Read the following passage and answer the questions.
A student uses the laboratory mains supply of 220 V (rms) and 50 Hz. First, she joins an electric heater of resistance 110 Ω across the mains. The heater acts as a pure resistor. An ac ammeter is joined in series with the heater. She draws the source voltage and the current as phasors.
Next, she removes the heater. In its place she joins a coil of negligible resistance. The coil acts as a pure inductor, and its inductive reactance at 50 Hz is also 110 Ω. She notes the ammeter reading again and draws the phasor diagram for the coil.
In the phasor diagram drawn for the coil, the current phasor I
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → is \(\pi/2\) behind the voltage phasor V
For a pure inductor, the source voltage \( v=v_m\sin\omega t \) drives the current \( i=i_m\sin(\omega t-\pi/2) \). So the current lags the voltage by a quarter of a cycle.
Option (1) points in the same direction as the voltage phasor V — Incorrect. Phasors in the same direction mean zero phase difference. This is the case for the heater, which is a pure resistor.
Option (2) is \(\pi/2\) behind the voltage phasor V — Correct. The current in the coil lags the voltage by \( \pi/2 \). As both phasors turn anticlockwise, I stays \( \pi/2 \) behind V. So the current reaches its peak a quarter period after the voltage.
Option (3) is \(\pi/2\) ahead of the voltage phasor V — Incorrect. A current phasor \( \pi/2 \) ahead of V belongs to a pure capacitor. There the current leads the voltage.
Option (4) points opposite to the voltage phasor V — Incorrect. Opposite phasors mean a phase difference of \( \pi \). For the coil the phase difference is only \( \pi/2 \).
NCERT (2026-27):
AC Voltage Applied to an Inductor – “… The current phasor I is π/2 behind the voltage phasor V.”