Problem 32 - Fundamentals of Physics

Question

Go An ac generator has emf \(\mathscr{E}=\mathscr{E}_{m} \sin \omega_{d} t,\) with \(\mathscr{E}_{m}=25.0 \mathrm{~V}\) and \(\omega_{d}=377 \mathrm{rad} / \mathrm{s} .\) It is connected to a \(12.7 \mathrm{H}\) inductor. (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) 5.22 mA, (b) 0 V, (c) 4.52 mA

Step 1: Understand the emf equation and initial variables

The given equation for the emf is \( \mathscr{E} = \mathscr{E}_{m} \sin(\omega_{d} t) \), where \( \mathscr{E}_{m} = 25.0 \mathrm{~V} \) and \( \omega_{d} = 377 \mathrm{rad/s} \). This is a sinusoidal function that represents the voltage over time generated by the AC generator.

Step 2: Calculate inductive reactance

The inductive reactance \( X_L \) is calculated using the formula \( X_L = \omega_{d} L \) where \( L = 12.7 \mathrm{~H} \). Therefore, \( X_L = 377 \times 12.7 = 4787.9 \mathrm{~\, \Omega} \).

Step 3: Find the maximum current (Question a)

The maximum current \( I_m \) in an inductor can be found using the formula \( I_m = \frac{\mathscr{E}_m}{X_L} \). Plugging in the values, \( I_m = \frac{25.0}{4787.9} \approx 0.00522 \mathrm{~A} \), or 5.22 \mathrm{~mA}.

Step 4: Calculate the emf when current is maximum (Question b)

For an inductor, the current lags the voltage by \( \frac{\pi}{2} \). Thus, when the current is maximum, the emf \( \mathscr{E} = \mathscr{E}_m \sin(\omega_{d} t - \frac{\pi}{2}) = \mathscr{E}_m \cos(\omega_{d} t) = 0 \). So, the emf is \( 0 \mathrm{~V} \) when the current is maximum.

Step 5: Find current when emf is -12.5 V and increasing (Question c)

Given \( \mathscr{E} = -12.5 \mathrm{~V} \), use \( \mathscr{E} = \mathscr{E}_{m} \sin(\omega_{d} t) \) to find \( \sin(\omega_{d} t) = \frac{-12.5}{25.0} = -0.5 \). Since this is increasing in magnitude, the angle must be larger than \( \frac{3\pi}{2} \) but less than \( 2\pi \). So, \( \omega_{d} t \approx \frac{7\pi}{6} \) or \( 2\pi - \frac{\pi}{6} \), thus \( \cos(\omega_{d} t) = \frac{\sqrt{3}}{2} \). Therefore, \( I = I_m \cdot \frac{\sqrt{3}}{2} \approx 0.00452 \mathrm{~A} \), or 4.52 \mathrm{~mA}.

Key Concepts

EMF Calculation

The term "EMF" stands for electromotive force, which in the context of an AC generator is the voltage generated across its terminals. In this exercise, our generator produces an emf represented by the equation \( \mathscr{E} = \mathscr{E}_{m} \sin(\omega_{d} t) \). This equation shows a sine wave, where \( \mathscr{E}_{m} \) is the maximum (or peak) emf voltage.
Here, the maximum voltage \( \mathscr{E}_{m} \) is \( 25.0 \text{ V} \), and the angular frequency \( \omega_{d} \) is \( 377 \text{ rad/s} \), which dictates how fast the voltage oscillates over time.

• The peak EMF \( \mathscr{E}_{m} = 25.0 \text{ V} \)
• The angular frequency \( \omega_{d} = 377 \text{ rad/s} \)

The sinusoidal nature implies that, as time progresses, the voltage will fluctuate between its peak and negative peak values. The angular frequency dictates the rate of these fluctuations – higher values mean faster oscillations. Understanding this can help in predicting how the voltage will behave at any point in time, which is crucial for calculations involving AC circuits.

Inductive Reactance

Inductive reactance is a crucial concept when dealing with inductors in AC circuits. It effectively measures the opposition that an inductor presents to the flow of alternating current.
The formula to calculate inductive reactance \( X_L \) is given by:\[X_L = \omega_{d} L\]where \( L \) is the inductance and \( \omega_{d} \) is the angular frequency.
In this particular problem, we have:

• Inductance \( L = 12.7 \text{ H} \)
• Angular frequency \( \omega_{d} = 377 \text{ rad/s} \)

Calculating the inductive reactance, we find:\[X_L = 377 \times 12.7 = 4787.9 \underline{\phantom{xxx}} \Omega\]This value tells us how much resistance the inductor effectively presents to the current in the circuit. In simple terms, the higher the inductive reactance, the less current will flow through the circuit for a given voltage. This makes the concept of inductive reactance essential for understanding and designing AC circuits as it affects how electricity flows through components.

Maximum Current

In an AC circuit containing an inductor, the maximum current is an important parameter. It represents the highest amount of current that can flow through the circuit under given conditions. To calculate the maximum current \( I_m \), the formula used is:\[I_m = \frac{\mathscr{E}_m}{X_L}\]where \( \mathscr{E}_m \) is the maximum voltage and \( X_L \) is the inductive reactance.
By substituting the known values from our problem:

• \( \mathscr{E}_m = 25.0 \text{ V} \)
• \( X_L = 4787.9 \underline{\phantom{xxx}} \Omega \)

The maximum current is calculated as:\[I_m = \frac{25.0}{4787.9} \approx 0.00522 \text{ A} \text{ (or 5.22 mA)}\]The above equation elucidates that maximum current inversely depends on inductive reactance, meaning high reactance results in lower current. This calculation is critical when designing circuits to ensure that components can handle the maximum expected current, providing safety and reliability in electrical systems.