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Chapter 31: Problem 11

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 \(t=2.00 \mathrm{ms} ?\)

Short Answer

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

Step by step solution

01

Understand the Problem

We are given a 0.180-H inductor in series with a 90.0-惟 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 \(t = 2.00\, \text{ms}\).
02

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 \(v_{R}(t)\).
03

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:\[\frac{di}{dt} = -66.67\sin(480t)\]
04

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:\[i(t) = 0.1389\cos(480t)\]
05

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 \(R = 90\,\Omega\):\[v_{R}(t) = 0.1389\cos(480t) \times 90 = 12.5\cos(480t)\]
06

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:\[v_{R}(t) \approx 12.5 \times 0.5736 \approx 7.17\,\text{V}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inductance
Inductance is a property of an electrical circuit that resists changes in electric current. It鈥檚 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.

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Most popular questions from this chapter

A Step-Up Transformer. A transformer connected to a \(120-\mathrm{V}(\mathrm{rms})\) ac line is to supply \(13,000 \mathrm{V}(\mathrm{rms})\) for a neon sign. To reduce shock hazard, a fuse is to be inserted in the primary circuit; the fuse is to blow when the rms current in the secondary circuit; exceeds 8.50 \(\mathrm{mA}\) . (a) What is the ratio of secondary to primary turns of the transformer? (b) What power must be supplied to the transformer when the rms secondary current is 8.50 \(\mathrm{mA}\) ? (c) What current rating should the fuse in the primary circuit have?

A series circuit consists of an ac source of variable frequency, a \(115-\Omega\) resistor, a \(1.25-\mu F\) capacitor, and a \(4.50-\mathrm{mH}\) inductor. Find the impedance of this circuit when the angular frequency of the ac source is adjusted to (a) the resonance angular frequency; (b) twice the resonance angular frequency; (c) half the resonance angular frequency.

(a) What is the reactance of a \(3.00-\mathrm{H}\) inductor at a frequency of 80.0 \(\mathrm{Hz}\) ? (b) What is the inductance of an inductor whose reactance is 120\(\Omega\) at 80.0 \(\mathrm{Hz}\) ? (c) What is the reactance of a \(4.00-\mu \mathrm{F}\) capacitor at a frequency of 80.0 \(\mathrm{Hz}\) ? (d) What is the capacitance of a capacitor whose reactance is 120\(\Omega\) at 80.0 \(\mathrm{Hz} ?\)

An \(L \cdot R-C\) series circuit has \(R=300 \Omega .\) At the frequency of the source, the inductor has reactance \(X_{L}=900 \Omega\) and the capacitor has reactance \(X_{C}=500 \Omega .\) The amplitude of the voltage across the inductor is 450 V. (a) What is the amplitude of the voltage across the resistor? (b) What is the amplitude of the voltage across the capacitor? (c) What is the voltage amplitude of the source? (d) What is the rate at which the source is delivering electrical energy to the circuit?

A Step-Down Transformer. A transformer connected to a \(120-\mathrm{V}(\mathrm{rms})\) ac line is to supply 12.0 \(\mathrm{V}(\mathrm{rms})\) to a portable electronic device. The load resistance in the secondary is 5.00\(\Omega .\) (a) What should the ratio of primary to secondary turns of the transformer be? (b) What rms current must the secondary supply? (c) What average power is delivered to the load? (d) What resistance connected directly across the \(120-\mathrm{V}\) line would draw the same power as the transformer? Show that this is equal to 5.00\(\Omega\) times the square of the ratio of primary to secondary turns.
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