Discussion
A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of the emf's is :
Answer: D
The ratio of the emfs is 3 : 2.
With the cells assisting, the balancing length measures \(E_1 + E_2\); with them opposing, it measures \(E_1 - E_2\).
\[\frac{E_1+E_2}{E_1-E_2} = \frac{50}{10} = 5\]
\[5E_1 - 5E_2 = E_1 + E_2 \Rightarrow 4E_1 = 6E_2 \Rightarrow \frac{E_1}{E_2} = \frac{3}{2}\]
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