Question E5B08:
What is the phase angle between the voltage across and the current through a series RLC circuit if XC is 300 ohms, R is 100 ohms, and XL is 100 ohms?  

Choices:  
A: 63 degrees with the voltage lagging the current  
B: 63 degrees with the voltage leading the current  
C: 27 degrees with the voltage leading the current  
D: 27 degrees with the voltage lagging the current    

Correct Answer is  A: 63 degrees with the voltage lagging the current  

To find the phase angle in a series RLC circuit, use the impedance angle:
$$\phi = \arctan\left(\frac{X_L - X_C}{R}\right)$$

Compute the net reactance and angle:
- Net reactance: $X = X_L - X_C = 100 - 300 = -200\ \Omega$ (capacitive)
- Resistance: $R = 100\ \Omega$
- Phase angle: 
  $$\phi = \arctan\left(\frac{-200}{100}\right) = \arctan(-2) \approx -63^\circ$$

Interpretation:
- Negative angle means the circuit is net capacitive, so the voltage lags the current by about $63^\circ$.
- Therefore, the correct answer is: 63 degrees with the voltage lagging the current (Choice A).

Why the others are incorrect:
- B: 63 degrees with the voltage leading the current — wrong sign; capacitive circuits make voltage lag, not lead.
- C: 27 degrees with the voltage leading the current — wrong magnitude and wrong sign (should be 63° lag).
- D: 27 degrees with the voltage lagging the current — wrong magnitude (should be about 63°).

Useful formulas and rules of thumb:
- Series RLC impedance: $$Z = R + j(X_L - X_C)$$
- Phase angle (voltage relative to current): $$\phi = \arctan\left(\frac{X_L - X_C}{R}\right)$$
- If $X_L > X_C$ (inductive, $X>0$): voltage leads current.
- If $X_L < X_C$ (capacitive, $X<0$): voltage lags current.