Question

A direct current of 5 Amp is superimposed on on AC of $I=5sin\omega t$ flowing through a wire. The effective value of the resulting current is

• A. $\frac{15}{2}Amp$
• B. $5\sqrt{3/2}$ Amp
• C. $5\sqrt{5}$ Amp
• D. $15Amp$

Answer 1

The correct option is B $5\sqrt{3/2}$ Amp

$I=5+5sin\omega t{I}_{rms}=\sqrt{\frac{{\int }_0^{\frac{2\pi }{\omega }}{I}^{2}dt}{{\int }_0^{\frac{2\pi }{\omega }}dt}}=\sqrt{\frac{{\int }_0^{\frac{2\pi }{\omega }}{(5+5sin\omega t)}^{2}dt}{{\int }_0^{\frac{2\pi }{\omega }}dt}}=\sqrt{\frac{{\int }_0^{\frac{2\pi }{\omega }}(25+25{sin}^2\omega t+50sin\omega t)dt}{{\int }_0^{\frac{2\pi }{\omega }}dt}}=\sqrt{\frac{\frac{50\pi }{\omega }+\frac{25\pi }{\omega }+0}{\frac{2\pi }{\omega }}}=\sqrt{\frac{75}{2}}=5\sqrt{\frac{3}{2}}$