Physics · Alternating Current · NEET
In a pure inductor the CURRENT lags the VOLTAGE by 90°. From NCERT: v = v_m sin(ωt) gives i = i_m sin(ωt − π/2). The minus π/2 (minus 90°) inside the current equation means the current reaches its peak a quarter-cycle AFTER the voltage. So the current is the one that is behind (lags). Use ELI: in the inductor L, E (voltage) leads, I (current) lags.
For a PURE inductor (zero resistance), the whole equation is L(di/dt) = v_m sin(ωt). To get the current you integrate sin(ωt), which gives −cos(ωt). And −cos(ωt) = sin(ωt − π/2). Integrating a sine always shifts it back by exactly 90°, so the phase difference is fixed at π/2 = 90°. If the inductor also had resistance, the lag would be less than 90°.
It means the current is a slow follower. When the source voltage is already at its peak, the inductor current is still at zero and only starting to rise. The current reaches its own peak one-quarter of a period later. Physically, the inductor's back-emf (Lenz's law) opposes the change in current, so the current cannot keep up with the fast-changing voltage — it is always a step behind.
A resistor obeys v = iR, so i = v/R — current and voltage rise and fall together (in phase, 0°). An inductor obeys v = L(di/dt): the voltage depends on the RATE of change of current, not the current itself. Voltage is largest when the current is changing fastest (at i = 0, the zero-crossing), and voltage is zero when the current is at its peak (rate of change = 0). This mismatch is the 90° lag.
Yes. ELI: in an inductor (L), E (voltage) is before I (current), so current lags. ICE: in a capacitor (C), I (current) is before E (voltage), so current leads. NCERT confirms both: 'in the case of an inductor, the current lags the voltage by π/2 and in the case of a capacitor, the current leads the voltage by π/2.' Just remember lag/lead is always about the CURRENT.
A small signal voltage V(t) = V₀ sin ωt is applied across an ideal capacitor C. Which statement is correct?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Exactly 90° (that is π/2 radians, or one-quarter of the full cycle). The current equation is i = i_m sin(ωt − π/2).
i = i_m sin(ωt − π/2), where i_m = v_m / (ωL) = v_m / X_L. Here X_L = ωL is the inductive reactance in ohms.
No. Because the current lags voltage by 90°, the average power P = V_rms·I_rms·cos(90°) = 0. The inductor stores energy in its magnetic field and returns it, so net energy over a cycle is zero.
Yes. The inductor sets up a back-emf that opposes any change in current (Lenz's law). This prevents the current from rising instantly with the voltage, so the current always arrives a quarter-cycle late.
Use ELI the ICE man. In an inductor (L): E before I → voltage leads, current lags. In a capacitor (C): I before E → current leads, voltage lags.
Inductive reactance X_L = ωL = 2πfL (in ohms) is the opposition an inductor gives to AC. It sets the current amplitude i_m = v_m / X_L, while the −90° phase sets the timing (the lag).
