d} t-\pi / 4\right),\) where \(\mathscr{E}_{m}=30.0 \mathrm{~V}\) and \(\omega_{d}=350 \mathrm{rad} / \mathrm{s} .\) The current produced in a connected circuit is \(i(t)=I \sin \left(\omega_{d} t-3 \pi / 4\right),\) where \(I=620 \mathrm{~m}\) A. At what time after \(t=0\) does (a) the generator emf first reach a maximum and (b) the current first reach a maximum? (c) The circuit contains a single element other than the generator. Is it a capacitor, an inductor, or a resistor? Justify your answer. (d) What is the value of the capacitance, inductance, or resistance, as the case may be?

Step-by-Step Solution

Answer

(a) 0.0067 s, (b) 0.0112 s, (c) Inductor, (d) 0.138 H

Step 1: Identify Maximum Emf Condition

To determine when the generator emf first reaches its maximum, recognize that the sinusoidal function reaches its maximum value, 1, when the argument is \((\omega_{d}t - \frac{\pi}{4}) = \frac{\pi}{2} + 2n\pi\), where \( n \) is an integer.

Step 2: Solve for Maximum Emf Time

Set the expression for the generator's sinusoidal argument to \(\frac{\pi}{2}\) to solve for time: \( \omega_{d}t - \frac{\pi}{4} = \frac{\pi}{2} \). Put known values: \( 350t - \frac{\pi}{4} = \frac{\pi}{2} \). Upon rearranging: \[ 350t = \frac{3\pi}{4} \]. Thus, \( t = \frac{3\pi}{4 \times 350} \) seconds. Calculate this to find \( t \approx 0.0067 \) seconds.

Step 3: Identify Maximum Current Condition

For the current to reach its maximum, its sinusoidal component \((\omega_{d}t - \frac{3\pi}{4})\) should also equal \(\frac{\pi}{2} + 2m\pi\). Again, \( m \) is an integer.

Step 4: Solve for Maximum Current Time

Equation is: \( \omega_{d}t - \frac{3\pi}{4} = \frac{\pi}{2} \). Simplify: \( 350t - \frac{3\pi}{4} = \frac{\pi}{2} \). Rearrange to \[ 350t = \frac{5\pi}{4} \]. Solve for \( t \): \( t = \frac{5\pi}{4 \times 350} \) seconds. This calculates to \( t \approx 0.0112 \) seconds.

Step 5: Determine Circuit Element Based on Phase Shift

Compare the phase shifts of the current and emf: \(\frac{\pi}{4}\) for the emf and \(\frac{3\pi}{4}\) for the current. The current lagging behind the voltage indicates the presence of an inductor.

Step 6: Calculate Inductance

The inductive reactance \(X_L\) is given by \(\omega_{d}L = \frac{\mathscr{E}_{m}}{I}\) and \( \omega_{d} = 350\) rad/s, \( I = 0.62 \) A. Hence, \( X_L = \frac{30}{0.62} = 48.39 \Omega \). Calculate \( L = \frac{X_L}{\omega_{d}} = \frac{48.39}{350} \approx 0.138 \) H.

Key Concepts

Sinusoidal Functions

A sinusoidal function is a mathematical function that describes a smooth, repetitive oscillation, such as a sine or cosine wave. These functions are vital in electrical engineering, particularly when analyzing alternating current (AC) circuits like an AC generator. An AC generator produces a voltage that varies over time in a sinusoidal pattern, represented by the equation \( \mathscr{E} = \mathscr{E}_{m} \sin(\omega_{d} t - \phi) \), where:

• \( \mathscr{E}_{m} \) is the maximum or peak voltage produced.
• \( \omega_{d} \) is the angular frequency, dictating how fast the wave oscillates.
• \( t \) is time.
• \( \phi \) is the phase angle, which shifts the entire wave in time.

The sinusoidal nature allows us to predict when the waveform will reach its maximum, cross zero, or hit any particular point. This behavior is essential for tuning systems and aligning waves for maximum efficiency and proper functioning of circuits. Understanding these elements helps in solving practical problems, like determining the time at which the generator emf reaches its maximum value.

Phase Shift

Phase shift refers to the horizontal shift between two sinusoidal waves. In the context of AC circuits, it is the difference in timing between the maximum values of voltage and current. This shift can be crucial in understanding how different components in the circuit affect performance. It is commonly expressed in radians or degrees, using the concept of angles because waves can be thought of as rotating vectors.In any AC circuit:

• If current is in phase with voltage (no phase shift), you typically have a resistor.
• If current lags voltage (current phase is more negative), this indicates an inductive component is present.
• If current leads voltage, a capacitive component is involved.

For example, with the problem at hand, we identify phase shifts between the generator emf \( \frac{\pi}{4} \) and current \( \frac{3\pi}{4} \). The fact that the current has a larger phase shift (lags behind) suggests that the circuit contains an inductor. Recognizing phase shifts help in determining the nature of the element in the circuit and in solving for other parameters like inductance in inductive setups.

Inductive Reactance

Inductive reactance is a concept used to describe the resistance that an inductor provides to the change in current in an AC circuit. This reactance is caused by the inductor's magnetic field, which opposes changes in current. The formula for inductive reactance \( X_L \) is \( X_L = \omega L \), where \( \omega \) is the angular frequency of the AC source, and \( L \) is the inductance.In the context of our example:

• The generator has a frequency \( \omega_{d} = 350 \, \text{rad/s} \).
• Using the peak current \( I = 0.62 \, \text{A} \) and peak emf \( \mathscr{E}_m = 30.0 \, \text{V} \), we calculate \( X_L = \frac{\mathscr{E}_m}{I} = 48.39 \, \Omega \).
• To find the inductance \( L \), we rearrange to \( L = \frac{X_L}{\omega} \), resulting in \( L \approx 0.138 \, \text{H} \).

Understanding inductive reactance allows us to comprehend how inductors affect circuit behavior, particularly in terms of impacting current flow and phase relationships in AC circuits.