Question

The inductance of a coil is 0.70 henry. An A.C. source of 120 volt is connected in parallel with it. If the frequency of A.C. is 60Hz, then the current which is flowing in inductance will be

• A. 4.55 A
• B. 0.355 A
• C. 0.455 A
• D. 3.55 A

Answer 1

Answer: (C)

0.455 A

$i=\frac{V_{rms}}{\omega L}$

$=\frac{120}{2\pi fL}$

$=\frac{120}{2\pi 60\times 0.7}$

$=0.455$A.