Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • A coil of inductance 0.12 H and resistance   is connected to an AC–source of 13 V, <img alt="\fr

A coil of inductance 0.12 H and resistance 5\Omega  is connected to an AC–source of 13 V, \frac{50}{\pi}Hz. The power factor of this circuit is

Option: 1

\mathrm{\frac{12}{5}}


Option: 2

\mathrm{\frac{12}{13}}


Option: 3

5


Option: 4

\frac{5}{13}


Answers (1)

best_answer

The power factor is given by,

\mathrm{\begin{aligned} & \cos \phi=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+(\omega \mathrm{L})^2}} \\ & =\frac{5}{\sqrt{5^2+(12)^2}}=\frac{5}{13} . \end{aligned}}

Posted by

chirag

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

NEET UG 2026 Tie-Breaking Rule Explained: Here’s how ranks are decided
anam-khan

NEET 2026: Odisha aspirant dies by suicide fearing failure
anam-khan

NEET 2026 Answer Key (Out) LIVE: NTA to open challenge window shortly; direct link, result date, counselling

Latest Question