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Chapter 21: Problem 35

A 6.20 -m \(H\) inductor is one of the elements in a simple RLC series circuit. When this circuit is connected to a \(1.60-\mathrm{kHz}\) sinusoidal source with an \(\mathrm{rms}\) voltage of \(960.0 \mathrm{V},\) an rms current of $2.50 \mathrm{A}\( lags behind the voltage by \)52.0^{\circ} .$ (a) What is the impedance of this circuit? (b) What is the resistance of this circuit? (c) What is the average power dissipated in this circuit?

Short Answer

Expert verified
Question: Calculate (a) the impedance (Z), (b) the resistance (R), and (c) the average power (P_avg) dissipated in an RLC series circuit with a frequency of 1.60 kHz, an inductor value of 6.20 mH, an rms voltage of 203 V, an rms current of 2.50 A, and a phase angle of 52.0 degrees between the voltage and current. Answer: (a) The impedance (Z) of the circuit is 80.67 Ω. (b) The resistance (R) of the circuit is 50.65 Ω. (c) The average power (P_avg) dissipated in the circuit is 316.56 W.

Step by step solution

01

Calculate the Inductive Reactance

To calculate the inductive reactance of the inductor, we use the formula \(X_L = 2\pi f H\), where \(f\) is the frequency and \(H\) is the inductor value. Given \(f = 1.60\,\mathrm{kHz}\) and \(H = 6.20\,\mathrm{mH}\), we can calculate \(X_L\) as follows: \(X_L = 2\pi \times 1.60\times 10^3\,\mathrm{Hz} \times 6.20\times 10^{-3}\,\mathrm{H} = 62.83\,\mathrm{\Omega}\).
02

Calculate the Impedance Using the Phase Angle

We are given that the phase angle (\(\phi\)) between the voltage and current is \(52.0^{\circ}\). Since \(\tan{\phi}=\frac{X_L - X_C}{R}\), and in this case, there is no capacitor, \(X_C = 0\), so the phase angle formula can be simplified to \(\tan{\phi}=\frac{X_L}{R}\). Using this equation, we can find the impedance (\(Z\)) of the circuit, given by \(Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + X_L^2}\), where \(R = \frac{X_L}{\tan{\phi}}\).
03

Calculate the Resistance

Now, to calculate the resistance (\(R\)) of the circuit, we use the formula \(R = \frac{X_L}{\tan{\phi}}\): \(R = \frac{62.83\,\mathrm{\Omega}}{\tan{52.0^{\circ}}} = 50.65\,\mathrm{\Omega}\). Now, we can find the impedance (\(Z\)) using the formula \(Z = \sqrt{R^2 + X_L^2}\): \(Z = \sqrt{(50.65\,\mathrm{\Omega})^2 + (62.83\,\mathrm{\Omega})^2} = 80.67\,\mathrm{\Omega}\). So, the impedance of the circuit is \(80.67\,\mathrm{\Omega}\).
04

Calculate the Average Power

Finally, to calculate the average power (\(P_{avg}\)) dissipated in the circuit, we use the formula \(P_{avg} = I_{rms}^2 \times R\) with given rms current (\(I_{rms} = 2.50\,\mathrm{A}\)) and calculated resistance (\(R = 50.65\,\mathrm{\Omega}\)): \(P_{avg} = (2.50\,\mathrm{A})^2 \times 50.65\,\mathrm{\Omega} = 316.56\,\mathrm{W}\). So, the average power dissipated in the circuit is \(316.56\,\mathrm{W}\).

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