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Chapter 21: Problem 38

A series \(R L C\) circuit has a \(0.20-\mathrm{mF}\) capacitor, a \(13-\mathrm{mH}\) inductor, and a \(10.0-\Omega\) resistor, and is connected to an ac source with amplitude \(9.0 \mathrm{V}\) and frequency \(60 \mathrm{Hz}\) (a) Calculate the voltage amplitudes \(V_{L}, V_{C}, V_{R},\) and the phase angle. (b) Draw the phasor diagram for the voltages of this circuit.

Short Answer

Expert verified
Question: Calculate the voltage amplitudes across the resistor, capacitor, and inductor, and the phase angle in a series RLC circuit with a source voltage of 9 V, a resistance of 10 ohms, an inductance of 13 mH, a capacitance of 0.20 mF, and a frequency of 60 Hz. Draw a phasor diagram for this circuit. Answer: The voltage amplitudes across the resistor, capacitor, and inductor are: - \(V_R = 6.97\,V\) - \(V_L = 3.41\,V\) - \(V_C = 9.26\,V\) The phase angle between the voltage and current is \(-38.9^\circ\). The phasor diagram can be drawn with arrow representations of the voltages, as mentioned in the step 5 description provided in the solution.

Step by step solution

01

Calculate the capacitive reactance (\(X_C\))

To calculate the capacitive reactance (\(X_C\)), use the following formula: \(X_C = \dfrac{1}{2\pi fC}\), where \(f\) is the frequency and \(C\) is the capacitance. Given values: \(f = 60 Hz\) and \(C = 0.20 mF = 0.20 * 10^{-3} F\). Now we can calculate the capacitive reactance: \(X_C = \dfrac{1}{2\pi (60)(0.20 * 10^{-3})} = 13.3\,\Omega\)
02

Calculate the inductive reactance (\(X_L\))

To calculate the inductive reactance (\(X_L\)), use the following formula: \(X_L = 2\pi fL\), where \(L\) is the inductance value. Given value: \(L = 13 mH = 13 * 10^{-3} H\). Now we can calculate the inductive reactance: \(X_L = 2\pi (60)(13 * 10^{-3}) = 4.9\,\Omega\)
03

Calculate the impedance (\(Z\) )

To calculate the impedance (\(Z\)), use the following formula: \(Z = \sqrt{R^2 + (X_L - X_C)^2}\), where \(R\) is the resistance value. Given value: \(R = 10\,\Omega\). Now we can calculate the impedance: \(Z = \sqrt{(10)^2 + (4.9 - 13.3)^2} = \sqrt{100 + 70.56} = 12.9\,\Omega\)
04

Calculate voltage amplitudes \(V_L, V_C, V_R\), and the phase angle

Calculate the voltage amplitudes by using Ohm's Law: \(V_R = I_R * R\), \(V_L = I_L * X_L\), \(V_C = I_C * X_C\) Given value: \(V = 9\,V\). First, calculate the current in the circuit: \(I = \dfrac{V}{Z} = \dfrac{9}{12.9} = 0.697\,A\) Now, find the voltage amplitudes: \(V_R = 0.697 * 10 = 6.97\,V\) \(V_L = 0.697 * 4.9 = 3.41\,V\) \(V_C = 0.697 * 13.3 = 9.26\,V\) Next, calculate the phase angle between the voltage and current: \(\phi = \arctan \dfrac{X_L - X_C}{R} = \arctan \dfrac{4.9 - 13.3}{10} = -38.9^\circ\)
05

Draw the phasor diagram

In the phasor diagram, we represent each voltage as an arrow with a length corresponding to its amplitude: - \(V_R\) has an amplitude of \(6.97\,V\), parallel to the horizontal axis (0º phase angle) - \(V_L\) has an amplitude of \(3.41\,V\), rotated 90º counter-clockwise (90º phase angle) - \(V_C\) has an amplitude of \(9.26\,V\), rotated 90º clockwise (-90º phase angle) The resultant voltage, \(V\), is the phasor sum of these three voltages. This can be represented graphically as a parallelogram with sides formed by the voltages \(V_R\), \(V_L\), and \(V_C\), with a total angle of \(-38.9^\circ\).

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Most popular questions from this chapter

A \(150-\Omega\) resistor is in series with a \(0.75-\mathrm{H}\) inductor in an ac circuit. The rms voltages across the two are the same. (a) What is the frequency? (b) Would each of the rms voltages be half of the rms voltage of the source? If not, what fraction of the source voltage are they? (In other words, \(V_{R} / \ell_{m}=V_{L} / \mathcal{E}_{m}=?\) ) (c) What is the phase angle between the source voltage and the current? Which leads? (d) What is the impedance of the circuit?
At what frequency is the reactance of a \(20.0-\mathrm{mH}\) inductor equal to \(18.8 \Omega ?\)
Finola has a circuit with a \(4.00-\mathrm{k} \Omega\) resistor, a \(0.750-\mathrm{H}\) inductor, and a capacitor of unknown value connected in series to a \(440.0-\mathrm{Hz}\) ac source. With an oscilloscope, she measures the phase angle to be \(25.0^{\circ} .\) (a) What is the value of the unknown capacitor? (b) Finola has several capacitors on hand and would like to use one to tune the circuit to maximum power. Should she connect a second capacitor in parallel across the first capacitor or in series in the circuit? Explain. (c) What value capacitor does she need for maximum power?
Two ideal inductors \((0.10 \mathrm{H}, 0.50 \mathrm{H})\) are connected in series to an ac voltage source with amplitude \(5.0 \mathrm{V}\) and frequency \(126 \mathrm{Hz}\). (a) What are the peak voltages across each inductor? (b) What is the peak current that flows in the circuit?
A capacitor and a resistor are connected in parallel across an ac source. The reactance of the capacitor is equal to the resistance of the resistor. Assuming that \(i_{\mathrm{C}}(t)=I \sin \omega t,\) sketch graphs of \(i_{\mathrm{C}}(t)\) and \(i_{\mathrm{R}}(t)\) on the same axes.
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