Problem 67 - Fundamentals of Physics

Question

An ac generator produces 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 in the circuit attached to the generator is \(i(t)=I \sin \left(\omega_{d} t+\pi / 4\right),\) where \(I=620 \mathrm{~mA}\). (a) At what time after \(t=0\) does the generator emf first reach a maximum? (b) At what time after \(t=0\) does 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) 0.0022 s; (c) Capacitor; (d) 60.3 µF.

Step 1: Determine the Maximum Condition for the Generator EMF

The emf is described by \( \mathscr{E} = \mathscr{E}_m \sin(\omega_d t - \pi / 4) \). The sine function reaches its maximum value (1) when the argument equals \( \frac{\pi}{2} \) plus any integer multiple of \( 2\pi \), i.e., \( \omega_d t - \frac{\pi}{4} = \frac{\pi}{2} + 2n\pi \). Solving for \( t \) and considering the smallest positive \( t \):\[ \omega_d t = \frac{3\pi}{4} \]\[ t = \frac{3\pi}{4\omega_d} \]Plugging in \( \omega_d = 350 \text{ rad/s} \):\[ t = \frac{3\pi}{4 \times 350} \approx 0.0067 \text{ s} \]

Step 2: Determine the Maximum Condition for the Current

The current is given by \( i(t) = I \sin(\omega_d t + \pi / 4) \). Similar to the emf condition, the current reaches its maximum when \( \omega_d t + \frac{\pi}{4} = \frac{\pi}{2} + 2n\pi \). Solving for \( t \) and considering the smallest positive \( t \):\[ \omega_d t = \frac{\pi}{4} \]\[ t = \frac{\pi}{4\omega_d} \]Substituting \( \omega_d = 350 \text{ rad/s} \):\[ t = \frac{\pi}{4 \times 350} \approx 0.0022 \text{ s} \]

Step 3: Determine the Element in the Circuit

Observing the phases, we see that the emf phase \(-\pi/4\) lagging behind the current phase \(\pi/4\) by \(\pi/2\) suggests it's a capacitive circuit. In a capacitive circuit, the current leads the voltage by \(\pi/2\). Thus, the circuit contains a capacitor.

Step 4: Calculate the Value of Capacitance

For a capacitive circuit, the capacitive reactance \(X_c\) is given by:\[ X_c = \frac{V_m}{I_m} \]Where \( V_m = 30 \text{ V} \) and \( I_m = 0.62 \text{ A} \).\[ X_c = \frac{30}{0.62} \approx 48.39 \Omega \]We know \( X_c = \frac{1}{\omega C} \). Solving for \( C \) gives:\[ C = \frac{1}{\omega_d X_c} \]Substitute \( \omega_d = 350 \text{ rad/s} \):\[ C = \frac{1}{350 \times 48.39} \approx 60.3 \mu F \]

Key Concepts

Electromotive Force (emf)

An alternating current (AC) generator produces an electromotive force (emf) that can be described using a sine wave. This is an essential concept in understanding how AC circuits operate. The emf generated by the AC generator can be mathematically represented as \( \mathscr{E} = \mathscr{E}_{m} \sin(\omega_{d} t - \frac{\pi}{4}) \). Here, \( \mathscr{E}_{m} \) represents the peak value of the voltage, and \( \omega_{d} \) is the angular frequency. In this exercise, the peak voltage is 30.0 V, and the angular frequency is 350 rad/s.

In general, the sine function reaches its maximum value of 1 when its argument is \( \frac{\pi}{2} + 2n\pi \), meaning that the voltage reaches its maximum value at this condition. Solving this equation helps determine the time at which maximum voltage occurs after time zero. This understanding is crucial in deciphering how and when AC voltage peaks relative to time.

Capacitive Reactance

Capacitive reactance is an important factor in AC circuits containing capacitors. When a circuit includes a capacitor, it opposes the change in voltage, creating a phase shift between the voltage and current. In AC terms, this opposition to changing current is known as capacitive reactance, symbolized by \( X_c \), and its value depends on the frequency of the AC supply and the capacitance. Mathematically, it is given by \( X_c = \frac{1}{\omega C} \).

In a capacitive circuit, the current phase leads the voltage phase by \( \frac{\pi}{2} \). This peculiarity is observable in the problem where the current reaches its peak before the voltage, reaffirming the presence of a capacitor. The capacitive reactance is also essential when calculating the capacitance in the circuit through the relation \( X_c = \frac{V_m}{I_m} \), where \( V_m \) is the maximum voltage, and \( I_m \) is the maximum current.

Circuit Components

In AC circuits, understanding the various components—resistors, capacitors, and inductors—helps in analyzing current and voltage behaviors. Each component reacts distinctly to AC:

• **Resistors**: Maintain the same phase relationship between voltage and current.
• **Inductors**: The current lags behind the voltage by a phase of \( \frac{\pi}{2} \).
• **Capacitors**: The current leads the voltage by a phase of \( \frac{\pi}{2} \).

For this particular circuit problem, the phase difference between the current and voltage is \( \pi/2 \), implying the presence of a capacitor. This discovery is a fundamental step in solving the problem, as it enables us to further calculate the capacitance value.

Sine Wave Analysis

Sine wave analysis is pivotal in AC circuit analysis, which includes understanding the time variation of voltage and current. By representing AC values with sine functions, we can predict how these quantities fluctuate over time. In the given exercise, both the voltage and current are expressed as sine functions, specifically:

\( \mathscr{E}(t) = \mathscr{E}_{m} \sin(\omega_{d} t - \frac{\pi}{4}) \) for voltage, and

\( i(t) = I \sin(\omega_{d} t + \frac{\pi}{4}) \) for current.

Analyzing these equations reveals the time at which their respective peaks occur, and the difference in their phases informs us about the type of components in the circuit. This phase difference is consistent with the presence of capacitors, which is why sine wave analysis is indispensable in understanding AC circuit behavior.