Question

The RMS value of current $i = 3 + 4\sin \left( {\omega t + \dfrac{\pi }{3}} \right)$ is
A. $5A$
B. $\sqrt {17} A$
C. $\dfrac{5}{{\sqrt 2 }}A$
D. $\dfrac{7}{{\sqrt 2 }}A$

Answer 1

Hint: To find the Root mean square value of the current, we have to use the concept of RMS current which is a statistical measure of the magnitude of a current varying from different values. The RMS current and voltage (for sinusoidal systems) are the peak current and voltage over the square root of two.

Complete step by step answer:
Given, the value of current is
$i = 3 + 4\sin \left( {\omega t + \dfrac{\pi }{3}} \right)$
To find the RMS value of current, we have to square the quantity and then find the mean value of the functions. Squaring both sides, we get
${i^2} = {\left[ {3 + 4\sin \left( {\omega t + \dfrac{\pi }{3}} \right)} \right]^2}$
$\Rightarrow {i^2} = 9 + 16{\sin ^2}\left( {\omega t + \dfrac{\pi }{3}} \right) + 24\sin \left( {\omega t + \dfrac{\pi }{3}} \right)$

Taking the mean value of the current, then
$\left\langle {{i^2}} \right\rangle = 9 + 16\left( {\dfrac{1}{2}} \right) + 24\left( 0 \right)$
Here, $\left\langle {{{\sin }^2}x} \right\rangle = \dfrac{1}{2}\& \left\langle {\sin x} \right\rangle = 0$
$\Rightarrow \left\langle {{i^2}} \right\rangle = 17$
$\therefore \sqrt {\left\langle {{i^2}} \right\rangle } = \sqrt {17} A$
The RMS value of the given current is $\sqrt {17} A$.

Hence, option B is correct.

Note: AC is an alternating current i.e. it changes its direction & magnitude periodically. Hence the average value of AC is always zero because +ve & -ve value cancel out each other. Therefore , we use RMS value to define AC. It is the ‘root mean square’ value of AC. We should know the mean values of the various functions to solve the questions.