Question

A 0.180-H inductor is connected in series with a 90.0-\(\Omega\) resistor and an ac source. The voltage across the inductor is \(v_L\) = -(12.0 V)sin[1480 rad/s)t]. (a) Derive an expression for the voltage \(v_R\) across the resistor. (b) What is \(v_R\) at t = 2.00 ms?

Short answer

Voltage across the resistor \( v_R(t) = 4.05 \cos(1480t) \). At \( t = 2.00 \, \text{ms} \), \( v_R = -3.98 \, \text{V} \).

Step-by-step solution

Understand the Given Information

We are given an inductor with inductance \( L = 0.180 \, \text{H} \) and a resistor with resistance \( R = 90.0 \, \Omega \) in series with an AC source. The voltage across the inductor is \( v_L(t) = -12.0 \, \text{V} \sin(1480 \, \text{rad/s} \cdot t) \). Our task is to derive an expression for the voltage \( v_R \) across the resistor, and then find \( v_R \) at \( t = 2.00 \, \text{ms} \).

Understand the Voltage Across the Inductor

The voltage across an inductor in a series RL circuit with angular frequency \( \omega \) is given as \( v_L(t) = -L \frac{di}{dt} \). Given \( v_L(t) = -12.0 \, \text{V} \sin(1480 \, \text{rad/s} \cdot t) \), we identify \( \omega = 1480 \, \text{rad/s} \). The current \( i(t) \) will be the product of this angular frequency. Thus, the current can be found using \( i(t) = I_m \cos(1480 \, t) \) where \( I_m \) is the current amplitude determined from \( v_L(t) \).

Find the Current Amplitude from Inductor Voltage

Since we know \( v_L(t) = -L \frac{di}{dt} \) and \( \omega = 1480 \, \text{rad/s} \), differentiate \( i(t) = I_m \cos(1480 \, t) \) with respect to \( t \). We have \( v_L(t) = -L\cdot(-I_m \cdot 1480 \, \sin(1480 \cdot t)) \). Given \( v_L(t) = -12.0 \sin(1480t) \), equate to \( L I_m 1480 \sin(1480t) \). Solve for \( I_m \), thus \( I_m = \frac{12.0}{1480 \cdot 0.180} \).

Calculate Current Amplitude

Calculate \( I_m \):\[I_m = \frac{12.0}{1480 \cdot 0.180} = \frac{12.0}{266.4} \approx 0.045 \text{ A}\]

Expression for Voltage Across the Resistor

The voltage across the resistor \( v_R(t) \) can be written as \( v_R(t) = i(t) \cdot R \), where \( i(t) = I_m \cos(1480t) \). Therefore, substitute the current amplitude and resistance into the expression:\[v_R(t) = 0.045 \cos(1480t) \cdot 90.0\] which simplifies to:\[v_R(t) = 4.05 \cos(1480t)\]

Plug in the Time to Find Voltage at t = 2.00 ms

To find \( v_R \) at \( t = 2.00 \, \text{ms} \), substitute \( t = 0.002 \) into the expression:\[v_R(0.002) = 4.05 \cos(1480 \cdot 0.002)\]Calculate \( 1480 \cdot 0.002 = 2.96 \, \text{radians} \), so:\[v_R(0.002) = 4.05 \cos(2.96)\]Using a calculator, compute \( \cos(2.96) \approx -0.984 \, \):\[v_R(0.002) \approx 4.05 \cdot (-0.984) \approx -3.98 \, \text{V}\]

Key concepts

Inductor in AC Circuit

The role of an inductor in an AC circuit is a fascinating subject. An inductor stores energy in its magnetic field when a current passes through it, resembling a spring that can be compressed or extended. In an AC circuit, an inductor's behavior becomes more dynamic due to the changing current.
When an alternating current flows through an inductor, the inductor resists changes in the current. This action results in a phase shift between the voltage across the inductor and the current through it. The voltage leads the current by 90 degrees, unlike resistors, where voltage and current are in phase. This phase shift is an important feature of inductors and affects the overall impedance in the circuit.
The voltage across the inductor can be described by the formula: \[ v_L(t) = -L \frac{di}{dt} \]where \( L \) is the inductance and \( \frac{di}{dt} \) is the rate of change of current. Understanding this relationship is crucial in analyzing AC circuits.

Voltage Across Resistor

In a series RL circuit, the voltage across the resistor, \( v_R \), is directly associated with the current flowing through the circuit. Ohm’s Law, which states \( v = iR \), applies here, but it must be conformed to the AC nature of the circuit.
The current in our AC circuit leads to a voltage across the resistor, which is in phase with the current. This is because resistors do not affect the phase of an electrical wave.
For our particular exercise, the expression for the voltage across the resistor can be derived from:\[ v_R(t) = i(t) \cdot R \]Substituting the values from the example, we obtain:\[ v_R(t) = 0.045 \cos(1480t) \cdot 90.0 \approx 4.05 \cos(1480t) \]This equation reflects how voltage across the resistor fluctuates in time, mirroring the behavior of the current.

Current Amplitude Calculation

To find the current amplitude, \( I_m \), in an RL circuit, understanding the relationship between the inductor's voltage and the current is essential. In the provided problem, this involves using the voltage across the inductor, \( v_L(t) \), which is given as a sinusoidal function of time.
We start with the given expression:\[ v_L(t) = -L \cdot \frac{di}{dt} \]Substituting the differentiation of the current equation with the known values allows us to solve for \( I_m \). By differentiating \( i(t) = I_m \cos(1480t) \) and matching it to \( v_L(t) \), you form:\[ -12.0 \sin(1480t) = -0.180 \cdot I_m \cdot 1480 \sin(1480t) \]This leads to the calculation:\[ I_m = \frac{12.0}{0.180 \cdot 1480} \approx 0.045 \, \text{A} \]The current amplitude, \( I_m = 0.045 \, \text{A} \), represents the peak value of the sinusoidal current in the circuit. Recognizing how \( I_m \) is derived helps in further calculations like voltage across other components.

Angular Frequency

Angular frequency, often denoted as \( \omega \), is a crucial element in AC circuit analysis that quantifies how fast the current or voltage waveform oscillates. Angular frequency is measured in radians per second and differs from standard frequency, which is measured in Hertz.
In our exercise, the angular frequency \( \omega \) was identified as 1480 rad/s from the expression for voltage across the inductor. This indicates that the waveform completes 1480/2π cycles per second.
Angular frequency helps determine the reactance of components like inductors, which affects how much they impede AC. For an inductor, the reactance is calculated as:\[ X_L = \omega L \]Reactance is frequency-dependent, effectively meaning that as the angular frequency increases, an inductor offers more resistance to AC, affecting both current amplitude and phase relationships in the circuit.
Understanding angular frequency allows one to comprehend deeply how component behavior changes with the frequency of the AC source.