Question

The rms value of alternating emf $E=(8\sin \omega t+6\sin 2\omega t)V$ is

• A. $7.07V$
• B. $14.14V$
• C. $10V$
• D. $20V$

Answer 1

The correct option is A $7.07V$

Given that,

$E=(8\sin \omega t+6\sin 2\omega t)v$

Now, the mean square value of e. m. f

${\overline{E}}^2=\frac{\underset{0}{\overset{T}{\int }}{E}^{2}dt}{\underset{0}{\overset{T}{\int }}dt}$

${\overline{E}}^2=\frac{\underset{0}{\overset{T}{\int }}{(8\sin \omega t+6\sin 2\omega t)}^{2}dt}{\underset{0}{\overset{T}{\int }}dt}$

${\overline{E}}^2=\frac{\underset{0}{\overset{T}{\int }}(64{\sin }^2\omega t+36{\sin }^{2}2\omega t+96\sin \omega t\sin 2\omega t)dt}{\underset{0}{\overset{T}{\int }}dt}$

We know that,

${\sin }^2\omega t=\frac{\underset{0}{\overset{T}{\int }}{\sin }^2\omega tdt}{\underset{0}{\overset{T}{\int }}dt}=\frac{1}{2}$

${\sin }^{2}2\omega t=\frac{\underset{0}{\overset{T}{\int }}{\sin }^{2}2\omega tdt}{\underset{0}{\overset{T}{\int }}dt}=\frac{1}{2}$

$\sin \omega t\sin 2\omega t=\frac{\underset{0}{\overset{T}{\int }}\sin \omega tsin2\omega tdt}{\underset{0}{\overset{T}{\int }}dt}=0$

Now, put the value of ${\sin }^2\omega t,{\sin }^{2}2\omega t,\sin \omega t\sin 2\omega t$

${\overline{E}}^2=64\times \frac{1}{2}+36\times \frac{1}{2}+96\times 0$

${\overline{E}}^2=32+18$

${\overline{E}}^2=50$

Now,

$E_{rms}=\sqrt{{\overline{E}}^2}$

$E_{rms}=\sqrt{50}$

$E_{rms}=7.07$

Hence, the rms value is $7.07$