Question

A current i = 2 + 3$\sin \left(2\pi {10}^{3}t\right)$ is flowing through a resistance of value $4\Omega$. Heat generated in kilo joules in 3 minutes across the resistance is (Write up to two digits after the decimal point)

Answer 1

$f={10}^3\Rightarrow T=\frac{1}{f}={10}^{-3}s$
$<i^2>=\frac{\int (2+3sin\theta {)}^{2}dt}{\int dt}=\frac{{\int }_0^T(4+9si{n}^2\theta +12sin\theta )dt}{{\int }_0^{T}dt}$
$<i^2>=4+\frac{9}{2}=\frac{17}{2}$
$i_{rms}=\sqrt{<i^2>}=\sqrt{\frac{17}{2}}$
Heat~lost ~in~one~cycle =
$i_{rms}^{2}R\times T$
$=\frac{17}{2}\times 4\times \frac{1}{{10}^3}$
Number of cycles in 3 minutes is,
$N=\frac{3\times 60}{T}=\frac{180}{{10}^{-3}}$
Therefore, heat lost in 3 minultes is
$=\frac{17}{2}\times 4\times \frac{1}{{10}^3}\times N$
= 6120 J = 6.12 kJ\)