Physics · Alternating Current · NEET
No. This is the most common mix-up. Wattless current is NOT zero. Real current flows through the wire and you can measure it with an AC ammeter. The word 'wattless' means it consumes zero average WATTS (power), not zero amperes. The energy just moves back and forth between the source and the inductor or capacitor each cycle, so the net energy used is zero.
Because average power P = V_rms x I_rms x cos(phi). For the wattless part, the current is 90 degrees out of phase with the voltage (phi = 90). Since cos 90 = 0, the average power is zero. Current is present, but its product with voltage averages to zero over a full cycle. So it moves charge but delivers no net energy — hence 'wattless'.
A pure inductor (L) and a pure capacitor (C). In both, the current is exactly 90 degrees out of phase with the applied voltage (current lags by 90 in an inductor, leads by 90 in a capacitor). Only a resistor consumes real power. So in an ideal L or C, the whole current is wattless.
Split the total current I into two parts using the phase angle phi. The power (active) component is I cos(phi) — this does work. The wattless (reactive) component is I sin(phi) — this does no work. So wattless current = I sin(phi). At phi = 90 (pure L or C), sin 90 = 1, so the entire current is wattless.
No. In a pure resistor, voltage and current are in phase, so phi = 0. The wattless part I sin(phi) = I sin 0 = 0. All the current is the power component I cos 0 = I. That is why a resistor dissipates maximum heat and has no wattless current.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the component of AC current, equal to I sin(phi), that is 90 degrees out of phase with voltage and therefore delivers zero average power.
When the phase difference between voltage and current is 90 degrees — that happens in an ideal pure inductor or pure capacitor. Then cos(phi) = 0, so P_avg = 0.
Power (active) current = I cos(phi) and does real work. Wattless (reactive) current = I sin(phi) and does no work. They are the two perpendicular components of the total current.
Yes. An AC ammeter reads the total rms current, which includes the wattless part. It is real, flowing current — only its average power contribution is zero.
It still heats the wires (I^2 R loss) even though it delivers no useful power to the load, so utilities try to keep the power factor cos(phi) close to 1 to reduce wasteful reactive current.
