Question

An ac source of $V=100sin\left(100\pi t\right)$ is connected to a resistor of $20\Omega$. The rms value of the current through the resistor is

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

Answer 1

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

$i_0=\frac{100}{20}=5A\therefore i_{rms}=\frac{5}{\sqrt{2}}A$