Question

An AC source of $V=100sin\left(100\pi t\right)$(in $Volts$) is connected to a resistor of $20\Omega$. The r.m.s. 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 $\frac{5}{\sqrt{2}}A$

Given,
$V=100sin\left(100\pi t\right)$ Volt
$R=20\Omega$

Ohm's law, $i=\frac{V}{R}$

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

where, $i_{rms}=\frac{i_0}{\sqrt{2}}$
$\therefore i_{\text{rms}}=\frac{5}{\sqrt{2}}A$