In a pure inductor, the current:

Solution

Physics sir · 4 steps

An inductor fights every change in current, so the current always arrives a quarter cycle after the voltage.

1. Voltage across L. For a pure inductor the applied voltage balances the induced emf, so v = Ldi/dt. The voltage depends on how fast the current changes, not on the current itself.
2. Integrate. Take v = V₀ sin ω t. Then di/dt = V₀/L sin ω t, and integrating gives the current.

   i = -V₀/(ω L)cos ω t = V₀/(ω L)sin(ω t - π/2)

3. Read the phase. The current has phase ω t - π/2, which is 90° behind the voltage's phase ω t. The voltage peaks first and the current peaks a quarter cycle later.
4. So. In a pure inductor the current lags the voltage by 90°, option (b).

The idea to keep. Remember ELI the ICE man: in L, voltage (E) comes before current (I); in C, current comes before voltage. Only a pure resistor keeps them in phase.

Answer from the NTA final answer key. Solution by PrepWiser.