Question

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

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

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: \(X_C = \dfrac{1}{2\pi (60)(0.20 * 10^{-3})} = 13.3\,\Omega\)

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: \(X_L = 2\pi (60)(13 * 10^{-3}) = 4.9\,\Omega\)

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: \(Z = \sqrt{(10)^2 + (4.9 - 13.3)^2} = \sqrt{100 + 70.56} = 12.9\,\Omega\)

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: \(\phi = \arctan \dfrac{X_L - X_C}{R} = \arctan \dfrac{4.9 - 13.3}{10} = -38.9^\circ\)

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 \(-38.9^\circ\).