Problem 14 - College Physics

Question

A \(\mathrm{A} 200 \Omega\) resistor is in series with a 0.100 \(\mathrm{H}\) inductor and a 0.500\(\mu \mathrm{F}\) capacitor. Compute the impedance of the circuit and draw the phasor diagram (a) at a frequency of \(500 \mathrm{Hz},\) (b) at a frequency of 1000 \(\mathrm{Hz}\) . In each case, compute the phase angle of the source voltage with respect to the current and state whether the source voltage lags or leads the current.

Step-by-Step Solution

Answer

For 500 Hz, impedance is 489.91 Ω with voltage lagging. For 1000 Hz, impedance is 374.17 Ω with voltage leading.

Step 1: Calculate the Reactance of the Inductor and Capacitor

To find the impedance, we first need to calculate the reactance of the inductor \(X_L\) and the reactance of the capacitor \(X_C\). The formulas are:\[ X_L = 2\pi f L \quad \text{and} \quad X_C = \frac{1}{2\pi f C} \]where \(f\) is the frequency, \(L\) is the inductance, and \(C\) is the capacitance. Plug in the values to find \(X_L\) and \(X_C\):- For \(f = 500\, \mathrm{Hz} \): \[ X_L = 2\pi (500)(0.100) = 314.16 \,\Omega \] \[ X_C = \frac{1}{2\pi (500)(0.5 \times 10^{-6})} = 636.62 \,\Omega \]- For \(f = 1000\, \mathrm{Hz} \): \[ X_L = 2\pi (1000)(0.100) = 628.32 \,\Omega \] \[ X_C = \frac{1}{2\pi (1000)(0.5 \times 10^{-6})} = 318.31 \,\Omega \]

Step 2: Calculate the Total Impedance of the Circuit

The impedance \(Z\) of an RLC series circuit is calculated using the formula:\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]Where \(R\) is the resistance:- For \(f = 500\, \mathrm{Hz} \): \[ Z = \sqrt{200^2 + (314.16 - 636.62)^2} \approx 489.91 \,\Omega \]- For \(f = 1000\, \mathrm{Hz} \): \[ Z = \sqrt{200^2 + (628.32 - 318.31)^2} \approx 374.17 \,\Omega \]

Step 3: Compute the Phase Angle

The phase angle \(\theta\) of the circuit is given by:\[ \theta = \tan^{-1}\left(\frac{X_L - X_C}{R}\right) \]Calculate \(\theta\) for each frequency:- For \(f = 500\, \mathrm{Hz} \): \[ \theta = \tan^{-1}\left(\frac{314.16 - 636.62}{200}\right) \approx -30.42^\circ \] (Voltage lags)- For \(f = 1000\, \mathrm{Hz} \): \[ \theta = \tan^{-1}\left(\frac{628.32 - 318.31}{200}\right) \approx 30.42^\circ \] (Voltage leads)

Step 4: Draw the Phasor Diagram

In a phasor diagram:1. Use arrows to represent the values of \(R\), \(X_L\), and \(X_C\).2. Draw the inductor reactance \(X_L\) and capacitor reactance \(X_C\) perpendicular to the resistor \(R\).3. For \(f = 500\, \mathrm{Hz}\), the phasor for \(X_C\) is longer, indicating that the voltage lags.4. For \(f = 1000\, \mathrm{Hz}\), the phasor for \(X_L\) is longer, indicating that the voltage leads.

Key Concepts

Reactance of Inductor

The reactance of an inductor is the opposition that an inductor presents to alternating current (AC). It depends on both the inductance and the frequency of the current passing through it. The formula for calculating the inductive reactance \(X_L\) is:
\[X_L = 2\pi f L\]
Where:

• \(f\) is the frequency of the AC signal, measured in hertz (Hz).
• \(L\) is the inductance, measured in henrys (H).

The reactance increases with both frequency and inductance. This means that at higher frequencies, the inductor impedes AC more significantly. It's crucial to understand this concept because it affects the overall impedance of the RLC circuit. For example, with \(f = 500 \, \mathrm{Hz}\) and \(L = 0.100 \, \mathrm{H}\), we find \(X_L = 314.16 \, \Omega\), illustrating that the inductor strongly opposes AC even at moderate frequencies.

Reactance of Capacitor

Capacitive reactance is the opposition a capacitor offers to changes in voltage. It is inversely proportional to the frequency and capacitance. Use the following formula to calculate the capacitive reactance \(X_C\):
\[X_C = \frac{1}{2\pi f C}\]
Where:

• \(f\) is the frequency in hertz (Hz)
• \(C\) is the capacitance in farads (F)

As frequency increases, \(X_C\) decreases, meaning the capacitor allows AC to pass through more readily at higher frequencies. In an RLC circuit, the balance between \(X_L\) and \(X_C\) determines how the circuit behaves. For instance, at \(f = 500 \, \mathrm{Hz}\) with \(C = 0.500 \, \mu\mathrm{F}\), \(X_C = 636.62 \, \Omega\), and at \(f = 1000 \, \mathrm{Hz}\), \(X_C = 318.31 \, \Omega\), showing how responsive the capacitor becomes.

Phasor Diagram

A phasor diagram visually represents the relationships between different components in an AC circuit, like resistance, inductive, and capacitive reactance.
In the RLC circuit, you draw:

• Resistance \(R\) horizontally.
• Inductive reactance \(X_L\) and capacitive reactance \(X_C\) perpendicularly to \(R\).

These phasors help determine the net reactance and the phase relationship in the circuit. The elements are represented by arrows in the diagram:

• If \(X_C\) is longer than \(X_L\), the circuit behaves more like a capacitor, making the voltage lag.
• If \(X_L\) is longer, the inductor's influence dominates, causing voltage to lead the current.

By practicing with phasor diagrams, you can better visualize how changes in frequency affect impedance and the timing relationship between voltage and current.

Frequency

Frequency is the rate at which current changes direction in an AC circuit, measured in hertz (Hz). It plays a pivotal role in determining the reactances of both inductors and capacitors. In an RLC circuit, frequency changes the balance between \(X_L\) and \(X_C\).
At lower frequencies:

• \(X_L\) is smaller, and \(X_C\) is larger, making the circuit more capacitive overall.

At higher frequencies:

• \(X_L\) becomes significant, potentially reversing the circuit's behavior to inductive.

Frequency is crucial because it directly affects the overall impedance and phase behavior, dictating how efficiently power is transferred and how devices perform in AC environments.

Phase Angle

Phase angle \(\theta\) is a measure of the shift between the current and the voltage in an AC circuit, calculated using the formula:
\[\theta = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)\]
This angle tells us about the lead or lag relationship of the voltage with respect to the current:

• A positive \(\theta\) indicates voltage leads the current, often seen in inductive circuits.
• A negative \(\theta\) shows voltage lags, which is typical for capacitive circuits.

The phase angle depends on the relative sizes of \(X_L\), \(X_C\), and \(R\). When designing circuits, it's essential to consider the phase angle for applications like power electronics, as it influences the power factor and the ability of the circuit to efficiently handle AC power. For example, at \(f = 500 \, \mathrm{Hz}\), \(\theta\) is \(-30.42^\circ\), indicating a lagging voltage, while at \(f = 1000 \, \mathrm{Hz}\), \(\theta\) becomes positive, showing a leading voltage.