Problem 11 - University Physics with Modern Physics

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 \(v_{L}=-(12.0 \mathrm{V}) \sin [(480 \mathrm{rad} / \mathrm{s}) t]\) . (a) Derive an expression for the voltage \(v_{R}\) across the resistor. (b) What is \(v_{R}\) at \(t=2.00 \mathrm{ms} ?\)

Step-by-Step Solution

Answer

(a) \(v_{R}(t)=12.5\cos(480t)\); (b) \(v_{R}(t = 2.00 \text{ ms}) \approx 7.17\,\text{V}\)."

Step 1: Understand the Problem

We are given a 0.180-H inductor in series with a 90.0-Ω resistor and a voltage source. The inductor's voltage is given by \(v_{L} = -12.0 \sin(480t)\). We need to find the expression for the voltage across the resistor, \(v_{R}\), and its value at \(t = 2.00\, \text{ms}\).

Step 2: Analyze the Circuit

In a series circuit, the total voltage \(v(t)\) is the sum of the voltage across the resistor \(v_{R}(t)\) and the inductor \(v_{L}(t)\): \[v(t) = v_{R}(t) + v_{L}(t)\] Since the voltage source is AC, we need to express the current first and then use it to find \(v_{R}(t)\).

Step 3: Find the Current Expression

The voltage across the inductor is related to the current through it by the formula:\[v_{L}(t) = L \cdot \frac{di}{dt}\]Using \(v_{L}(t) = -12\sin(480t)\) and \(L = 0.180\,\text{H}\), solve the differential equation for \(i(t)\):\[-12\sin(480t) = 0.180 \cdot \frac{di}{dt}\]Leading to:\[\frac{di}{dt} = -66.67\sin(480t)\]

Step 4: Integrate to Find i(t)

Integrate the expression from Step 3 to find \(i(t)\):\[i(t) = \int -66.67\sin(480t) dt = \frac{66.67}{480} \cos(480t) + C\]To simplify, assuming \(C=0\) as there is no initial current, we have:\[i(t) = 0.1389\cos(480t)\]

Step 5: Find the Voltage Across the Resistor, v_R(t)

The voltage across the resistor \(v_{R}(t)\) is given by Ohm's Law:\[v_{R}(t) = i(t) \times R\]Substitute \(i(t) = 0.1389\cos(480t)\) and \(R = 90\,\Omega\):\[v_{R}(t) = 0.1389\cos(480t) \times 90 = 12.5\cos(480t)\]

Step 6: Find v_R at t = 2.00 ms

Using the expression for \(v_{R}(t)\), evaluate at \(t = 2.00\, \text{ms} = 0.002\,\text{s}\):\[v_{R}(t) = 12.5\cos(480 \times 0.002) = 12.5\cos(0.96)\]Calculating:\[v_{R}(t) \approx 12.5 \times 0.5736 \approx 7.17\,\text{V}\]

Key Concepts

Inductance

Inductance is a property of an electrical circuit that resists changes in electric current. It’s important in AC circuits where varying currents are typical. Inductance is measured in henrys (H). In our problem, we have a 0.180-H inductor. Inductors store energy in a magnetic field when current flows through them. This stored energy is represented by the equation:\[ v_L(t) = L \cdot \frac{di}{dt} \]Here, \(L\) is the inductance, \(\frac{di}{dt}\) represents the rate of change of current. Inductors in AC circuits oppose changes in current, functioning like energy buffers. This characteristic is crucial for understanding how the voltage across an inductor, \(v_L(t)\), is generated.- Inductors cause a phase shift in the current's sine wave.- They resist changes in current, creating counter-emf (electromotive force).- The frequency of AC signals affects the behaviour of inductors.

Ohm's Law

Ohm's Law is a fundamental principle used to calculate the relationship between voltage, current, and resistance in an electrical circuit. It is mathematically expressed as:\[ V = I \cdot R \]Where \(V\) stands for voltage, \(I\) for current, and \(R\) for resistance. In our exercise, Ohm's Law helps determine the voltage across the resistor:- The current flowing through the resistor and inductor is the same, as they're in series.- Ohm's Law simplifies finding the voltage across the resistor when current and resistance are known.

Voltage across Resistor

The voltage across a resistor in an AC circuit can be computed once you know the current through the resistor and its resistance. In the exercise, after finding the current expression, we used it to calculate the voltage across the resistor using Ohm's Law:- Determining \(i(t)\) was crucial for this calculation, found as \(0.1389\cos(480t)\)- Using \ i(t) imes R \, the voltage \(v_R(t)\) becomes \(12.5\cos(480t)\)This expression shows how the voltage fluctuates in time with the current in an AC circuit. At a specific moment, like \(t = 2.00 \, \text{ms}\), you can find the voltage's precise value by substituting directly into \(v_R(t)\). Calculating this gives insight into how resistors behave in time-varying circuits. - This concept highlights the oscillatory nature of voltage and current in AC.- A phase relationship is crucial, including how in-time resistance voltage follows the current's lead.