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 largest value of the instantaneous power taken by the heater during a cycle is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 880 W
The power taken by a resistor at any instant is \( p=i_m^{2}R\sin^{2}\omega t \). It is largest when \( \sin^{2}\omega t=1 \), so the largest value is \( i_m^{2}R \). The mains value of 220 V is an rms value. So we find the rms current first and then the peak current \( i_m \).
Given:
\( V=220\ \mathrm{V} \)
\( R=110\ \Omega \)
\( \begin{aligned} I &= \frac{V}{R} \\ &= \frac{220}{110} \\ &= 2\ \mathrm{A} \end{aligned} \)
\( \begin{aligned} i_m &= \sqrt{2}\,I \\ &= 2\sqrt{2}\ \mathrm{A} \end{aligned} \)
\( \begin{aligned} p_{\max} &= i_m^{2}R \\ &= (2\sqrt{2})^{2}\times110 \\ &= 8\times110 \\ &= 880\ \mathrm{W} \end{aligned} \)
This value is reached twice in every cycle, when the current is at its peak in either direction.
NCERT (2026-27):
AC Voltage Applied to a Resistor – “The instantaneous power dissipated in the resistor is”
\( p=i^{2}R=i_m^{2}R\sin^{2}\omega t \)\( \qquad (7.4) \)
AC Voltage Applied to a Resistor – “… It is called, root mean square (rms) or effective current … and is denoted by Irms or I. … It is defined by”
\( I=\sqrt{\overline{i^{2}}}=\sqrt{\frac{1}{2}i_m^{2}}= \)\( \frac{i_m}{\sqrt{2}} \)\( =0.707\,i_m \qquad (7.6) \)