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Chapter 21: Problem 88

An RLC series circuit is connected to a \(240-\mathrm{V}\) ms power supply at a frequency of \(2.50 \mathrm{kHz}\). The elements in the circuit have the following values: \(R=12.0 \Omega, C=\) \(0.26 \mu \mathrm{F},\) and $L=15.2 \mathrm{mH},$ (a) What is the impedance of the circuit? (b) What is the rms current? (c) What is the phase angle? (d) Does the current lead or lag the voltage? (e) What are the rms voltages across each circuit element?

Short Answer

Expert verified
Answer: The impedance of the circuit is \(12.9\Omega\), the rms current is \(18.6\mathrm{A}\), the phase angle is \(-9.4^\circ\), the current leads the voltage, and the rms voltages across the resistor, capacitor, and inductor are \(223\mathrm{V}, 4.57\times10^3\mathrm{V}\), and \(4.43\times10^3\mathrm{V}\) respectively.

Step by step solution

01

Calculate angular frequency, inductive reactance, and capacitive reactance.

First, we need to find the angular frequency, inductive reactance, and capacitive reactance using the given values. Angular frequency: \(\omega = 2\pi f = 2\pi (2.50 \times 10^3 \mathrm{s^{-1}}) = 15.7 \times 10^3\mathrm{s^{-1}}\) Inductive reactance: \(X_L = \omega L = (15.7 \times 10^3\mathrm{s^{-1}}) (15.2 \times 10^{-3}\mathrm{H}) = 238\Omega\) Capacitive reactance: \(X_C = \frac{1}{\omega C} = \frac{1}{(15.7 \times 10^3\mathrm{s^{-1}}) (0.26 \times 10^{-6}\mathrm{F})} = 246\Omega\)
02

Calculate impedance.

Now, find the impedance using the resistance, inductive reactance, and capacitive reactance. \(Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{(12.0 \Omega)^2 + (238\Omega - 246\Omega)^2} = 12.9\Omega\) The impedance of the circuit is \(12.9\Omega\).
03

Calculate rms current.

Next, find the rms current using the given voltage and calculated impedance. \(I = \frac{V}{Z} = \frac{240\mathrm{V}}{12.9\Omega} = 18.6\mathrm{A}\) The rms current is \(18.6\mathrm{A}\).
04

Calculate phase angle.

Now, find the phase angle using the resistance and reactances. \(\phi = \tan^{-1}(\frac{X_L - X_C}{R}) = \tan^{-1}(\frac{238\Omega - 246\Omega}{12.0\Omega}) = -9.4^\circ\) The phase angle is \(-9.4^\circ\).
05

Determine if current leads or lags the voltage.

Since the phase angle is negative, the current leads the voltage in this circuit.
06

Calculate rms voltages across each circuit element.

To find the rms voltages across the resistor, capacitor, and inductor, use the rms current and the resistance/reactances. \(V_R = IR = (18.6\mathrm{A})(12.0\Omega) = 223\mathrm{V}\) \(V_C = IX_C = (18.6\mathrm{A})(246\Omega) = 4.57\times10^3\mathrm{V}\) \(V_L = IX_L = (18.6\mathrm{A})(238\Omega) = 4.43\times10^3\mathrm{V}\) The rms voltages across the resistor, capacitor, and inductor are \(223\mathrm{V}, 4.57\times10^3\mathrm{V}\), and \(4.43\times10^3\mathrm{V}\) respectively.

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