Question

Average heat produced per unit time during a cycle of $AC$ is the same as produced by $DC$ with current $i=$

• A. $i_{avg}$
• B. $i_{rms}$
• C. $i_{avg}^2$
• D. $i_{peak}$

Answer 1

The correct option is B $i_{rms}$

For $AC$, $i=i_0\sin \omega t$

Therefore, the instantaneous value of heat produced in time $dt$ across a resistance $R$ is

$dH=i^{2}Rdt=i_0^{2}R{\sin }^2\omega tdt$

$\therefore$ Average value of heat produced per unit of time during a cycle of $AC$,

$H_{avg}=\frac{{\int }_0^{T}dH}{{\int }_0^{T}dt}$

$=\frac{{\int }_0^{{}^{\frac{2\pi }{\omega }}}\left(i_0^{2}R{\sin }^2\omega t\right)dt}{{\int }_0^{\frac{2\pi }{\omega }}dt}$

$=\frac{i_0^{2}R{\int }_0^{{}^{\frac{2\pi }{\omega }}}\left({\sin }^2\omega t\right)dt}{{\int }_0^{\frac{2\pi }{\omega }}dt}$

$[\because {\int }_0^{{}^{\frac{2\pi }{\omega }}}\left({\sin }^2\omega t\right)dt=\frac{1}{2}\times \frac{2\pi }{\omega }]$


$=\frac{i_0^2}{2}R=i_{rms}^{2}R$

Heat produced per unit time when a $DC$ current $i$ flows through a resistance $R$ is,

$H=i^{2}R$

On comparing the heat produced during a cycle of $AC$ and $DC$ we get,

$i=i_{rms}$

Hence, option $\left(B\right)$ is correct.