Q111 / 125 · BITSAT 2015 · Physics
A direct current of $5 \mathrm{~A}$ is superposed on an alternating current $I=10 \sin \omega t$ flowing through the wire. The effective value of the resulting current will be
- A$(15 / 2) \mathrm{A}$
- Correct answerB$5 \sqrt{3} \mathrm{~A}$
- C$5 \sqrt{5} \mathrm{~A}$
- D$15 \mathrm{~A}$
Solution
Total carrent, $1=(5+10 \sin \omega \mathrm{t})$
$$
\begin{aligned}
\Rightarrow I_{\text {eff }} &=\left[\frac{\int_{0}^{\mathrm{T}} \mathrm{I}^{2} \mathrm{dt}}{\int_{0}^{\mathrm{T}} \mathrm{dt}}\right]^{1 / 2} \\
&=\left[\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}}(5+10 \sin \omega \mathrm{t})^{2} \mathrm{dt}\right]^{1 / 2} \\
&=\left[\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}}\left(25+100 \sin \omega \mathrm{t}+100 \sin ^{2} \omega \mathrm{t}\right)\right]^{1 / 2} \\
\text { But, } \quad \frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}} \sin \omega \mathrm{t} \mathrm{dt}=0
\end{aligned}
$$
and
$$
\begin{array}{l}
\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}} \sin ^{2} \omega \mathrm{t} \mathrm{dt}=\frac{1}{2} \\
\text { So, } \mathrm{I}_{\mathrm{eff}}=\left[25+\frac{1}{2} \times 100\right]^{1 / 2}=5 \sqrt{3} \mathrm{~A}
\end{array}
$$
Hence, the correct answer is option: 'B'.