Question

When a solenoid is connected to a 48.0-V dc battery that has negligible internal resistance, the current in the solenoid is 5.50 A. When this solenoid is connected to an ac source that has voltage amplitude 48.0 V and angular frequency 20.0 rad/s, the current in the solenoid is 3.60 A. What is the inductance of this solenoid?

Short answer

The inductance of the solenoid is approximately 0.504 H.

Step-by-step solution

Identify Given Values

First, let's identify the values given in the problem. We have:- DC Voltage (\( V_{dc} \) = 48.0 V),- DC Current (\( I_{dc} \) = 5.50 A),- AC Voltage Amplitude (\( V_{ac} \) = 48.0 V),- AC Current (\( I_{ac} \) = 3.60 A),- Angular Frequency (\( \omega \) = 20.0 rad/s).We need to find the inductance (\( L \)) of the solenoid.

Calculate Resistance Using DC Circuit

Using Ohm's law in the DC circuit, where resistance (\( R \)) is given by\[ R = \frac{V_{dc}}{I_{dc}} \]we can substitute the known values:\[ R = \frac{48.0 \text{ V}}{5.50 \text{ A}} = 8.727 \text{ ohms} \].This is the resistance of the solenoid.

Use AC Circuit to Find Impedance

In an AC circuit, impedance (\( Z \)) is given by\[ Z = \frac{V_{ac}}{I_{ac}} \]Substituting the known values:\[ Z = \frac{48.0 \text{ V}}{3.60 \text{ A}} = 13.333 \text{ ohms} \].

Relate Impedance to Inductive Reactance

Impedance in an RL circuit is defined by:\[ Z = \sqrt{R^2 + (\omega L)^2} \]Substitute the values found previously:\[ 13.333 = \sqrt{(8.727)^2 + (20.0L)^2} \].

Solve for Inductance

Square both sides and solve for \( L \):\[ 13.333^2 = (8.727)^2 + (20.0L)^2 \177.689 = 76.153 + 400L^2 \101.536 = 400L^2 \L^2 = \frac{101.536}{400} \L = \sqrt{0.2538} \L \approx 0.504 \text{ H} \].

Finalize

Thus, the inductance of the solenoid is approximately \( 0.504 \text{ H} \).

Key concepts

Ohm's Law

Ohm's Law is a fundamental principle in electrical engineering that allows us to relate voltage, current, and resistance within a circuit. It is represented by the formula: \( V = IR \). Here, \( V \) represents the voltage across the circuit, \( I \) is the current flowing through it, and \( R \) is the resistance. This simple formula helps to determine one of these values if the other two are known.

In the context of our exercise, we utilized Ohm’s Law to find the resistance of the solenoid using the given DC circuit values. By substituting the known voltage (48.0 V) and current (5.50 A) into the formula, we calculated the resistance to be 8.727 ohms. This resistance is an essential step in further calculations involving the RL circuit.

Impedance

Impedance is a broader concept similar to resistance but applicable to AC circuits. It measures the opposition that a circuit offers to the flow of alternating current and is represented as \( Z \). Impedance combines both resistance (R) and reactance (X), given as \( Z = R + jX \), where \( j \) is the imaginary unit.

In the context of RL circuits, impedance is defined as \( Z = \sqrt{R^2 + (\omega L)^2} \). In the exercise, we calculated impedance using the AC voltage amplitude (48.0 V) and current (3.60 A). The impedance was found to be 13.333 ohms. Calculating impedance was key to eventually determining the inductance of the solenoid because it incorporates both the resistive and inductive elements of the circuit.

RL Circuit

An RL circuit includes a resistor (R) and an inductor (L) connected in series or parallel to an AC source. These circuits are used to filter signals, manage power flow, and in tuning applications. The unique feature of an RL circuit lies in how it handles alternating current. The inductor creates a magnetic field as the current passes through, resisting changes in the current flow.

In our problem, we dealt with an RL series circuit. The inductance was determined by solving an equation derived from the impedance formula \( Z = \sqrt{R^2 + (\omega L)^2} \). By substituting the known values of impedance (13.333 ohms), resistance (8.727 ohms), and angular frequency (20.0 rad/s), we solved for the inductance, establishing a basis for understanding how RL circuits manage electrical currents.

Inductive Reactance

Inductive reactance is the opposition that an inductor provides to the change in current in an AC circuit. It is denoted by \( X_L \) and calculated using the formula \( X_L = \omega L \), where \( \omega \) is the angular frequency and \( L \) is the inductance. Inductive reactance causes the current to lag behind the voltage in an RL circuit, which is an important characteristic in AC systems.

For our exercise, we combined this concept with the impedance formula to find the inductance of the solenoid. We rearranged the impedance equation to solve for the inductive reactance and eventually the inductance. Inductive reactance plays a crucial role in determining how much the inductor in an RL circuit resists the flow of alternating current, which directly contributes to calculating the inductance value needed to solve the exercise.