Question

A resistor of resistance $20\Omega$ is connected to a source of alternating current rated $110\text{V},50\text{Hz}$. Find the maximum instantaneous current in the resistor.

• A. $7.7\text{A}$
• B. $8.7\text{A}$
• C. $9.7\text{A}$
• D. $6.7\text{A}$

Answer 1

The correct option is A $7.7\text{A}$

Given:

$E_{rms}=110\text{V}$

So, $i_{rms}=\frac{E_{rms}}{R}=\frac{110}{20}=5.5\text{A}$

The maximum instantaneous current in the resistor,

$i_0=\sqrt{2}{i}_{rms}=1.4\times 5.5=7.7\text{A}$