Three resistances 2Ω, 4Ω, 5Ω are combined in series and this combination is connected to a battery of 12 V emf and negligible internal resistance. The potential drop across these resistances are
- a)(5,45,4.36,2.18)V
- b)(2,18,5.45,4.36)V
- c)(4.36,2.18,5.45)V
- d)(2.18,4.36,5.45)V
Correct answer is option 'D'. Can you explain this answer?
Related Test
Answers
|
Bs Academy
Feb 13, 2022 |
|
Let current in the circuit is I. Then total resistance in the circuit.
R = R1 + R2 + R3 = 2 + 4 + 5 = 11Ω
∴
The potential drop across 2Ω resistance
The potential drop across 4Ω resistance
The potential drop across 5Ω resistance
Hence(V1, V2, V3) = (2.18, 4.36, 5.45)V
Let current in the circuit is I. Then total resistance in the circuit.R = R1 + R2 + R3 = 2 + 4 + 5 = 11Ω∴The potential drop across 2Ω resistanceThe potential drop across 4Ω resistanceThe potential drop across 5Ω resistanceHence(V1, V2, V3) = (2.18, 4.36, 5.45)V
Top Courses for NEET
Related Content
 |
|
 |
|
 |
|
 |
|
Do you know? How Toppers prepare for NEET Exam
With help of the best NEET teachers & toppers, We have prepared a guide for student who are
preparing for NEET : 15 Steps to clear NEET Exam
Download free EduRev App
Track your progress, build streaks, highlight & save important lessons and more!
Let current in the circuit is I. Then total resistance in the circuit.R = R1 + R2 + R3 = 2 + 4 + 5 = 11Ω∴The potential drop across 2Ω resistanceThe potential drop across 4Ω resistanceThe potential drop across 5Ω resistanceHence(V1, V2, V3) = (2.18, 4.36, 5.45)V
|
© EduRev
|
Follow Us
|
