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There is a `5 Omega` resistance in an `AC`, circuit. Impedence of `0.1 H` is connected with it in series. If equation of `AC` emf is `5 sin 50 t` then the phase difference between current and e.m.f. isA. `(pi)/2`B. `(pi)/6`C. `(pi)/4`D. `0`

Answer» Correct Answer - C
`cos varphi=R/Z=R/(sqrt(R^(2)+omega^(2)L^(2)))=5/(sqrt(25+(50)^(2)xx(0.1)^(2)))`
`=5/(sqrt(25+25))=1/(sqrt(2)) implies varphi=pi//4`


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