Question

The emf of an A.C. source is given by $E=8sin\omega t+6cos\omega t$ volt. The rms value of emf is

• A. $5\sqrt{2}V$
• B. $7\sqrt{2}V$
• C. 10 V
• D. $10\sqrt{2}V$

Answer 1

The correct option is A $5\sqrt{2}V$

$E_{rms}=\sqrt{\frac{1}{T}{\int }_0^{T}E(t{)}^{2}dt}$

$(Erms{)}^2=\frac{{\int }_0^{\frac{2\pi }{w}}(64Si{n}^{2}wt+36Co{s}^{2}wt+96SinwtCoswt)dt}{{\int }_0^{\frac{2\pi }{w}}dt}$

$=\frac{{\int }_0^{\frac{2\pi }{w}}\left(\frac{64}{2}\right(1-Cos2wt)+\frac{36}{2}(1+Cos2wt\left)\right)+48Sin2wt)dt}{\left(\frac{2\pi }{w}\right)}$

$=\frac{\frac{64+36}{2}(\frac{2\pi }{w}-0)+{\int }_0^{\frac{2\pi }{w}}(-32Cos2wt+18cos2wt+48Sin2wt)dt}{\frac{2\pi }{w}}$

$E^{2}rms=50+zero$

$Erms=5\sqrt{2}$