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
14.2k views
in Physics by (51.7k points)
edited by
How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?

1 Answer

0 votes
by (266k points)
selected by
 
Best answer

(a) To get total resistance 4 Ω, connect 3 Ω and 6 Ω resistors in parallel and 2 Ω resistance in series with the resultant.

⟹12=2/1=2 Ω Now, the resultant 12 and 2 Ω resistors are in series. So the net resistance =12+2 Ω=2+2=4 Ω (b). To get total resistance 1 Ω, connect 2 Ω, 3 Ω and 6 Ω resistors in parallel.
The net resistance in parallel is given by

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

...