An alternating current \(i=4\sin\omega t\) (in ampere) flows through a resistor. The average value of \(i^{2}\) over one complete cycle is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(8\ \mathrm{A^{2}}\)
Here \( i^{2}=i_m^{2}\sin^{2}\omega t \). The value of \( \sin^{2}\omega t \) moves between 0 and 1. It is never negative, and its average over a full cycle is \( 1/2 \).
\( i_m=4\ \mathrm{A} \)
\( \begin{aligned} \langle i^{2}\rangle &= i_m^{2}\,\langle\sin^{2}\omega t\rangle \\ &= 4^{2}\times\frac{1}{2} \\ &= 8\ \mathrm{A^{2}} \end{aligned} \)
The heat produced in the resistor depends on this mean square value.
NCERT (2026-27):
AC Voltage Applied to a Resistor – “… Using the trigonometric identity, sin2 ωt = 1/2 (1– cos 2ωt), we have < sin2 ωt > = (1/2) (1– < cos 2ωt >) and since < cos2ωt > = 0 …, we have,”
\( \displaystyle \langle\sin^{2}\omega t\rangle=\frac{1}{2} \)