Question 128: AC fundamentals

Question

The electric current in the circuit is given as $i=i_{\mathrm{o}}(t / T)$. The r.m.s current for the period $t=0$ to $t=T$ is $\_\_\_\_$ .

Accepted Answer

The root mean square (r.m.s) value of a time-varying current $i(t)$ over a period $T$ is defined by the formula :
$$ \mathrm{i}_{\mathrm{rms}}=\sqrt{\frac{1}{\mathrm{~T}} \int\limits_0^{\mathrm{T}} \mathrm{i}^2 \mathrm{dt}} $$
We are given $\mathrm{i}=\mathrm{i}_{\mathrm{o}}\left(\frac{\mathrm{t}}{\mathrm{T}}ight)$.
So, using the rms formula,
$$ \mathrm{i}_{\mathrm{rms}}^2=\frac{1}{\mathrm{~T}} \int\limits_0^{\mathrm{T}}\left(\mathrm{i}_{\mathrm{o}}^2 \frac{\mathrm{t}^2}{\mathrm{~T}^2}ight) \mathrm{dt} $$
Since $i_o$ and $T$ are constants, we can move them outside the integral :
$\mathrm{i}_{\mathrm{rms}}^2=\frac{\mathrm{i}_{\mathrm{o}}^2}{\mathrm{~T}^3} \int_0^{\mathrm{T}} \mathrm{t}^2 \mathrm{dt}=\frac{\mathrm{i}_{\mathrm{o}}^2}{\mathrm{~T}^3}\left[\frac{\mathrm{t}^3}{3}ight]_0^{\mathrm{T}}=\frac{\mathrm{i}_{\mathrm{o}}^2}{\mathrm{~T}^3} 	imes \frac{\mathrm{T}^3}{3}$
$\Rightarrow \mathrm{i}_{\mathrm{rms}}=\sqrt{\frac{\mathrm{i}_{\mathrm{o}}^2}{3}}$
$$\Rightarrow $$ $$ \mathrm{i}_{\mathrm{rms}}=\frac{\mathrm{i}_{\mathrm{o}}}{\sqrt{3}} $$
Hence, the correct option is (B).

Answer

$\frac{i_{\mathrm{o}}}{\sqrt{2}}$

Answer

$\frac{i_o}{\sqrt{3}}$

Answer

$i_{\mathrm{o}}$

Answer

$\frac{i_o}{\sqrt{6}}$

AC fundamentals

The electric current in the circuit is given as i=i_{\mathrm{o}}(t / T). The r.m.s current for the period t=0 to t=T is \_\_\_\_ .

Choose the correct answer: