What is the formula to find wattless current in AC?

In AC circuit analysis, the total RMS current I can be decomposed into an active (in-phase) component that does real work and a wattless (quadrature) component that stores and returns energy each cycle without net consumption. If the phase angle between voltage and current is φ, the wattless (reactive) current is:

I_w = I \u00D7 \sin φ

The complementary active component is I_a = I \u00D7 \cos φ, and the apparent current obeys I^2 = I_a^2 + I_w^2. In a purely inductive or capacitive circuit, φ = 90°, so \sin φ = 1 and the entire current is wattless; average power is zero even though current flows. Real installations rarely operate at 90°, but low power factor implies a large wattless component that inflates current, conductor size, I²R losses, and utility charges.

To reduce wattless current in practice, engineers deploy power factor correction (e.g., shunt capacitor banks for inductive loads). This brings φ closer to zero, decreasing I_w, improving efficiency, and increasing capacity without new cabling. For students, remember the triangle: active along the base (I \cos φ), reactive vertical (I \sin φ), hypotenuse equals I. Framing problems with this geometry makes exam calculations fast and dependable.

Exam Preparation tip: Identify φ from impedance: \tan φ = X/R. Then compute I_w directly via I \sin φ to avoid round-off from multi-step power calculations.