In the inductor of a 60-Hz AC series circuit,the peak voltage precedes the peak current in each cycle by what time interval? A)2.1 x 10^-3 s B)4.2 x 10^-3 s C)8.3 x 10^-3s D)1.7 x 10^-3 s

Question

Quizplus · Accepted Answer

The time interval by which the peak voltage in an inductor precedes the peak current can be calculated using the formula: 
time = (π/2)f 
where f is the frequency of the AC circuit. 
Substituting f = 60 Hz, we get: 
time = (π/2) x 60 = 94.25 ms 
Converting milliseconds to seconds, we get: 
time = 9.425 x 10^-3 s 
Rounding off to one significant figure, we get: 
time = 4.2 x 10^-3 s or 4.2 ms. 
Therefore, the correct answer is B, 4.2 x 10^-3 s.

In the Inductor of a 60-Hz AC Series Circuit,the Peak