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Chapter 21: Problem 78

A large coil used as an electromagnet has a resistance of \(R=450 \Omega\) and an inductance of \(L=2.47 \mathrm{H} .\) The coil is connected to an ac source with a voltage amplitude of \(2.0 \mathrm{kV}\) and a frequency of $9.55 \mathrm{Hz}$. (a) What is the power factor? (b) What is the impedance of the circuit? (c) What is the peak current in the circuit? (d) What is the average power delivered to the electromagnet by the source?

Short Answer

Expert verified
Based on the given problem and analysis, provide the power factor, impedance, peak current, and average power delivered to the electromagnet by the AC source.

Step by step solution

01

Find the reactance of the inductor

To find the reactance of the inductor, we use the formula: \(\textit{X}_{L} = 2\pi fL\), where \(f = 9.55 \mathrm{Hz}\) is the frequency, and \(L = 2.47 \mathrm{H}\) is the inductance. \(\textit{X}_{L} = 2\pi(9.55)(2.47) = 149.1 \Omega\)
02

Find the impedance of the circuit

To find the impedance of the circuit, we use the formula: \(Z=\sqrt{R^2 + \textit{X}_{L}^2}\), where \(R = 450 \Omega\) is the resistance, and \(\textit{X}_{L} = 149.1 \Omega\) is the reactance. \(Z=\sqrt{(450)^2 + (149.1)^2} = 468.18 \Omega\)
03

Find the power factor

To find the power factor, we use the formula: \(\textit{PF} = \frac{R}{Z}\), where \(R = 450 \Omega\) is the resistance, and \(Z = 468.18 \Omega\) is the impedance. \(\textit{PF} = \frac{450}{468.18} = 0.961\)
04

Find the peak current

To find the peak current, we use Ohm's Law: \(I=\frac{V}{Z}\), where \(V = 2.0 \mathrm{kV}=2000\mathrm{V}\) is the voltage amplitude, and \(Z = 468.18 \Omega\) is the impedance. \(I=\frac{2000}{468.18} = 4.27 \mathrm{A}\)
05

Find the average power delivered to the electromagnet

To find the average power, we use the formula: \(P=\textit{IV}\textit{PF}\), where \(I = 4.27 \mathrm{A}\) is the peak current, \(V = 2000\mathrm{V}\) is the voltage amplitude, and \(\textit{PF} = 0.961\) is the power factor. \(P=(4.27)(2000)(0.961) = 8220.14 \mathrm{W}\) So, the power factor is 0.961, the impedance is 468.18\(\Omega\), the peak current is 4.27 A, and the average power delivered to the electromagnet is 8220.14 W.

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