Phase relationship in a series RL circuit: Considering standard phasor conventions, what is the correct relationship between the resistor voltage V_R and the inductor voltage V_L in a series RL branch driven by a sinusoidal source?

Introduction / Context:Understanding voltage phasors in series RL circuits is essential for predicting waveform timing, interpreting oscillograph traces, and constructing accurate phasor diagrams. While the current is common in a series branch, the individual element voltages are shifted relative to that current and, therefore, relative to one another.

Given Data / Assumptions:

• Series RL circuit with sinusoidal steady-state excitation.
• Standard passive sign convention.
• Ideal components (no parasitic capacitance or resistance in the inductor beyond modeled R).

Concept / Approach:In a resistor, voltage and current are in phase. In an inductor, voltage leads current by 90 degrees (v_L = L * di/dt corresponds to a +90° phase shift of voltage relative to current in phasor form). Because the series current is the same through both elements, V_R aligns with the current, while V_L leads the current by +90°. Therefore, V_L leads V_R by +90°, or equivalently, V_R lags V_L by 90°.

Step-by-Step Solution:

Verification / Alternative check:Construct a phasor diagram: draw I along the horizontal axis; V_R on the same axis; V_L upward (positive imaginary axis). The source voltage phasor V_S is the vector sum of V_R and V_L, sitting at an angle between them, verifying the lag relationship.

Why Other Options Are Wrong:

• V_R leads V_L: reverses the correct order.
• V_R is in phase with V_L: not in an RL branch; they are 90° apart.
• All in phase: impossible when an inductor is present with nonzero reactance.
• V_R leads by 180°: no such inversion occurs in linear passive RL.

Common Pitfalls:Confusing voltage-current relationships of R and L; thinking all voltages in one series branch must be in phase; mixing up the sign of the 90° shift.

Final Answer:V_R lags V_L