Question E5B07:
What is the phase angle between the voltage across and the current through a series RLC circuit if XC is 500 ohms, R is 1 kilohm, and XL is 250 ohms?  

Choices:  
A: 68.2 degrees with the voltage leading the current  
B: 14.0 degrees with the voltage leading the current  
C: 14.0 degrees with the voltage lagging the current  
D: 68.2 degrees with the voltage lagging the current    

Correct Answer is  C: 14.0 degrees with the voltage lagging the current  

To find the phase angle in a series RLC circuit, use the impedance angle:
$$\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)$$
Given: $X_C = 500\,\Omega$, $X_L = 250\,\Omega$, $R = 1000\,\Omega$.

1) Why the correct answer is correct  
- Net reactance:  
  $$X = X_L - X_C = 250 - 500 = -250\,\Omega \quad (\text{capacitive})$$
- Phase angle:  
  $$\phi = \tan^{-1}\left(\frac{-250}{1000}\right) = \tan^{-1}(-0.25) \approx -14.0^\circ$$
- Negative angle means the voltage lags the current in a capacitive circuit.  
Therefore: 14.0 degrees with the voltage lagging the current (Choice C).

2) Why the other choices are incorrect  
- A: 68.2° leading — wrong sign and magnitude. The circuit is net capacitive (should lag), and 68.2° corresponds to $|X|/R \approx 2.5$, not 0.25.  
- B: 14.0° leading — correct magnitude but wrong sign; capacitive circuits lag.  
- D: 68.2° lagging — correct lagging sense but wrong magnitude; actual angle is about 14°.

3) Useful formulas and rules of thumb  
- Series impedance:  
  $$Z = R + j(X_L - X_C)$$
- Phase angle (voltage relative to current):  
  $$\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)$$
- Reactances:  
  $$X_L = 2\pi f L,\quad X_C = \frac{1}{2\pi f C}$$
- Lead/lag rule:  
  - If $X_L > X_C$ (inductive), $\phi > 0$ and voltage leads current.  
  - If $X_C > X_L$ (capacitive), $\phi < 0$ and voltage lags current.  
- Magnitude check: $|X|/R = 0.25$ implies a small angle (~14°); $|X|/R = 1$ would give 45°.