Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
9.9k views
in Physics by (130k points)

Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency.

1 Answer

0 votes
by (93.9k points)
edited by
 
Best answer

An inductor (L), a capacitor (C), and a resistor (R) is connected in parallel with each other in a circuit where,

L = 5.0 H 

C = 80 μF = 80 × 10−6 F R

 = 40 Ω 

Potential of the voltage source, V = 230 V 

Impedance (Z) of the given parallel LCR circuit is given as:

Hence, the magnitude of Z is the maximum at 50 rad/s. As a result, the total current is minimum. rms current flowing through inductor L is given as:

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

...