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

Answer 1

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

$i_0=\frac{100}{20}=5A$

$\therefore i_{\text{rms}}=\frac{5}{\sqrt{2}}A$