Problem 13 - College Physics

Question

(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 \(1000 \mathrm{rad} / \mathrm{s}\) to \(500 \mathrm{rad} / \mathrm{s}\). (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

Impedance at 1000, 750, 500 rad/s are 1008.1 Ω, 761.0 Ω, and 503.5 Ω respectively. Current amplitude increases as frequency decreases. Phase angle at 1000 rad/s is 80.8°.

Step 1: Understanding Impedance Calculation

To compute the impedance of a series \( R-L-C \) circuit, use the formula \( Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} \), where \( \omega \) is the angular frequency, \( L \) is the inductance, and \( C \) is the capacitance.

Step 2: Calculate Impedance at 1000 rad/s

Substitute \( \omega = 1000 \), \( R = 200 \ \Omega \), \( L = 0.9 \ \text{H} \) and \( C = 2 \times 10^{-6} \ \text{F} \) into the impedance formula. Calculate: \[ Z = \sqrt{200^2 + (1000 \times 0.9 - \frac{1}{1000 \times 2 \times 10^{-6}})^2} \]. Upon calculation, \( Z \approx 1008.1 \ \Omega \).

Step 3: Calculate Impedance at 750 rad/s

Substitute \( \omega = 750 \), with the same \( R, L, C \) values. Calculate: \[ Z = \sqrt{200^2 + (750 \times 0.9 - \frac{1}{750 \times 2 \times 10^{-6}})^2} \]. Upon calculation, \( Z \approx 761.0 \ \Omega \).

Step 4: Calculate Impedance at 500 rad/s

Substitute \( \omega = 500 \), with the same \( R, L, C \) values. Calculate: \[ Z = \sqrt{200^2 + (500 \times 0.9 - \frac{1}{500 \times 2 \times 10^{-6}})^2} \]. Upon calculation, \( Z \approx 503.5 \ \Omega \).

Step 5: Determine Current Amplitude Variation

The impedance \( Z \) changes as \( \omega \) decreases. As \( Z \) decreases from \( 1000 \ \Omega \) to \( 503.5 \ \Omega \), the current amplitude \( I = \frac{V}{Z} \) increases since \( Z \) is in the denominator.

Step 6: Calculate Phase Angle at 1000 rad/s

The phase angle \( \phi \) is calculated using \( \tan(\phi) = \frac{\omega L - \frac{1}{\omega C}}{R} \). Substitute \( \omega = 1000 \), \( L = 0.9 \), \( C = 2 \times 10^{-6} \), and \( R = 200 \) to get \[ \phi = \arctan\left( \frac{1000 \times 0.9 - \frac{1}{1000 \times 2 \times 10^{-6}}}{200} \right) \approx 80.8^\circ \].

Step 7: Construct Phasor Diagram at 1000 rad/s

Draw vectors representing \( V_R = I R \), \( V_L = I \omega L \), and \( V_C = \frac{I}{\omega C} \). The net voltage \( V \), obtained by vector addition, leads \( V_R \) by the calculated phase angle \( 80.8^\circ \). The phasor \( V_L \) is perpendicular and leading \( V_R \), while \( V_C \) is perpendicular but lagging.

Key Concepts

Impedance Calculation

In an R-L-C circuit, impedance refers to how much the circuit resists the flow of electrical current. Impedance combines both resistance and reactance components and is represented by the symbol \( Z \). The formula for calculating the impedance in an R-L-C series circuit is:
\[ Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} \]
Here, \( R \) is resistance, \( \omega \) is the angular frequency, \( L \) is inductance, and \( C \) is capacitance.

The impedance is a critical factor in determining how the circuit responds to different frequencies. As the angular frequency \( \omega \) changes, both the inductive reactance \( \omega L \) and the capacitive reactance \( \frac{1}{\omega C} \) change, thereby affecting the overall impedance.

Angular Frequency

Angular frequency \( \omega \) is a measure of how fast the alternating current oscillates. It is expressed in radians per second and is related to the frequency \( f \) of the wave by \( \omega = 2\pi f \).

In circuit analysis, angular frequency plays a crucial role by influencing the reactance values:

• Inductive reactance \( X_L = \omega L \) increases with \( \omega \).
• Capacitive reactance \( X_C = \frac{1}{\omega C} \) decreases as \( \omega \) increases.

Understanding angular frequency helps predict how a circuit behaves at various frequencies, which is vital for tuning circuits to desired outcomes such as resonance.

Current Amplitude Variation

The current amplitude in a series R-L-C circuit varies inversely with the impedance \( Z \). As the impedance decreases, the current amplitude \( I = \frac{V}{Z} \) increases, given a constant source voltage \( V \).

When the angular frequency changes from high to low, such as from 1000 rad/s to 500 rad/s, we typically observe:

• A decrease in impedance due to the decreasing frequency causing less dominance of inductive reactance.
• Consequently, an increase in current amplitude since the circuit becomes less resistive to current flow.

This relationship is crucial for applications requiring control over the circuit's current flow, like radio tuners and oscillators.

Phase Angle

The phase angle \( \phi \) in an R-L-C circuit indicates the phase difference between the source voltage and the current. Calculated using:
\[ \tan(\phi) = \frac{\omega L - \frac{1}{\omega C}}{R} \]
The phase angle reveals how much the voltage leads or lags the current.

• If \( \phi > 0 \), the circuit is inductive, meaning the voltage leads the current.
• If \( \phi < 0 \), the circuit is capacitive, where the current leads the voltage.
• A phase angle of 0 indicates a purely resistive circuit, with voltage and current perfectly in phase.

Correct phase angle calculation is crucial in designing circuits to ensure optimal performance.

Phasor Diagram

A phasor diagram provides a visual representation of the relationships between voltage and current in an R-L-C circuit. In the diagram:

• Vectors represent the voltage across resistors \( V_R \), inductors \( V_L \), and capacitors \( V_C \).
• The resistive component \( V_R \) is along the x-axis.
• The inductive voltage \( V_L \) is perpendicular to \( V_R \), drawn above the x-axis.
• The capacitive voltage \( V_C \) is also perpendicular but below the x-axis.

The resultant phasor shows the total voltage \( V \), which usually leads or lags the resistive voltage by the phase angle \( \phi \). Phasor diagrams are helpful tools to easily comprehend how different components influence the circuit's timing and response.