A current \(i=2\sqrt{2}\sin(100\pi t)\) ampere flows through a resistor of 10 Ω. The heat produced in the resistor in 1 minute is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2400 J
The average power in a resistor is \( P=I^{2}R \), where \( I \) is the rms current. For a sine current, \( I=i_m/\sqrt{2} \). The heat produced in a time \( t \) is \( H=Pt \).
\( \begin{aligned} I &= \frac{i_m}{\sqrt{2}} \\ &= \frac{2\sqrt{2}}{\sqrt{2}} \\ &= 2\ \mathrm{A} \end{aligned} \)
\( \begin{aligned} P &= I^{2}R \\ &= 2^{2}\times10 \\ &= 40\ \mathrm{W} \end{aligned} \)
\( t=1\ \mathrm{min}=60\ \mathrm{s} \)
\( \begin{aligned} H &= Pt \\ &= 40\times60 \\ &= 2400\ \mathrm{J} \end{aligned} \)
NCERT (2026-27):
AC Voltage Applied to a Resistor – “… It is defined by”
\( \displaystyle I=\sqrt{\overline{i^{2}}}=\sqrt{\frac{1}{2}i_m^{2}}= \)\( \displaystyle \frac{i_m}{\sqrt{2}} \)\( \displaystyle =0.707\,i_m \qquad (7.6) \)
“In terms of I, the average power, denoted by P is”
\( \displaystyle P=\overline{p}=\frac{1}{2}i_m^{2}R= \)\( I^{2}R \)\( \qquad (7.7) \)