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.
Consider the following statements about the two circuits:
(A) For the heater, the angle between the voltage phasor and the current phasor is zero.
(B) If the mains frequency were raised, the current through the heater would not change.
(C) For the heater, the average current over one complete cycle is equal to the rms current.
(D) The coil draws a larger rms current from the mains than the heater.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (B) only
The heater is a pure resistor, so its current is in phase with the voltage and is fixed by \( R \) alone. The coil limits the current through its reactance \( X_L=\omega L \).
(A) For the heater, the angle between the voltage phasor and the current phasor is zero. — True. In a pure resistor the voltage and the current are in phase. Both phasors turn together and always point the same way.
(B) If the mains frequency were raised, the current through the heater would not change. — True. For a resistor \( i_m=v_m/R \). The frequency does not appear in this relation, so the heater current stays the same.
(C) For the heater, the average current over one complete cycle is equal to the rms current. — False. The current is positive for half a cycle and negative for the other half, so its average over a full cycle is zero. The rms value \( I=i_m/\sqrt{2} \) is not zero.
(D) The coil draws a larger rms current from the mains than the heater. — False. The reactance of the coil at 50 Hz is 110 Ω, equal to the resistance of the heater. So both draw the same rms current:
\( \begin{aligned} I &= \frac{V}{X_L} \\ &= \frac{220}{110} \\ &= 2\ \mathrm{A} \end{aligned} \)
NCERT (2026-27):
Representation of AC Current and Voltage by Rotating Vectors — Phasors – “… we see that phasors V and I for the case of a resistor are in the same direction. This is so for all times. This means that the phase angle between the voltage and the current is zero.”
AC Voltage Applied to a Resistor – “… where the current amplitude im is given by”
\( i_m=\frac{v_m}{R} \)\( \qquad (7.3) \)
AC Voltage Applied to a Resistor – “… Thus, the sum of the instantaneous current values over one complete cycle is zero, and the average current is zero. …”
AC Voltage Applied to an Inductor – “… The amplitude of the current is, then”
\( i_m=\frac{v_m}{X_L} \)\( \qquad (7.14) \)