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#wattless current

2 public questions tagged with this topic.

Why does an AC circuit with only an inductor exhibit a "wattless" current?

**Average power in LCR** P_avg = V_rms I_rms cos φ, cos φ = R/Z power factor, φ phase between V and I, tan φ = (X_L - X_C)/R. For R=80 Ω, X_L=100 Ω, X_C=40 Ω, X_L-X_C=60 Ω, Z=√(80²+60²)=100 Ω, cos φ=0.8, V_rms=240 V, I_rms=2.4 A, P=240×2.4×0.8=460.8 W, only R dissipates. In a purely inductive AC circuit, the current lags the voltage by 90°. The power factor ( cos 90° = 0 ) is zero, meaning no average power is dissipated; the current is "wattless" because it only stores and releases energy without consuming it. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² +

Ref: NCERT > Physics Book > Alternating Currents > Power in AC Circuits - Power Factor and Wattless Current

What is the physical significance of the term "wattless current" in an AC circuit?

**Average power** in resistor P_avg = V_rms I_rms = V_rms²/R = I_rms² R, for 254.6 V peak, V_rms =180 V, R=90 Ω, P=180²/90=360 W, for 226.3 V peak, V_rms=160 V, R=80 Ω, P=320 W, illustrating rms use for power. "Wattless current" refers to the component of current in a purely reactive (inductive or capacitive) AC circuit that is 90° out of phase with the voltage. It does not contribute to average power dissipation, as power is zero when cos 90° = 0 , hence "wattless.". Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms

Ref: NCERT > Physics Book > Alternating Currents > AC Fundamentals - RMS, Average and Peak Values