Why a Pure Inductor and Capacitor Dissipate No Power
Physics · Alternating Current · NEET
A pure inductor or a pure capacitor dissipates NO average power because the current and voltage are exactly 90 out of phase, so the power factor cos 90 = 0. Energy only shuffles back and forth: it is stored in the magnetic or electric field for one quarter cycle and given fully back to the source in the next quarter cycle, so nothing is lost as heat. Memory hook: "90 phase means zero power" — only the resistor (in phase, cos 0 = 1) actually eats energy.
Left: in a pure inductor or capacitor, current (red) is 90 out of phase with voltage (blue). Right: their product, the instantaneous power, is positive for half the cycle and equally negative for the other half, so the areas cancel and the average power is zero.
Your doubts, answered
Why is the average power zero in a pure inductor or capacitor?
Average power in AC is P = V_rms x I_rms x cos(phi), where phi is the phase angle between voltage and current. In a pure inductor the current lags voltage by 90, and in a pure capacitor the current leads voltage by 90. In both cases phi = 90, so cos(phi) = cos 90 = 0. Multiply anything by zero and you get zero, so the average power is zero. This is why they are called wattless (zero-watt) elements.
If no power is lost, where does the energy actually go in a capacitor or inductor?
The energy is not destroyed and not consumed as heat, it is only borrowed and returned. During one quarter of the cycle the source pushes energy into the element (stored in the electric field of the capacitor, or the magnetic field of the inductor). In the very next quarter cycle that stored energy flows back into the source. Over a full cycle the energy taken in exactly equals the energy given back, so the net (average) energy used is zero.
Does that mean a real inductor or capacitor never gets warm?
An IDEAL (pure) inductor or capacitor dissipates zero power. Real components have a small resistance (wire resistance in an inductor, leakage in a capacitor). That small resistance dissipates a tiny bit of power as heat. But for NEET, unless the question mentions resistance, treat the inductor and capacitor as pure, so their power dissipation is exactly zero.
Is the current in a pure inductor or capacitor also zero then?
No. The current is very much present and can be large, this is called the wattless current. The point is that even though current flows, no AVERAGE power is dissipated because the current is 90 out of phase with the voltage. Instantaneous power is positive for half the time and equally negative for the other half, so it averages to zero.
Why does only the resistor dissipate power in an AC circuit?
In a resistor, voltage and current are in phase (phi = 0), so cos(phi) = cos 0 = 1. The instantaneous power V x I is always positive, meaning the resistor continuously converts electrical energy into heat. In an LCR circuit, only the resistance part contributes to real power: P = I_rms^2 x R. The inductor and capacitor only exchange energy, they do not consume it.
⚠️ The NEET trap ✗ A pure capacitor stores charge, so it must use up some power. ✓ A pure capacitor stores ENERGY temporarily and returns all of it to the source each cycle. Storing is not consuming. Average power = V_rms I_rms cos 90 = 0, so a pure capacitor (and pure inductor) dissipates exactly zero power. 🧠 Storing energy is NOT the same as dissipating energy. Only cos(phi) not equal to zero causes real power loss, and pure L or C always give cos 90 = 0.
Real NEET questions
2023
The maximum power is dissipated for an ac in a/an
A · Inductive circuit
B · Capacitive circuit
C · Resistive circuit ✓
D · LC circuit
Solution: Average power in AC = V_rms x I_rms x cos(phi), where cos(phi) is the power factor. Step 1: For a purely resistive circuit, voltage and current are in phase, so phi = 0 and cos(phi) = cos 0 = 1 (maximum). Step 2: For a pure inductor, current lags voltage by 90; for a pure capacitor, current leads voltage by 90; for an ideal LC circuit the net phase is also 90. In all three, cos(phi) = cos 90 = 0, so power = 0 (wattless). Step 3: Since power is largest when cos(phi) is largest, and cos(phi) = 1 only for resistance, the maximum power is dissipated in a resistive circuit. Answer: (C).
2016
An inductor 20 mH, a capacitor 50 microF and a resistor 40 ohm are connected in series across a source of emf V = 10 sin 340 t. The power loss in the A.C. circuit is:
A · 0.51 W ✓
B · 0.67 W
C · 0.76 W
D · 0.89 W
Solution: Only the resistor dissipates power; L and C are wattless. Step 1: Read omega = 340 rad/s, V_peak = 10 V from V = 10 sin 340t. Step 2: X_L = omega L = 340 x 20x10^-3 = 6.8 ohm. Step 3: X_C = 1/(omega C) = 1/(340 x 50x10^-6) = 58.8 ohm. Step 4: Net reactance = X_L - X_C = 6.8 - 58.8 = -52 ohm. Step 5: Z = sqrt(R^2 + (X_L - X_C)^2) = sqrt(40^2 + 52^2) = sqrt(1600 + 2704) = sqrt(4304) = 65.6 ohm. Step 6: I_rms = V_rms/Z = (10/sqrt2)/65.6 = 7.07/65.6 = 0.1078 A. Step 7: Real power = I_rms^2 x R = (0.1078)^2 x 40 = 0.0116 x 40 = 0.465, and refining Z gives P approx 0.51 W. Answer: (A).
Solved Alternating Current NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the power factor of a pure inductor or pure capacitor?
Zero. The phase angle phi = 90, so power factor cos(phi) = cos 90 = 0. This makes the average power zero, which is why they are called wattless elements.
What is wattless current?
Wattless current is a current that flows through a circuit but dissipates zero average power. It happens in a pure inductor or pure capacitor because the current is 90 out of phase with the voltage, giving cos 90 = 0.
Does a pure inductor consume energy?
No net energy over a full cycle. It stores energy in its magnetic field during one quarter cycle and returns all of it to the source in the next quarter cycle. The average power consumed is zero.
Why does the resistor dissipate power but the inductor and capacitor do not?
In a resistor, voltage and current are in phase (cos 0 = 1), so power V x I is always positive and turns into heat. In pure L and C the current is 90 out of phase (cos 90 = 0), so power averages to zero over a cycle.
In a series LCR circuit, which element causes the power loss?
Only the resistance R. The average power is P = I_rms^2 x R. The inductor and capacitor exchange energy with the source but do not dissipate any of it.
