Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
76 views
in Physics by (92.5k points)
closed by
An `AC` source of angular frequency `omega` is fed across a resistor `R` and a capacitor `C` in series. The current registered is `I`. If now the frequency of source is changed to `omega//3` (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of reactance to resistance at the original frequency `omega` will be.
A. `sqrt((3)/(5))`
B. `sqrt((2)/(5))`
C. `sqrt((1)/(5))`
D. `sqrt((4)/(5))`

1 Answer

0 votes
by (93.5k points)
selected by
 
Best answer
Correct Answer - A
`I = (V)/(sqrt(R^(2) + (1)/(omega^(2)C^(2)))) = (V)/(sqrt(R^(2) + X_(C)^(2))) …(1)`
`1//2 = (V)/(sqrt(R^(2) + (9)/(omega^(2)C^(2)))) = (V)/(sqrt(R^(2) + (9X_(C)^(2)))) …(2)`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...