Question

The r.m.s. value of current for a variable current $i={{i}_{1}}\cos \omega t+{{i}_{2}}$ :
A. $\frac{1}{\sqrt{2}}\left( {{i}_{1}}+{{i}_{2}} \right)$
B. $\frac{1}{\sqrt{2}}{{\left( {{i}_{1}}+{{i}_{2}} \right)}^{2}}$
C. $\frac{1}{\sqrt{2}}{{\left( i_{1}^{2}+i_{2}^{2} \right)}^{\frac{1}{2}}}$
D. ${{\left( \frac{i_{1}^{2}}{2}+i_{2}^{2} \right)}^{\frac{1}{2}}}$

Answer 1

Hint: The r.m.s. value of current can be found by the relation ${{i}_{rms}}={{\left( \frac{\int{{{i}^{2}}dt}}{\int{dt}} \right)}^{\frac{1}{2}}}$ where the integration takes place under the limits of initial and final time period of a full cycle of current. Using the above relation find the r.m.s. value of the current for the given variable current.

Complete answer:
Given that the current is varying and the equation of the variable current is given as $i={{i}_{1}}\cos \omega t+{{i}_{2}}$
To find the r.m.s. value of the variable current, we are having the relation for the r.m.s. value of the current as follows:
${{i}_{rms}}={{\left( \frac{\int{{{i}^{2}}dt}}{\int{dt}} \right)}^{\frac{1}{2}}}$
Where the integration takes place under initial and final value of the time period of a complete cycle of the current.
Now put the give variable current in the above formula and integrate it
${{i}_{rms}}={{\left( \frac{\int{{{\left( {{i}_{1}}\cos \omega t+{{i}_{2}} \right)}^{2}}dt}}{\int{dt}} \right)}^{\frac{1}{2}}}$
Expand the formula in the above equation
${{i}_{rms}}={{\left( \frac{\int\limits_{0}^{T}{\left( i_{1}^{2}{{\cos }^{2}}\omega t+i_{2}^{2}+2\cdot {{i}_{1}}{{i}_{2}}\cos \omega t \right)dt}}{\int\limits_{0}^{T}{dt}} \right)}^{\frac{1}{2}}}$
For a complete cycle, that is under the limits of 0 to T the integration of $\cos \omega t$will becomes zero while the integration of square of it will results half $\left( i.e.\text{ }\int\limits_{0}^{T}{\cos \omega t}=0;\int\limits_{0}^{T}{{{\cos }^{2}}\omega t}=\frac{1}{2} \right)$
Put these values in the above equation
${{i}_{rms}}={{\left( \frac{\int\limits_{0}^{T}{\left( i_{1}^{2}\left( \frac{1}{2} \right)+i_{2}^{2} \right)dt}}{T} \right)}^{\frac{1}{2}}}$
${{i}_{rms}}={{\left( \frac{T\left( i_{1}^{2}\left( \frac{1}{2} \right)+i_{2}^{2} \right)}{T} \right)}^{\frac{1}{2}}}$
So finally we obtained the r.m.s. value of the given variable current is
${{i}_{rms}}={{\left( i_{1}^{2}\left( \frac{1}{2} \right)+i_{2}^{2} \right)}^{\frac{1}{2}}}$

Hence option (D) is the correct answer.

Note:
We should take care while integrating with the trigonometric terms, sometimes the integration looks like zero but it is not. Also when we got the terms which cannot be integrated directly or the terms that do not have the direct formula for integration then we should convert them into other trigonometric terms which can be directly integrated.