Parallel RL circuit behavior: Which statement correctly describes the currents and phase relationships?

Introduction / Context:In a parallel RL circuit, the branch currents do not add arithmetically because they are out of phase. Understanding phasor addition is crucial for computing total current, power factor, and reactive power.

Given Data / Assumptions:

• Parallel RL, steady-state sinusoidal excitation.
• Same voltage across both branches.
• I_R is in phase with V; I_L lags V by 90 degrees.

Concept / Approach:Total current is the phasor sum: I_T = I_R + I_L∠(-90°). Since the two branch currents are orthogonal in phase, their magnitudes combine by the Pythagorean relationship: |I_T| = sqrt(I_R^2 + I_L^2). This magnitude is strictly less than the arithmetic sum I_R + I_L.

Step-by-Step Solution:Recognize that currents add vectorially (phasors), not arithmetically.Compute magnitude: |I_T| = sqrt(I_R^2 + I_L^2).Conclude: |I_T| < I_R + I_L because of orthogonality.Phase: I_T lags the voltage by an angle between 0 and 90 degrees for RL (not leads).

Verification / Alternative check:Draw a right triangle with legs I_R (horizontal) and I_L (vertical downward for lag). The hypotenuse is I_T. Geometry confirms magnitude and lagging phase.

Why Other Options Are Wrong:Arithmetic sum: ignores phase; incorrect.Same amplitude and phase: true for series current, not parallel branches with reactive elements.Leads voltage: an RL network current lags, not leads; leading occurs with capacitive effects.None: incorrect since option (c) is correct.

Common Pitfalls:Adding magnitudes instead of phasors; mixing RL (lagging) with RC (leading); misreading “parallel” versus “series.”

Final Answer:The total current is less than the sum of the currents through the resistor and the inductor (phasor sum)