Question

In an \(L-R-C\) series circuit, the phase angle is 40.0\(^\circ\), with the source voltage leading the current. The reactance of the capacitor is 400 \(\Omega\), and the resistance of the resistor is 200 \(\Omega\). The average power delivered by the source is 150 W. Find (a) the reactance of the inductor, (b) the rms current, (c) the rms voltage of the source.

Short answer

(a) 568 Ω, (b) 1.25 A, (c) 328 V.

Step-by-step solution

Understanding the Problem

We are given an L-R-C circuit where the phase angle \( \phi \) between the source voltage and the current is 40.0° with the voltage leading the current. Also, we are given the reactance of the capacitor \( X_C = 400 \text{ } \Omega \), and the resistance \( R = 200 \text{ } \Omega \). The average power \( P = 150 \text{ } W \). We need to find the reactance of the inductor \( X_L \), the rms current \( I_{rms} \), and the rms voltage of the source \( V_{rms} \).

Calculate the Inductive Reactance

The phase angle \( \phi \) is given by \( \tan^{-1}\left(\frac{X_L - X_C}{R}\right) = 40.0^\circ \). First, convert the phase angle to radians: \( \phi = 40.0^\circ = \frac{40}{180}\pi \text{ rad} \).\Using the tangent function, the formula becomes: \[ \tan(\phi) = \frac{X_L - X_C}{R} \]Substituting the known values:\[ \tan\left(\frac{40\pi}{180}\right) = \frac{X_L - 400}{200} \]\Solving for \( X_L \):\[ X_L = 200 \tan\left(\frac{40\pi}{180}\right) + 400 \approx 568 \text{ } \Omega \]

Finding the RMS Current

The average power \( P \) is given by:\[ P = I_{rms}^2 R \cos(\phi) \]Rearranging to find \( I_{rms} \):\[ I_{rms} = \sqrt{\frac{P}{R \cos(\phi)}} \]Substitute the known values:\[ I_{rms} = \sqrt{\frac{150}{200 \cos\left(\frac{40\pi}{180}\right)}} \approx 1.25 \text{ } A \]

Calculate the RMS Voltage of the Source

Using the known formula for voltage in an L-R-C circuit:\[ V_{rms} = I_{rms} \sqrt{R^2 + (X_L - X_C)^2} \]Fill in the known values:\[ V_{rms} = 1.25 \times \sqrt{200^2 + (568 - 400)^2} \]Calculate the expression:\[ V_{rms} = 1.25 \times \sqrt{200^2 + 168^2} \approx 328 \text{ } V \]

Key concepts

Phase Angle in L-R-C Circuits

The phase angle in an L-R-C circuit represents the difference in phase between the source voltage and the current flowing through the circuit. It is a crucial concept in AC (Alternating Current) circuit analysis. The phase angle indicates how much the voltage leads (or lags) the current:

• A positive phase angle means the voltage leads the current.
• A negative phase angle means the current leads the voltage.

In this exercise, the phase angle is 40 degrees, meaning the voltage leads the current. This angle is related to the reactive components of the circuit—the inductor and capacitor—and their reactances. By using trigonometric functions, we derive the phase angle from the L-R-C circuit parameters. Specifically, the tangent of the phase angle can be calculated by the formula: \[\tan(\phi) = \frac{X_L - X_C}{R}\]where \(X_L\) is the inductive reactance, \(X_C\) is the capacitive reactance, and \(R\) is the resistance. Understanding this relationship allows us to calculate the missing parameters in the circuit.

Inductive Reactance (X_L)

Inductive reactance represents the "opposition" to the change in current by an inductor in an AC circuit. It is denoted as \(X_L\) and is measured in ohms (\(\Omega\)). The formula to calculate inductive reactance is:\[X_L = 2\pi f L\]where \(f\) is the frequency of the AC source and \(L\) is the inductance of the coil. In the context of the given problem, we determine the inductive reactance by using the known phase angle and the given circuit parameters like capacitance and resistance using the relation:\[\tan(\phi) = \frac{X_L - X_C}{R}\]By rearranging and solving this equation, we find that the inductive reactance \(X_L\) is 568 \(\Omega\). Inductive reactance plays a vital role in defining how an inductor affects the overall impedance of the circuit, hence influencing both the phase angle and the total power factor.

RMS Current (I_rms)

RMS stands for Root Mean Square, a statistical measure used widely in dealing with sinusoidal waves like those in AC circuits. The RMS current \(I_{rms}\) is a representation of the effective value of the alternating current, which amounts to the equivalent DC current that would produce the same power. The formula to determine the RMS current in an L-R-C circuit using the average power and component parameters is:\[I_{rms} = \sqrt{\frac{P}{R \cos(\phi)}}\]where \(P\) is the average power, \(R\) is the resistance, and \(\cos(\phi)\) is the power factor. In this problem, the RMS current comes out to be approximately 1.25 A. The significance of \(I_{rms}\) lies in its ability to simplify the calculation of AC parameters like power, no matter how complex the waveform of the current is.

RMS Voltage (V_rms)

The RMS voltage \(V_{rms}\) is a key factor in AC circuits and represents the effective voltage, similar to what RMS current does for current. It is used to quantify the net voltage level of an AC supply. The formula to calculate \(V_{rms}\) in a series L-R-C circuit is:\[V_{rms} = I_{rms} \times \sqrt{R^2 + (X_L - X_C)^2}\]This formula incorporates the impedance of the circuit, which includes both resistance and reactance. By substituting the previously found RMS current and reactances into this formula, we calculate that \(V_{rms}\) is approximately 328 V for the given circuit. Understanding RMS voltage helps in designing systems that rely on AC power, ensuring they can withstand the equivalent voltages they are subjected to.