Question

An alternating current is given by the equation $i=i_{1}cos\omega t+i_{2}sin\omega t$. The r.m.s. current is given by:

• A. $\frac{1}{\sqrt{2}}(i_1+i_2)$
• B. $\frac{1}{\sqrt{2}}(i_1+i_2{)}^2$
• C. $\frac{1}{\sqrt{2}}(i_1^2+i_2^2{)}^{1/2}$
• D. $\frac{1}{2}(i_1^2+i_2^2{)}^{1/2}$

Answer 1

The correct option is C $\frac{1}{\sqrt{2}}(i_1^2+i_2^2{)}^{1/2}$

The r.m.s current is defined as $I_{rms}=\sqrt{<i^2>}$

Now, $i^2=i_1^{2}co{s}^2\omega t+i_2^2{\sin }^2\omega t+2i_1{i}_{2}cos\omega t\sin \omega t$

The mean value of $i^2$ $=<i^2>=\frac{1}{T}{\int }_0^T{i}^{2}dt$ where $T=\frac{2\pi }{\omega }$

So, $<i^2>=\frac{i^2}{T}{\int }_0^{T}co{s}^2\omega tdt+\frac{i_2^2}{T}{\int }_0^T{\sin }^2\omega tdt+\frac{i_1{i}_2}{T}{\int }_0^{T}2cos\omega t\sin \omega tdt$

Integration for first term,

${\int }_0^{T}co{s}^2\omega tdt=\frac{1}{2}{\int }_0^{T}2co{s}^2\omega tdt$

$=\frac{1}{2}{\int }_0^T[1+cos2\omega t]dt$ using formula $2co{s}^{2}A=1+cos2A$

$=\frac{1}{2}[t+\frac{\sin 2\omega t}{2}{]}_0^T=\frac{1}{2}[T+\frac{\sin 4\pi }{2}-0-\sin 0]$ (use $\omega T=2\pi$)

$=\frac{T}{2}$

Integration for second term,

${\int }_0^T{\sin }^2\omega tdt=\frac{1}{2}{\int }_0^{T}2{\sin }^2\omega tdt$

$=\frac{1}{2}{\int }_0^T[1-cos2\omega t]dt$ using formula $2{\sin }^{2}A=1-cos2A$

$=\frac{1}{2}[t-\frac{\sin 2\omega t}{2}{]}_0^T=\frac{1}{2}[T-\frac{\sin 4\pi }{2}-0+\sin 0]$ (use $\omega T=2\pi$)

$=\frac{T}{2}$

Integration for third term,

${\int }_0^{T}2\sin \omega tcos\omega tdt={\int }_0^T\sin 2\omega tdt=[-\frac{cos2\omega t}{2}{]}_0^T=-\frac{1}{2}[1-1]=0$

$<i^2>=\frac{i_1^2}{2}+\frac{i_2^2}{2}+0$

Thus, $i_{rms}=\sqrt{i_1^2/2+i_2^2/2}=\frac{1}{\sqrt{2}}(i_1^2+i_2^2{)}^{1/2}$