First slide

Phasors and Phasor Diagrams

Question

A 120 volt AC source is connected across a pure inductor of inductance 0.70 henry. If the frequency of the source is 60 Hz, the current passing through the inductor is

Easy

A

4.55 A
B

0.355 A
C

0.455 A
D

3.55 A

Solution

$Z = X_{L} = 2 π \times 60 \times 0 .7$
$∴ i = \frac{120}{Z} = \frac{120}{2 π \times 60 \times 0 .7} = 0 .455 A$

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