Problem 39 - College Physics

Question

\(\bullet\) In an \(R-L-C\) series circuit the magnitude of the phase angle is \(54.0^{\circ},\) with the source voltage lagging the current. The reactance of the capacitor is \(350 \Omega,\) and the resistor resistance is 180\(\Omega .\) The average power delivered by the source is 140 \(\mathrm{W}\) . Find (a) the reactance of the inductor; (b) the rms current; (c) the rms voltage of the source.

Step-by-Step Solution

Answer

(a) 598 Ω; (b) 1.153 A; (c) 351 V.

Step 1: Understanding the Problem

We have an RLC circuit with given parameters: the phase angle is \(54^{\circ}\), the capacitive reactance \(X_C = 350 \, \Omega\), the resistance \(R = 180 \, \Omega\), and the average power \(P = 140 \, \text{W}\). The voltage lags the current, indicating an inductive circuit.

Step 1: Determine the Inductive Reactance

Since the voltage lags, the algebraic sum of inductive and capacitive reactance should satisfy: \(\tan \phi = \frac{X_L - X_C}{R}\), where \(X_L\) is the inductive reactance. Substituting \(\phi = 54^{\circ}\) and \(X_C = 350\, \Omega\), \(R = 180\, \Omega\): \[\tan 54^{\circ} = \frac{X_L - 350}{180}\]Solve for \(X_L\) to find:\[X_L = 180\tan 54^{\circ} + 350\approx 180\times 1.3764 + 350\approx 598 \Omega\].

Step 2: Calculate the RMS Current

The average power \(P\) in an AC circuit is given by: \[P = I_{rms}^2 R\cos \phi\]Rearranging for \(I_{rms}\) gives:\[I_{rms} = \sqrt{\frac{P}{R \cos \phi}}\]Substituting \(P = 140 \, \text{W}, R = 180 \, \Omega, \cos 54^{\circ} = 0.5878\):\[I_{rms} = \sqrt{\frac{140}{180 \times 0.5878}} \approx \sqrt{1.3294} \approx 1.153 \, \text{A}\].

Step 3: Find the RMS Voltage of the Source

The source voltage \(V_{rms}\) relates to current and impedance \(Z\) as \(V_{rms} = I_{rms} Z\). Impedance is \(Z = \sqrt{R^2 + (X_L - X_C)^2}\). Substitute known values:\[Z = \sqrt{180^2 + (598 - 350)^2}\approx \sqrt{180^2 + 248^2} \approx \sqrt{32400 + 61504} \approx 305\, \Omega\].Thus, calculate:\[V_{rms} = 1.153 \times 305 \approx 351 \, \text{V}\].

Key Concepts

Phase Angle in RLC Circuits

In an RLC circuit, the phase angle (\( \phi \)) is the angle by which the source voltage lags or leads the current. This angle provides insight into the relationship between voltage and current in a circuit that contains resistors (R), inductors (L), and capacitors (C). A positive phase angle, where voltage lags current, indicates that the circuit is inductive. Conversely, a negative phase angle indicates a capacitive circuit.

• The phase angle is calculated using the formula: \( \tan \phi = \frac{X_L - X_C}{R} \), where \( X_L \) and \( X_C \) are the inductive and capacitive reactance, respectively, and \( R \) is the resistance.
• The given phase angle in our problem is \( 54^{\circ} \), indicating a strong inductive characteristic, as the voltage lags behind the current.

Understanding the phase angle helps in identifying whether the circuit behaves more like an inductor or a capacitor, which is crucial for analyzing and simulating RLC circuits effectively.

Capacitive and Inductive Reactance

The reactance in an RLC circuit is a measure of how much the circuit resists the change of current with respect to alternating current (AC).

• Inductive reactance (\( X_L \)) is due to the inductor in the circuit and is calculated using \( X_L = 2\pi f L \), where \( f \) is the frequency and \( L \) is the inductance.
• Capacitive reactance (\( X_C \)) arises from the capacitor and is calculated using \( X_C = \frac{1}{2\pi f C} \), where \( C \) is the capacitance.

In the given exercise, the voltage lags the current due to the inductive reactance being higher than the capacitive reactance, specifically with \( X_C = 350 \Omega \) and a calculated \( X_L = 598 \Omega \).
The difference (\( X_L - X_C \)) affects the net reactance and phase angle, showing the dominance of inductance in this situation.

Average Power in AC Circuits

Average power in AC circuits, particularly in RLC combinations, reflects the power consumed over a complete cycle. This power is related directly to both the resistive and reactive components.

• The formula for calculating the average power (\( P \)) is \( P = I_{rms}^2 R \cos \phi \), where \( I_{rms} \) is the root mean square current, \( R \) is resistance, and \( \cos \phi \) is the power factor.
• A power factor of \( \cos 54^{\circ} = 0.5878 \) suggests that 58.78% of the apparent power is being effectively used while the rest is reactive power.

This exercise provided an average power delivery by the source of 140 W, calculated by accommodating both resistance and phase angle.
This demonstrates the real power available to do work in the circuit.

RMS Current and Voltage Calculations

The RMS (root mean square) method is critical in determining effective values of voltage and current in AC systems.

• The RMS Current (\( I_{rms} \)) is calculated using the power formula: \( I_{rms} = \sqrt{\frac{P}{R \cos \phi}} \). In this exercise, it equates to about 1.153 A.
• Similarly, RMS Voltage (\( V_{rms} \)) is found by multiplying RMS current by the impedance (\( Z \)): \( V_{rms} = I_{rms} Z \), where the impedance takes into account both the resistive and reactive effects.

The RMS values are particularly useful as they represent direct current equivalence in terms of heating and power capability.
This allows for direct comparison and interoperability of equipment across AC and DC systems, with an RMS voltage of 351 V calculated here in the circuit.