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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

Expert verified
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 Ω), inductance (L = 3.60 mH), capacitance (C = 5.00 µF), 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 µF, 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 Ω, 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) What is the reactance of a \(10.0-\mathrm{mH}\) inductor at the frequency \(f=250.0 \mathrm{Hz} ?\) (b) What is the impedance of a series combination of the \(10.0-\mathrm{mH}\) inductor and a \(10.0-\Omega\) resistor at $250.0 \mathrm{Hz} ?$ (c) What is the maximum current through the same circuit when the ac voltage source has a peak value of \(1.00 \mathrm{V} ?\) (d) By what angle does the current lag the voltage in the circuit?
A computer draws an rms current of \(2.80 \mathrm{A}\) at an \(\mathrm{rms}\) voltage of \(120 \mathrm{V}\). The average power consumption is 240 W. (a) What is the power factor? (b) What is the phase difference between the voltage and current?
A variable inductor can be placed in series with a lightbulb to act as a dimmer. (a) What inductance would reduce the current through a \(100-\) W lightbulb to \(75 \%\) of its maximum value? Assume a \(120-\mathrm{V}\) rms, \(60-\mathrm{Hz}\) source. (b) Could a variable resistor be used in place of the variable inductor to reduce the current? Why is the inductor a much better choice for a dimmer?
An x-ray machine uses \(240 \mathrm{kV}\) rms at \(60.0 \mathrm{mA}\) ms when it is operating. If the power source is a \(420-\mathrm{V}\) rms line, (a) what must be the turns ratio of the transformer? (b) What is the rms current in the primary? (c) What is the average power used by the \(x\) -ray tube?
A series combination of a \(22.0-\mathrm{mH}\) inductor and a \(145.0-\Omega\) resistor are connected across the output terminals of an ac generator with peak voltage \(1.20 \mathrm{kV},\) (a) \(\mathrm{At}\) \(f=1250 \mathrm{Hz},\) what are the voltage amplitudes across the inductor and across the resistor? (b) Do the voltage amplitudes add to give the source voltage (i.e., does \(\left.V_{R}+V_{L}=1.20 \mathrm{kV}\right) ?\) Explain. (c) Draw a phasor diagram to show the addition of the voltages.
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