Questions: What is the phase angle of an AC series circuit that is constructed of a 10.0-Ω resistor along with 12.00 Ω inductive reactance and 7.00 Ω capacitive reactance? a. 26.6° b. 18.4° c. 87.0° d. 63.4°

Solution Steps

The net reactance \( X \) in an AC series circuit is the difference between the inductive reactance \( X_L \) and the capacitive reactance \( X_C \). Given:

• Inductive reactance, \( X_L = 12.00 \, \Omega \)
• Capacitive reactance, \( X_C = 7.00 \, \Omega \)

The net reactance is calculated as: \[ X = X_L - X_C = 12.00 \, \Omega - 7.00 \, \Omega = 5.00 \, \Omega \]

The impedance \( Z \) of the circuit is given by the formula: \[ Z = \sqrt{R^2 + X^2} \] where \( R = 10.0 \, \Omega \) is the resistance. Substituting the values: \[ Z = \sqrt{(10.0 \, \Omega)^2 + (5.00 \, \Omega)^2} = \sqrt{100.0 + 25.0} = \sqrt{125.0} = 11.1803 \, \Omega \]

The phase angle \( \phi \) is given by: \[ \tan \phi = \frac{X}{R} \] Substituting the known values: \[ \tan \phi = \frac{5.00 \, \Omega}{10.0 \, \Omega} = 0.5 \] To find the phase angle \( \phi \), take the arctangent: \[ \phi = \tan^{-1}(0.5) \approx 26.5651^{\circ} \]

Final Answer