Question

A pure inductor of self inductance 1 H is connected across an alternating voltage of 115 V and frequency 60 Hz. The reactance $X_L$ and peak current respectively are

• A. 37.1$\Omega$, 0.43 A
• B. 337.1$\Omega$, 0.43 A
• C. 377.1$\Omega$, 0.43 A
• D. 3.7$\Omega$, 0.42 A

Answer 1

The correct option is C 377.1$\Omega$, 0.43 A

$X_L=\omega L$
$=2\pi fL$
$=2\pi 60\times 1$
$=377.1\Omega$
peak current = $\frac{\text{peak voltage}}{X_L}$

$=\frac{115\times \sqrt{2}}{377.1}=0.43A$

So, $X_L=377.1$

peak current, $I=0.43A$.