Problem 40 - College Physics

Question

A coil with a resistance of \(30 \Omega\) and an inductance of \(0.15 \mathrm{H}\) is connected to \(\mathrm{a} 120-\mathrm{V}, 60\) -Hz source. \((\mathrm{a})\) Is the phase angle of this circuit (1) positive, (2) zero, or (3) negative? Why? (b) What is the phase angle of the circuit? (c) How much rms current is in the circuit? (d) What is the average power delivered to the circuit?

Step-by-Step Solution

Answer

(a) Positive; (b) 62.34°; (c) 1.854 A; (d) 103.3 W.

Step 1: Determine Circuit Reactance

The reactance due to the inductance (known as inductive reactance), denoted as \(X_L\), is calculated using the formula \(X_L = 2\pi f L\), where \(f = 60 \text{ Hz}\) and \(L = 0.15 \text{ H}\). Thus, \(X_L = 2\pi \times 60 \times 0.15 = 56.55 \Omega\).

Step 2: Calculate Impedance

Impedance \(Z\) in an RL circuit is given by \(Z = \sqrt{R^2 + X_L^2}\), where \(R = 30 \Omega\). Substitute the values: \[ Z = \sqrt{30^2 + 56.55^2} = \sqrt{900 + 3196.3} = 64.73 \Omega \].

Step 3: Identify Phase Angle Type

The phase angle \(\phi\) is determined by the relationship \(\tan \phi = \frac{X_L}{R}\). Since \(X_L > 0\) and there is no capacitive reactance, the phase angle is positive (1).

Step 4: Calculate the Phase Angle

The phase angle \(\phi\) is calculated using \(\tan \phi = \frac{X_L}{R}\). Substitute the values to find \(\phi\): \[ \tan \phi = \frac{56.55}{30} = 1.885 \]Use the inverse tangent function to find \(\phi\): \(\phi = \tan^{-1}(1.885) = 62.34^\circ\).

Step 5: Find RMS Current

The RMS current \(I_{rms}\) is calculated using Ohm's Law for AC circuits: \(I_{rms} = \frac{V_{rms}}{Z}\). Here, \(V_{rms} = 120 \text{ V}\) and \(Z = 64.73 \Omega\). Therefore:\[ I_{rms} = \frac{120}{64.73} = 1.854 \text{ A} \].

Step 6: Determine Average Power

The average power \(P\) delivered to the circuit is given by \(P = I^2_{rms} R\). Use the calculated current and resistance:\[ P = (1.854)^2 \times 30 = 103.3 \text{ W} \].

Key Concepts

Inductive Reactance

Inductive reactance is a measure of the opposition that an inductor presents to alternating current (AC) due to its inductance. Unlike resistance, which depends on the resistive material, inductive reactance depends on the frequency of the AC and the inductance of the coil itself.
The formula to determine inductive reactance (\(X_L\)) is given by the expression:

• \(X_L = 2\pi f L\)

where:

• \(X_L\) is the inductive reactance (in ohms, \(\Omega\))
• \(f\) is the frequency of the AC (in hertz, Hz)
• \(L\) is the inductance of the coil (in henries, H)

In this exercise, with a frequency of 60 Hz and inductance of 0.15 H, the inductive reactance was calculated as 56.55 \(\Omega\). Inductive reactance increases with frequency. This is why higher frequencies make a greater impact in circuits with inductance.

Phase Angle Calculation

The phase angle is a crucial concept in AC circuits with inductors and resistors. It describes the difference in phase between the voltage across and the current through the circuit.
In an RL circuit, the phase angle \(\phi\) is calculated using the tangent of the phase angle:

• \(\tan \phi = \frac{X_L}{R}\)

where:

• \(\phi\) is the phase angle (in degrees)
• \(R\) is the resistance (in ohms, \(\Omega\))
• \(X_L\) is the inductive reactance (in ohms, \(\Omega\))

To find the actual phase angle, take the inverse tangent of the calculated value:

• \(\phi = \tan^{-1}(\frac{X_L}{R})\)

In this particular problem, with an inductive reactance of 56.55 \(\Omega\) and resistance of 30 \(\Omega\), the phase angle is calculated to be approximately 62.34 degrees. The positive phase angle indicates that the voltage leads the current.

RMS Current

RMS (Root Mean Square) Current is a statistical measure of the magnitude of a varying AC current. It provides a meaningful average value of AC currents, allowing for the calculation of power in these circuits.
In AC circuits, the RMS current \(I_{rms}\) is determined using the voltage across the circuit and the total impedance:

• \(I_{rms} = \frac{V_{rms}}{Z}\)

where:

• \(I_{rms}\) is the root mean square current (in amperes, A)
• \(V_{rms}\) is the root mean square voltage (in volts, V)
• \(Z\) is the impedance (in ohms, \(\Omega\))

In this example, with a voltage of 120 V and an impedance of 64.73 \(\Omega\), the RMS current calculated is approximately 1.854 A. RMS values are crucial for determining effective, usable power in AC circuits.

Average Power in AC Circuit

Average power in an AC circuit reveals how much power, on average, is being used over time. This is important for understanding how much energy is consumed or required by electrical appliances.
The average power \(P\) delivered to an RL circuit is given by the formula:

• \(P = I^2_{rms} R\)

where:

• \(P\) is the average power (in watts, W)
• \(I_{rms}\) is the root mean square current (in amperes, A)
• \(R\) is the resistance (in ohms, \(\Omega\))

For the given exercise, with the RMS current of approximately 1.854 A and a resistance of 30 \(\Omega\), the average power is 103.3 W. Understanding average power helps in designing components like fuses and in reducing energy waste.