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Chapter 21: Problem 44

An ac circuit contains a \(12.5-\Omega\) resistor, a \(5.00-\mu \mathrm{F}\) capacitor, and a \(3.60-\mathrm{mH}\) inductor connected in series to an ac generator with an output voltage of \(50.0 \mathrm{V}\) (peak) and frequency of \(1.59 \mathrm{kHz}\). Find the impedance,

Short Answer

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Question: Determine the impedance of an AC circuit containing a resistor, a capacitor, and an inductor connected in series given the values of resistance (R = 12.5 惟), inductance (L = 3.60 mH), capacitance (C = 5.00 碌F), voltage, and frequency (f = 1.59 kHz). Answer: The impedance of the AC circuit is approximately \(13.4 \Omega\).

Step by step solution

01

Calculate the angular frequency

Given the frequency f = 1.59 kHz, we can calculate the angular frequency, \(\omega\), as follows: \(\omega = 2 \pi f = 2 \pi (1.59 \times 10^3) = 10000 \pi\)
02

Calculate the inductive reactance

Using the value of \(\omega\) and the given inductance, L = 3.60 mH, we can calculate the inductive reactance, \(X_L\), as follows: \(X_L = \omega L = 10000 \pi (3.6 \times 10^{-3}) = 36 \pi\)
03

Calculate the capacitive reactance

Using the value of \(\omega\) and the given capacitance, C = 5.00 碌F, we can calculate the capacitive reactance, \(X_C\), as follows: \(X_C = \frac{1}{\omega C} = \frac{1}{10000 \pi (5 \times 10^{-6})} = \frac{1}{50 \pi} = \frac{2 \pi}{100}\)
04

Calculate the impedance

With the resistance, R = 12.5 惟, and the calculated values for \(X_L\) and \(X_C\), we can now calculate the total impedance, Z, using the formula: \(Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{(12.5)^2 + (36\pi - \frac{2\pi}{100})^2} = 13.4 \Omega\) So the impedance of the AC circuit is approximately \(13.4 \Omega\).

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Most popular questions from this chapter

A certain circuit has a \(25-\Omega\) resistor and one other component in series with a \(12-\mathrm{V}\) (rms) sinusoidal ac source. The rms current in the circuit is 0.317 A when the frequency is \(150 \mathrm{Hz}\) and increases by \(25.0 \%\) when the frequency increases to \(250 \mathrm{Hz}\). (a) What is the second component in the circuit? (b) What is the current at $250 \mathrm{Hz} ?$ (c) What is the numerical value of the second component?
A 40.0 -mH inductor, with internal resistance of \(30.0 \Omega\) is connected to an ac source $$\varepsilon(t)=(286 \mathrm{V}) \sin [(390 \mathrm{rad} / \mathrm{s}) t]$$ (a) What is the impedance of the inductor in the circuit? (b) What are the peak and rms voltages across the inductor (including the internal resistance)? (c) What is the peak current in the circuit? (d) What is the average power dissipated in the circuit? (e) Write an expression for the current through the inductor as a function of time.
What is the rins current flowing in a \(4.50-\mathrm{kW}\) motor connected to a \(220-\mathrm{V}\) rms line when (a) the power factor is 1.00 and (b) when it is \(0.80 ?\)
A 25.0 -mH inductor, with internal resistance of \(25.0 \Omega\) is connected to a \(110-\mathrm{V}\) ms source. If the average power dissipated in the circuit is \(50.0 \mathrm{W},\) what is the frequency? (Model the inductor as an ideal inductor in series with a resistor.)
Make a figure analogous to Fig. 21.4 for an ideal inductor in an ac circuit. Start by assuming that the voltage across an ideal inductor is \(v_{\mathrm{L}}(t)=V_{\mathrm{L}}\) sin \(\omega t .\) Make a graph showing one cycle of \(v_{\mathrm{L}}(t)\) and \(i(t)\) on the same axes. Then, at each of the times \(t=0, \frac{1}{8} T, \frac{2}{8} T, \ldots, T,\) indicate the direction of the current (or that it is zero), whether the current is increasing, decreasing, or (instantaneously) not changing, and the direction of the induced emf in the inductor (or that it is zero).
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