alleryStart learning →A 0.2 H inductor is joined to a 50 Hz source of peak emf 210 V. What is the peak current?
A6.68 A
B3.34 A
C1.05 A
D10.5 A
Answer & Solution
Correct answer: B. 3.34 A
XL = 2π × 50 × 0.2 = 62.8 Ω, so i0 = 210/62.8 = 3.34 A.
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