In a series RLC circuit, the approximate phase angle when VC = 117 V, VR = 14.5 V, and VL = 3.3 V is closest to which of the following?

Introduction:
This problem assesses phasor reasoning in a series RLC circuit using the individual component voltage drops. The sign and magnitude of the phase angle come from the balance between inductive and capacitive reactances relative to the resistive drop.

Given Data / Assumptions:

• Series RLC circuit under sinusoidal steady state.
• Measured voltages: VC = 117 V, VR = 14.5 V, VL = 3.3 V.
• We seek the overall phase angle phi of the current relative to the source voltage.

Concept / Approach:
In series RLC, voltages across R, L, and C are mutually perpendicular in the phasor diagram: VR is in phase with current, VL leads current by 90 degrees, and VC lags current by 90 degrees. The reactive phasor is VL − VC, so the tangent of the phase angle is (VL − VC) / VR (sign indicates inductive or capacitive behavior).

Step-by-Step Solution:

Verification / Alternative check:
The source voltage magnitude would be Vs ≈ sqrt(VR^2 + (VL − VC)^2) ≈ sqrt(14.5^2 + 113.7^2), consistent with a very large reactive component compared with resistive drop, supporting a phase angle magnitude near 90 degrees but not exactly 90 degrees.

Why Other Options Are Wrong:

• –45.0 degrees: Would imply comparable reactive and resistive drops; contradicted by the much larger |VL − VC| compared to VR.
• –90.0 degrees: Ideal limit for purely capacitive with negligible resistance; here VR is small but nonzero, so angle is slightly less in magnitude.
• –172.7 degrees: Physically inconsistent for a simple series RLC driven by a single-frequency source.

Common Pitfalls:
Adding VL and VC instead of subtracting them, ignoring the sign (which determines lead/lag), or computing tan(phi) using component reactances directly without relating them to the measured voltage drops.

Final Answer:
–82.7 degrees.