Discussion
The difference between maximum and minimum resistance that can be obtained by connecting three resistances of \(\frac{1}{3}\ \Omega\) each is :
Answer: B
The difference is \(\frac{8}{9}\ \Omega\).
The maximum resistance comes from connecting all three in series and the minimum from connecting all three in parallel:
\[ \begin{aligned} R_{max} &= 3 \times \tfrac{1}{3} = 1\ \Omega \\ \frac{1}{R_{min}} &= 3 + 3 + 3 = 9 \Rightarrow R_{min} = \tfrac{1}{9}\ \Omega \\ R_{max} - R_{min} &= 1 - \tfrac{1}{9} = \tfrac{8}{9}\ \Omega \end{aligned} \]
1 Ω and 1/9 Ω are the maximum and minimum themselves, not their difference, and 2/3 Ω does not follow from either arrangement.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development