Problem 13 - College Physics

Question

\(\bullet\) (a) Compute the impedance of a series \(R-L-C\) circuit at angular frequencies of \(1000,750,\) and 500 \(\mathrm{rad} / \mathrm{s} .\) Take \(R=\) \(200 \Omega, L=0.900 \mathrm{H},\) and \(C=2.00 \mu \mathrm{F}\) . (b) Describe how the current amplitude varies as the angular frequency of the source is slowly reduced from 1000 rad/s to 500 rad/s. (c) What is the phase angle of the source voltage with respect to the current when \(\omega=1000 \mathrm{rad} / \mathrm{s} ?\) (d) Construct a phasor diagram when \(\omega=1000 \mathrm{rad} / \mathrm{s}\)

Step-by-Step Solution

Answer

Impedances are 500, 200.17, and 550.91 Ω at specified frequencies; current peaks at 750 rad/s. Phase angle is 63.43°.

Step 1: Understanding Impedance in a Series R-L-C Circuit

Impedance in a series R-L-C circuit is a combination of resistance (R), inductive reactance \( X_L \), and capacitive reactance \( X_C \). It is given by the formula \( Z = \sqrt{R^2 + (X_L - X_C)^2} \), where \( X_L = \omega L \) and \( X_C = \frac{1}{\omega C} \). We'll calculate this for different angular frequencies \( \omega \).

Step 2: Calculation of Impedance for Different Angular Frequencies

Compute \( Z \) for \( \omega = 1000, 750, \) and \( 500 \; \mathrm{rad/s} \). For each frequency:1. Calculate \( X_L = \omega L \).2. Calculate \( X_C = \frac{1}{\omega C} \).3. Find \( Z = \sqrt{R^2 + (X_L - X_C)^2} \).- \( \omega = 1000 \, \mathrm{rad/s} \): \( X_L = 900 \, \Omega \), \( X_C = 500 \, \Omega \), so \( Z = \sqrt{200^2 + (900-500)^2} = \sqrt{200^2 + 400^2} = \sqrt{200000} = 500 \, \Omega \).- \( \omega = 750 \, \mathrm{rad/s} \): \( X_L = 675 \, \Omega \), \( X_C = 666.67 \, \Omega \), so \( Z = \sqrt{200^2 + (675-666.67)^2} = \sqrt{200^2 + 8.33^2} \approx 200.17 \, \Omega \).- \( \omega = 500 \, \mathrm{rad/s} \): \( X_L = 450 \, \Omega \), \( X_C = 1000 \, \Omega \), so \( Z = \sqrt{200^2 + (450-1000)^2} = \sqrt{200^2 + (-550)^2} = \sqrt{302500} \approx 550.91 \, \Omega \).

Step 3: Analyzing Current Amplitude Variations

Current amplitude \( I \) in a circuit is inversely related to impedance \( Z \), given by \( I = \frac{V}{Z} \) where \( V \) is the voltage amplitude. As \( \omega \) decreases from 1000 to 500 rad/s, impedance first decreases then increases, suggesting that current amplitude initially increases then decreases. Specifically, maximum current occurs nearest resonance, seen at \( \omega = 750 \, \mathrm{rad/s} \) where \( Z \) is minimum.

Step 4: Calculating Phase Angle at \( \omega = 1000 \, \mathrm{rad/s} \)

The phase angle \( \phi \) between the source voltage and current is given by \( \tan^{-1}\left(\frac{X_L - X_C}{R}\right) \). For \( \omega = 1000 \, \mathrm{rad/s} \), calculate:\(\phi = \tan^{-1}\left(\frac{900 - 500}{200}\right) = \tan^{-1}(2) \approx 63.43^\circ\).This positive angle indicates that the circuit is inductive at this frequency.

Step 5: Constructing the Phasor Diagram for \( \omega = 1000 \, \mathrm{rad/s} \)

The phasor diagram represents current \( I \) as the reference vector horizontally to the right. The voltage phasors for the resistor \( V_R = I \times R \), the inductor \( V_L = I \times \omega L \), and the capacitor \( V_C = I \times \frac{1}{\omega C} \) are then plotted. Since \( \phi \approx 63.43^\circ \) indicates a larger inductive component:1. \( V_R \) is along \( I \).2. \( V_L \) leads \( I \) by \( 90^\circ \), add vertically upwards.3. \( V_C \) lags \( I \) by \( 90^\circ \), add vertically downwards.Overall, \( V \) (source voltage) leads \( I \) by \( \phi \).

Key Concepts

Phase Angle in AC Circuits

In alternating current (AC) circuits, the phase angle \( \phi \) is a vital indicator of the relationship between the voltage and the current. This angle helps determine whether the circuit behaves inductively or capacitively at a given frequency. A positive phase angle signifies that the circuit is more inductive, meaning that the current lags behind the voltage. Conversely, a negative phase angle reveals a capacitive nature, where the current precedes the voltage.

To calculate the phase angle \( \phi \) in an RLC circuit, we use the formula:\[\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)\]where:

• \( X_L = \omega L \) is the inductive reactance.
• \( X_C = \frac{1}{\omega C} \) is the capacitive reactance.
• \( R \) is the resistance.

For example, at \( \omega = 1000 \, \mathrm{rad/s} \), if \( X_L = 900 \, \Omega \) and \( X_C = 500 \, \Omega \), with \( R = 200 \, \Omega \), the phase angle becomes:\[\phi = \tan^{-1}\left(\frac{900 - 500}{200}\right) = \tan^{-1}(2) \approx 63.43^\circ\]This illustrates an inductive behavior, implying that the source voltage leads the current by this angle.

Current Amplitude in RLC Circuits

In RLC circuits, the current amplitude \( I \) fluctuates based on the impedance \( Z \) as the frequency changes. The relationship is explained by the formula:\[I = \frac{V}{Z}\]where \( V \) is the voltage amplitude across the circuit. Understanding this, we see that the current amplitude is inversely proportional to the impedance.

As the angular frequency \( \omega \) is decreased from 1000 to 500 rad/s, the trend in the current amplitude becomes clearer. Initially, the impedance \( Z \) falls, leading to increased current amplitude. However, beyond a certain point, the impedance rises again, reducing the current. This pattern implies that the maximum current amplitude occurs at the circuit's resonant frequency, which is often found where the impedance is at its minimum. For the given problem, this resonance was observed around \( \omega = 750 \, \mathrm{rad/s} \).

Therefore, understanding current amplitude in RLC circuits provides insights into peak performance and efficiency at different frequencies.

Phasor Diagrams

Phasor diagrams offer a graphical representation of the phase relationships between voltage and current within AC circuits. By using these diagrams, students can visualize how components like resistors, capacitors, and inductors interact within a circuit. In a phasor diagram, vectors represent the magnitudes and phases of voltages or currents.

When constructing a phasor diagram at \( \omega = 1000 \, \mathrm{rad/s} \) for a series RLC circuit:

• The current \( I \) vector is typically placed along the horizontal axis, as the reference.
• The resistor voltage \( V_R = I \times R \) aligns directly with \( I \).
• For the inductor, \( V_L = I \times \omega L \) is drawn perpendicular to \( I \), pointing upwards to indicate it is leading by \( 90^\circ \).
• The capacitor's voltage \( V_C = I \times \frac{1}{\omega C} \) also appears perpendicular, but downwards, demonstrating it lags \( 90^\circ \).

These vectors together establish the net voltage, \( V \), which forms the hypotenuse of the phasor triangle, showing the phase difference \( \phi \) with the current. This insightful tool aids in comprehending how voltage components vs. current behavior shift across different circuit elements.