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From the two e.m.f. Equation e(1)=E(0) s...

From the two e.m.f. Equation `e_(1)=E_(0) sin (100 pi t)` and `e_(2)=E_(0)sin (100 pi t+(pi)/3)`, we find that

A

`e_(1) "leads" e_(2) "by" 60^(@)`

B

`e_(2) "lags behind" e_(1) "by" 60^(@)`

C

`e_(2)` achieves its maximum value `1/300` second before `e_(1)`

D

`e_(1)` achieves its maximum value `1/300` second before `e_(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
For both, `omega t = 2 pi f t = 100 pi t`
`:. F = 50 Hz :. T = 1/50 s`
For a phase diff. = `2 pi`, time required = `T = 1/50 s`
`:.` For a phase diff. of `(pi)/3`,
time = `(pi)/3 xx 1/50 xx 1/(2 pi) = 1/300 s`.
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MARVEL PUBLICATION-ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS -TEST YOUR GRASP - 16
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