Phase relation in an R–C series circuit: For R = 100 Ω in series with a capacitor having reactance Xc = 10 Ω, what is the approximate phase angle between current and applied voltage?

Introduction / Context:
Series resistor–capacitor (R–C) circuits exhibit a characteristic phase shift between current and voltage due to the reactive nature of the capacitor. This question assesses your ability to compute the phase angle using given resistance and capacitive reactance values and to interpret whether current leads or lags the applied voltage.

Given Data / Assumptions:

• R = 100 Ω (resistance).
• Xc = 10 Ω (capacitive reactance at the operating frequency).
• Sinusoidal steady-state conditions.

Concept / Approach:

For a series R–C circuit, the impedance is Z = R − j Xc. The current leads the applied voltage by an angle θ where tan θ = Xc / R (positive lead for capacitive circuits). The magnitude of θ is small when Xc ≪ R and large when Xc ≫ R.

Step-by-Step Solution:

Verification / Alternative check:

Phasor diagram: current I taken as reference; voltage across R is in phase with I, voltage across C lags I by 90°. The resultant applied voltage lags I by θ, so I leads V by θ. With R ≫ Xc, the angle is indeed small (≈ 6°).

Why Other Options Are Wrong:

• In phase: only true when Xc = 0 (purely resistive).
• Lead by 84°: would require Xc ≫ R, not the case here.
• Lagging options: incorrect sign; capacitive circuits cause current to lead voltage.

Common Pitfalls:

• Using θ = arctan(R / Xc), which is the complement and gives the wrong magnitude.
• Confusing lead/lag conventions between current and voltage.

Final Answer:

the current leads the voltage by about 6°