Problem 68 - Fundamentals of Physics

Question

A series \(R L C\) circuit is driven by a generator at a frequency of \(2000 \mathrm{~Hz}\) and an emf amplitude of \(170 \mathrm{~V}\). The inductance is \(60.0 \mathrm{mH},\) the capacitance is \(0.400 \mu \mathrm{F},\) and the resistance is \(200 \Omega .\) (a) What is the phase constant in radians? (b) What is the current amplitude?

Step-by-Step Solution

Answer

(a) Phase constant is approximately 0.359 radians. (b) Current amplitude is approximately 0.795 A.

Step 1: Calculate Angular Frequency

The frequency of the generator is given as \( f = 2000 \mathrm{~Hz} \). We calculate the angular frequency \( \omega \) using the formula: \[ \omega = 2 \pi f \]Substituting the given value, we have:\[ \omega = 2 \pi \times 2000 \]Thus, \( \omega = 4000 \pi \) rad/s.

Step 2: Find Reactance of Inductance and Capacitance

The inductive reactance \( X_L \) and capacitive reactance \( X_C \) are given by:\[ X_L = \omega L \quad \text{and} \quad X_C = \frac{1}{\omega C} \]Substituting the values:\[ X_L = 4000 \pi \times 0.060 = 240\pi \Omega \]For capacitance:\[ X_C = \frac{1}{4000 \pi \times 0.0000004} = \frac{1}{0.0016\pi} \approx \frac{625}{\pi} \Omega \approx 198.94 \Omega \]

Step 3: Calculate the Phase Constant

The phase constant \( \phi \) can be calculated using the formula:\[ \phi = \tan^{-1} \left( \frac{X_L - X_C}{R} \right) \]Substitute the values from the previous steps:\[ \phi = \tan^{-1} \left( \frac{240\pi - 198.94}{200} \right) \approx \tan^{-1} \left( \frac{75.68}{200} \right) \]Thus:\[ \phi \approx \tan^{-1}(0.3784) \approx 0.359 \text{ radians} \]

Step 4: Calculate the Current Amplitude

The current amplitude \( I_0 \) for an \( RLC \) circuit is given by:\[ I_0 = \frac{\text{emf amplitude}}{\sqrt{R^2 + (X_L - X_C)^2}} \]Substitute the values:\[ I_0 = \frac{170}{\sqrt{200^2 + 75.68^2}} \]Calculate the value:\[ I_0 = \frac{170}{\sqrt{40000 + 5723.4624}} \approx \frac{170}{\sqrt{45723.4624}} \approx \frac{170}{213.79} \approx 0.795 \text{ A} \]

Key Concepts

Phase Constant

When dealing with an RLC circuit, the phase constant is an important parameter that determines the phase difference between the voltage supplied by the generator and the current passing through the circuit. In simple terms, it helps us understand how much the current wave lags or leads the voltage wave.

To calculate the phase constant, we use:

• Inductive reactance ( \(X_L\) )
• Capacitive reactance ( \(X_C\) )
• Resistance ( \(R\) )

The formula used is:\[ \phi = \tan^{-1} \left( \frac{X_L - X_C}{R} \right) \]Here, \(\phi\) is the phase constant in radians. The difference between \(X_L\) and \(X_C\) determines if the circuit is predominantly inductive or capacitive—if \(X_L > X_C\), the circuit is inductive, and if \(X_C > X_L\), it is capacitive.

Knowing the phase constant can help better understand the power distribution in the circuit and is crucial for applications requiring precise control of electrical signals.

Inductive Reactance

Inductive reactance is the opposition that an inductor presents to the change in current flowing through it. It's a crucial aspect for AC circuits, such as RLC circuits, since it affects the amplitude and phase of the current.

The formula for calculating inductive reactance is:\[ X_L = \omega L \]Where:

• \(X_L\) is the inductive reactance, measured in ohms (\(\Omega\))
• \(\omega\) is the angular frequency ( \(\omega = 2\pi f\) )
• \(L\) is the inductance

Inductive reactance increases with higher frequencies and larger inductance values. This is because inductors resist changes in current more at high frequencies, thereby increasing their overall reactance.

Understanding this concept allows you to predict the behavior of your circuit under varying frequencies, enabling better control and design.

Capacitive Reactance

Capacitive reactance is the opposition to the change of voltage across an element. It's particularly relevant in AC circuits, impacting how the circuit reacts to different frequencies.

The capacitive reactance is calculated by:\[ X_C = \frac{1}{\omega C} \]Where:

• \(X_C\) is the capacitive reactance (\(\Omega\))
• \(\omega\) is the angular frequency
• \(C\) is the capacitance

As frequency increases, \(X_C\) decreases, which means capacitors offer less resistance to high-frequency signals. In contrast to inductors, capacitors store energy in the electric field created by a voltage across their plates.

Evaluating capacitive reactance helps in analyzing how a capacitor will influence the phase and amplitude of the current in an RLC circuit.

Current Amplitude

Current amplitude in an RLC circuit reflects the maximum current value that the circuit will experience, and it is closely tied to the reactive elements present in the circuit.

The current amplitude, \(I_0\), is determined using the equation:\[ I_0 = \frac{\text{emf amplitude}}{\sqrt{R^2 + (X_L - X_C)^2}} \]Here:

• \(\text{emf amplitude}\) is the maximum voltage from the generator
• \(R\) is the resistance
• \(X_L - X_C\) reflects the net reactance

Current amplitude depends on the balance between resistive and reactive components in the circuit. A larger difference between \(X_L\) and \(X_C\) leads to a decrease in \(I_0\), due to increased opposition to current flow.

Understanding current amplitude is essential for ensuring electrical safety, assessing circuit performance, and optimizing energy efficiency.