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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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Most popular questions from this chapter

A \(0.400-\mu \mathrm{F}\) capacitor is connected across the terminals of a variable frequency oscillator. (a) What is the frequency when the reactance is \(6.63 \mathrm{k} \Omega ?\) (b) Find the reactance for half of that same frequency.
The instantaneous sinusoidal emf from an ac generator with an mos emf of $4.0 \mathrm{V}$ oscillates between what values?
A 0.48 - \(\mu\) F capacitor is connected in series to a $5.00-\mathrm{k} \Omega\( resistor and an ac source of voltage amplitude \)2.0 \mathrm{V}$ (a) At \(f=120 \mathrm{Hz},\) what are the voltage amplitudes across the capacitor and across the resistor? (b) Do the voltage amplitudes add to give the amplitude of the source voltage (i.e., does \(V_{\mathrm{R}}+V_{\mathrm{C}}=2.0 \mathrm{V}\) )? Explain. (c) Draw a phasor diagram to show the addition of the voltages.
An alternator supplics a peak current of \(4.68 \mathrm{A}\) to a coil with a negligibly small internal resistance. The voltage of the alternator is \(420-\mathrm{V}\) peak at \(60.0 \mathrm{Hz}\) When a capacitor of $38.0 \mu \mathrm{F}$ is placed in series with the coil, the power factor is found to be \(1.00 .\) Find (a) the inductive reactance of the coil and (b) the inductance of the coil.
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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