Question

A $10\Omega$ electric iron is connected to $120V,60Hz$ wall outlet. What is the peak current through the electric iron?

• A. $12A$
• B. $12\sqrt{2}A$
• C. $10\sqrt{2}A$
• D. 10 A

Answer 1

The correct option is B $12\sqrt{2}A$

Given $V_{rms}=120V$
Since only resistance is connected in the a.c circuit,
$i_{rms}=\frac{V_{rms}}{10\Omega }=12A$
and current and voltage are in phase.

We know $i_{rms}=\frac{i_0}{\sqrt{2}}$
$\Rightarrow i_0=i_{rms}\times \sqrt{2}=12\sqrt{2}A$