Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • In an a.c.circuit , V & I are given by  <img alt="\mathrm{\begin{aligned} & V=100 \sin (100 t) \text { volt. } \\ & I=100 \sin (100 t+\pi / 3) \mathrm{mA} . \end{aligned}

In an a.c.circuit , V & I are given by 

\mathrm{\begin{aligned} & V=100 \sin (100 t) \text { volt. } \\ & I=100 \sin (100 t+\pi / 3) \mathrm{mA} . \end{aligned}}

The power dissipate in the ciicuit  is :

Option: 1

\mathrm{10^{+}watt}


Option: 2

\mathrm{ 10 \: watt}


Option: 3

\mathrm{2.5 \: watt}


Option: 4

\mathrm{5 \: watt}


Answers (1)

best_answer

\mathrm{\begin{aligned} \mathrm{P}=\mathrm{V}_{n=m} & \mathrm{I}_{n=} \cos \phi \\ & =\frac{100}{\sqrt{2}} \cdot\left(\frac{100}{\sqrt{2}} \times 10^{-3}\right) \cos 60^{\circ} \\ & =\frac{10}{2} \times \frac{1}{2}=2.5 \mathrm{watt} . \end{aligned}}

Posted by

Pankaj Sanodiya

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

NTA Exams 2025 Live: NEET UG date, registration form soon at neet.nta.nic.in; UGC NET revised schedule out
anam-khan

Is it difficult to get into NIT Durgapur? JEE Main cut-offs for BTech programmes
anam-khan

JEE Main Deleted Syllabus 2025: Topics to skip this year; why no optional questions?

Latest Question