Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • An ideal choke takes a current of 8 A when connected to an AC source of 100 V and 50 Hz. A pure resistor under the same conditions takes a current of 10 A. If two are connected in series to an AC s

An ideal choke takes a current of 8 A when connected to an AC source of 100 V and 50 Hz. A pure resistor under the same conditions takes a current of 10 A. If two are connected in series to an AC supply of 100 V and 40 Hz, then the current in the series combination of above resistor and inductor is:

Option: 1

10 A


Option: 2

5 A


Option: 3

5 \sqrt{2} \mathrm{~A}


Option: 4

10 \sqrt{2} \mathrm{~A}


Answers (1)

best_answer

\mathrm{X}_{\mathrm{L}}=\frac{100}{8} \Omega=12.5 \Omega  with 50 Hz frequency and with 40 Hz frequency

\mathrm{ \mathrm{X}_{\mathrm{L}}^{\prime}=12.5 \times \frac{40}{50}=10 \Omega \quad\left[\right. as \left.\mathrm{X}_{\mathrm{L}} \propto \omega\right]}
\mathrm{ \begin{aligned} & \mathrm{R}=\frac{100}{10}=10 \Omega \\ & \therefore \quad \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}^{\prime}\right)^2}=10 \sqrt{2} \Omega \\ & \quad \mathrm{I}=\frac{100}{10 \sqrt{2}}=5 \sqrt{2} \mathrm{~A} \end{aligned}}

                       

Posted by

manish

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

JEE Main 2026: NTA reopens Session 2 registration; apply by March 13 at jeemain.nta.nic.in
anam-khan

JEE Mains 2026 LIVE: NTA session 2 registration last day, city slip soon; here's how to prepare
anam-khan

JEE Main 2026 session 2 application correction begins on jeemain.nta.nic.in; edit details by tomorrow

Latest Question