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.
The reading of the ac ammeter in the heater circuit gives
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → the rms value of the current
The current in the heater keeps changing in size and direction. An ac meter still has to show one steady value. The value used is the one that decides the heating effect of the current.
Option (1) the rms value of the current — Correct. Ac quantities are measured and stated as rms values. The rms current is the steady dc current that would heat the heater at the same average rate.
Option (2) the peak value of the current — Incorrect. The current reaches its peak only for an instant, twice in each cycle. The peak value is \( \sqrt{2} \) times the rms value.
Option (3) the average value of the current over one cycle — Incorrect. The current is positive for half a cycle and negative for the other half. Its average over a full cycle is zero, so such a meter would always read zero.
Option (4) the average value of the current over half a cycle — Incorrect. The average over half a cycle is not zero, but it is a different number from the rms value. The heating effect depends on the square of the current, so ac meters are set to show the rms value.
NCERT (2026-27):
AC Voltage Applied to a Resistor – “It is customary to measure and specify rms values for ac quantities. For example, the household line voltage of 220 V is an rms value with a peak voltage of”
\( v_m=\sqrt{2}\,V=(1.414)(220\ \mathrm{V})=311\ \mathrm{V} \)