Problem 4 - College Physics

Question

\(\bullet\) (a) Compute the reactance of a 0.450 \(\mathrm{H}\) inductor at frequencies of 60.0 \(\mathrm{Hz}\) and 600 \(\mathrm{Hz}\) . (b) Compute the reactance of a 2.50\(\mu \mathrm{F}\) capacitor at the same frequencies. (c) At what frequency is the reactance of a 0.450 \(\mathrm{H}\) inductor equal to that of a 2.50\(\mu \mathrm{F}\) capacitor?

Step-by-Step Solution

Answer

The reactance of the inductor is 169.65 ohms at 60 Hz and 1696.5 ohms at 600 Hz. The reactance of the capacitor is 1061.03 ohms at 60 Hz and 106.103 ohms at 600 Hz. At a frequency of 150.53 Hz, both reactances are equal.

Step 1: Inductive Reactance Formula

The formula to calculate the inductive reactance \( X_L \) is given by \( X_L = 2\pi f L \), where \( f \) is the frequency in hertz and \( L \) is the inductance in henrys.

Step 2: Calculate Reactance at 60 Hz

Using the formula \( X_L = 2\pi f L \), for \( f = 60 \) Hz and \( L = 0.450 \) H, we can calculate the inductive reactance: \[ X_L = 2 \pi \times 60 \times 0.450 = 169.65 \text{ ohms} \].

Step 3: Calculate Reactance at 600 Hz

Using the same formula \( X_L = 2\pi f L \), for \( f = 600 \) Hz and \( L = 0.450 \) H, the inductive reactance is: \[ X_L = 2 \pi \times 600 \times 0.450 = 1696.5 \text{ ohms} \].

Step 4: Capacitive Reactance Formula

The formula to calculate the capacitive reactance \( X_C \) is \( X_C = \frac{1}{2\pi f C} \), where \( C \) is the capacitance in farads.

Step 5: Calculate Capacitive Reactance at 60 Hz

Convert the capacitance \( C = 2.50 \mu F = 2.50 \times 10^{-6} \) F. Then use \( X_C = \frac{1}{2\pi f C} \) for \( f = 60 \) Hz: \[ X_C = \frac{1}{2\pi \times 60 \times 2.50 \times 10^{-6}} = 1061.03 \text{ ohms} \].

Step 6: Calculate Capacitive Reactance at 600 Hz

Using \( f = 600 \) Hz, calculate \( X_C \): \[ X_C = \frac{1}{2\pi \times 600 \times 2.50 \times 10^{-6}} = 106.103 \text{ ohms} \].

Step 7: Equating Reactances to Find Frequency

To find the frequency where the reactance of the inductor equals that of the capacitor, set \( X_L = X_C \): \( 2 \pi f L = \frac{1}{2\pi f C} \). Solving for \( f \), the frequency is \( f = \frac{1}{2\pi \sqrt{LC}} \).

Step 8: Calculate Frequency for Equal Reactance

Plug \( L = 0.450 \) H and \( C = 2.50 \times 10^{-6} \) F into \( f = \frac{1}{2\pi \sqrt{LC}} \): \[ f = \frac{1}{2\pi \sqrt{0.450 \times 2.50 \times 10^{-6}}} = 150.53 \text{ Hz} \].

Key Concepts

Inductive Reactance

Inductive reactance denotes the opposition that an inductor provides to the flow of alternating current (AC). It's a crucial concept when dealing with AC circuits. The formula for calculating inductive reactance, represented as \( X_L \), is \( X_L = 2\pi f L \). Here, \( f \) stands for the frequency in hertz, and \( L \) is the inductance measured in henrys. Understanding this formula helps grasp how reactance changes with frequency: as frequency increases, so does the reactance. ### Key Insights:- Inductive reactance is directly proportional to the frequency of the AC source.- Expressed in ohms, higher frequency means higher inductive reactance.Knowing this, when you compute the reactance for an inductor of 0.450 H at different frequencies like 60 Hz and 600 Hz, you can see a significant increase as the frequency rises from 169.65 ohms to 1696.5 ohms.

Capacitive Reactance

Capacitive reactance is the opposition a capacitor offers to the flow of AC. Unlike inductors, capacitors store energy in an electric field, which affects how they behave under AC. The capacitive reactance \( X_C \) is calculated with the formula \( X_C = \frac{1}{2\pi f C} \). Here, \( C \) represents capacitance in farads. This formula shows that capacitive reactance decreases with increasing frequency, highlighting a key difference from inductive reactance.### Highlights:- It is inversely proportional to both frequency and capacitance.- Higher frequency results in lower capacitive reactance, reflecting a capacitive "preference" for higher frequencies.Using this formula, at a frequency of 60 Hz, a capacitor of 2.50 \( \mu \text{F} \) yields a reactance of 1061.03 ohms. At 600 Hz, this reactance substantially lowers to 106.103 ohms, showing how capacitors behave with varying frequencies.

Frequency Dependence

Frequency dependence is crucial when analyzing reactance because an inductor's and a capacitor's reactance vary differently with changes in frequency. - **Inductive Reactance:** Increases linearly with an increase in frequency. This is because of the direct relationship in the formula \( X_L = 2\pi f L \).- **Capacitive Reactance:** Lowers as frequency increases, as seen in the formula \( X_C = \frac{1}{2\pi f C} \).### Understanding Impact:- When designing circuits, one must consider how each component's reactance will change with frequency.- Devices that operate under a broad range of frequencies, like radios and oscillators, rely on understanding frequency dependence thoroughly.The real test of frequency dependence is determining how components will behave in circuits transmitting signals across those frequencies.

Inductor Capacitor Reactance Equality

Inductor capacitor reactance equality occurs when the reactance of inductive and capacitive components in a circuit are equal, creating a resonant state. This is important in applications like filters and oscillators. At resonance, the entire circuit's impedance is at a minimum.The condition for equal reactance is given by equating the formulas for inductance and capacitance: \( 2\pi f L = \frac{1}{2\pi f C} \). Solving this equation determines the frequency at which both reactances are equal:\[ f = \frac{1}{2\pi \sqrt{LC}} \] ### Practical Application:- It demonstrates how circuits can be tuned to resonate at a desired frequency.- In the exercise solution, for a 0.450 H inductor and a 2.50 \( \mu \text{F} \) capacitor, the resonant frequency is computed to be 150.53 Hz, putting both elements precisely in balance.This understanding is fundamental when developing frequency-based technologies, ensuring devices operate optimally at their specified frequencies.