Problem 34 - Fundamentals of Physics

Question

An ac generator with emf \(\mathscr{E}=\mathscr{E}_{m} \sin \omega_{d} t,\) where \(\mathscr{E}_{m}=25.0 \mathrm{~V}\) and \(\omega_{d}=377 \mathrm{rad} / \mathrm{s},\) is connected to a \(4.15 \mu \mathrm{F}\) capacitor. (a) What is the maximum value of the current? (b) When the current is a maximum, what is the emf of the generator? (c) When the emf of the generator is \(-12.5 \mathrm{~V}\) and increasing in magnitude, what is the current?

Step-by-Step Solution

Answer

(a) 0.0391 A. (b) 0 V. (c) 0.0339 A.

Step 1: Understanding the given formula

The generator's electromotive force (emf) is given by the formula \( \mathscr{E} = \mathscr{E}_{m} \sin \omega_{d} t \), where \( \mathscr{E}_{m} = 25.0 \mathrm{~V} \) is the amplitude of the emf, and \( \omega_{d} = 377 \mathrm{rad/s} \) is the angular frequency. The system includes a capacitor with capacitance \( C = 4.15 \mu \mathrm{F} \).

Step 2: Finding the maximum current - Calculating capacitive reactance

First, we calculate the capacitive reactance \( X_{C} \) using the formula: \( X_{C} = \frac{1}{\omega_{d} C} \). Convert capacitance \( 4.15 \mu \mathrm{F} \) into farads: \( 4.15 \times 10^{-6} \mathrm{~F} \). Then, \[ X_{C} = \frac{1}{377 \times 4.15 \times 10^{-6}} \approx 639 \space \Omega. \]

Step 3: Finding the maximum current - Calculating maximum current

The maximum current \( I_{m} \) is given by \( I_{m} = \frac{\mathscr{E}_{m}}{X_{C}} \). Substitute the values: \( \mathscr{E}_{m} = 25.0 \mathrm{~V} \) and \( X_{C} = 639 \Omega: \) \[ I_{m} = \frac{25.0}{639} \approx 0.0391 \mathrm{~A} \].

Step 4: Determining the emf when current is maximum

In a capacitive circuit, the current leads the voltage by 90 degrees (or \( \frac{\pi}{2} \) radians). Thus, when the current is maximum, the emf (voltage) is zero. This is because \( \sin(\omega_{d} t + \frac{\pi}{2}) = 1 \), meaning \( \sin \omega_{d} t = 0 \).

Step 5: Calculating the current when emf is \(-12.5 \mathrm{~V}\)

Use the relation \( \mathscr{E} = \mathscr{E}_{m} \sin \omega_{d} t \). Given \( \mathscr{E} = -12.5 \mathrm{~V} \), solve for \( \sin \omega_{d} t = -0.5 \). Then, the current \( I \) is \( I = I_{m} \cos \omega_{d} t \). Since \( \omega_{d} t = \arcsin(-0.5) \), the corresponding cosine value is \( \cos(\arcsin(-0.5)) = \sqrt{1 - (-0.5)^2} = \sqrt{0.75} \). Therefore, \[ I = 0.0391 \times 0.866 = 0.0339 \mathrm{~A} \].

Key Concepts

Capacitive Reactance

When alternating current (AC) passes through a capacitor, it doesn't face resistance like in a regular conductor. Instead, it experiences something called capacitive reactance, which can be thought of as opposition to the current flow resulting from the capacitor.
The capacitive reactance, denoted as \(X_C\), depends on both the frequency of the AC supply and the capacitance of the capacitor. It is calculated using the formula:
\[ X_C = \frac{1}{\omega_d C} \]where \( \omega_d \) is the angular frequency and \( C \) is the capacitance.

• A higher frequency or smaller capacitance will lead to lower reactance, which means less opposition to the current.
• Conversely, lower frequency or larger capacitance leads to higher reactance.

In our example, a capacitor of \( 4.15 \mu F \) is connected to an AC source with a frequency corresponding to \( \omega_d = 377 \, \text{rad/s} \). This calculation gives us a reactance of about \( 639 \, \Omega \). This value indicates how much the capacitor impedes the flow of AC at the given frequency.
Understanding capacitive reactance is vital as it affects how the capacitor behaves in different circuits, impacting things like the maximum current that the circuit can handle.

Maximum Current in AC Circuits

In AC circuits, the maximum current is an essential factor, especially when considering how capacitors respond to alternating voltages.
The maximum current \( I_m \) in a capacitive circuit can be calculated using the relation:
\[ I_m = \frac{\mathscr{E}_m}{X_C} \]where \( \mathscr{E}_m \) is the peak voltage of the AC source and \( X_C \) is the capacitive reactance.

• In our exercise, the peak voltage is \( 25.0 \, \text{V} \), and the capacitive reactance is \( 639 \, \Omega \).
• This results in a maximum current of \( 0.0391 \, \text{A} \) or \( 39.1 \, \text{mA} \).

This maximum current occurs because the AC voltage changes direction periodically, and the capacitor charges and discharges. In capacitive circuits, the maximum current occurs when the voltage is momentarily zero but in the process of changing direction.
The relationship between voltage and current in an AC circuit with a capacitor is phase-shifted, which uniquely affects when the maximum current occurs compared to resistive circuits where the voltage and current are in phase.

Phase Difference in Capacitive Circuits

In capacitive AC circuits, it is crucial to understand that the current and voltage do not align perfectly; instead, they are out of phase by a specific angle. This phase difference is a key aspect of how these circuits function.

• For a purely capacitive circuit, the current leads the voltage by \( 90 \) degrees, or \( \frac{\pi}{2} \) radians. This means by the time the voltage reaches its peak, the current has already peaked and is on a decline.
• This characteristic phase difference is why when the current is at its maximum, the voltage from the generator is zero.

The phrase "leading current" manifests in how the capacitor charges up at the start of the voltage cycle, peaking in stored energy just as the input voltage begins returning to zero.
Understanding phase difference is vital when designing AC circuits, especially in applications involving resonance and filtering, where the timing of current versus voltage is critical. Handling these phase differences correctly can lead to more efficient and functional circuit designs, impacting power delivery, signal processing, and much more.