Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

When and AC source of emf \mathrm{e=E_0 \sin (100 t) } is connected across a circuit, the phase difference between the emf e and the current l in the circuit is observed to be \mathrm{ \frac{\pi}{4}}, as shown in the diagram. If the circuit consists possibly only of R–C or R–L or L–C in series, find the relationship between the two elements :
                                                          

Option: 1

R=1 k \Omega, C=10 \mu F


Option: 2

R=1 k \Omega, C=1 \mu F


Option: 3

R=1 \mathrm{k} \Omega, L=10 \mathrm{H}


Option: 4

R=1 \mathrm{k} \Omega, L=1 \mathrm{H}


Answers (1)

best_answer

As the current i leads the emf e by \frac{\pi}{4}, it is an R-C circuit.

\mathrm{\begin{aligned} \operatorname{Tan} \phi & =\frac{X_C}{R} \\ \text { or } \quad \tan \frac{\pi}{4} & =\frac{\frac{1}{\omega C}}{R} \\ \therefore \quad \omega C R & =1 \\ \text { As } \quad \omega & =100 \mathrm{rad} / \mathrm{s} \end{aligned} }
The product of C-R should be \mathrm{\frac{1}{100} \mathrm{~s}^{-1} }
\mathrm{\therefore } correct answer is (a)

Posted by

himanshu.meshram

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: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question