
All wires have the same resistance and equivalent resistance between A and B is ${R_y}$. Now keys are closed, then the equivalent resistance will become

A. \[\frac{{7{R_0}}}{3}\]
B. \[\frac{{7{R_0}}}{9}\]
C. \[7{R_0}\]
D. \[\frac{{{R_0}}}{3}\]
Answer
232.8k+ views
Hint: We understand how to solve resistor series and parallel combinations in any circuit. However, in this case, we have resistors in both parallel and series configurations. To solve this type of problem, we must divide it into series components and solve it repeatedly. The resistance in these circuits is estimated by resolving them piece by piece.
Formula used:
Parallel resistance’s equivalent resistance formula:
\[\frac{1}{{{{\rm{R}}_{{\rm{eq }}}}}} = \frac{1}{{{{\rm{R}}_{{\rm{1 }}}}}} + \frac{1}{{{{\rm{R}}_{{\rm{2 }}}}}} + \frac{1}{{{{\rm{R}}_{{\rm{3 }}}}}} + .....\]
Complete answer:
We have been provided in the given data that,
All wires have the same resistance and equivalent resistance between A and B is Ry.
Let us restructure the given circuit by breaking the circuit to simpler circuits for easy understanding, we have


Now, we have to calculate the equivalent resistance, for that we can write as
\[\frac{{{\rm{7}}{{\rm{R}}_{{\rm{0 }}}}}}{3} = {{\rm{R}}_{{\rm{eq }}}}\]
As, from the given circuit it is understood that,
\[{{\rm{R}}_{{\rm{0 }}}} = \frac{{\rm{R}}}{3}\]
Now, we have substitute the value obtained before, we get
\[{{\rm{R}}_{{\rm{eq }}}} = \frac{{7{\rm{R}}}}{9}\]
Therefore, when keys are closed, then the equivalent resistance will become \[{{\rm{R}}_{{\rm{eq }}}} = \frac{{7{\rm{R}}}}{9}\]
Hence, option B is the correct answer
Note:All wires have the same resistance and equivalent resistance between A and B is \[Ry\], then when keys are closed, the equivalent resistance will become \[\frac{{7{R_0}}}{9}\].
To generate the same impact on the circuit by allowing the same amount of current, one resistor can be used in place of two or more. This is referred to as efficient resistance.
Formula used:
Parallel resistance’s equivalent resistance formula:
\[\frac{1}{{{{\rm{R}}_{{\rm{eq }}}}}} = \frac{1}{{{{\rm{R}}_{{\rm{1 }}}}}} + \frac{1}{{{{\rm{R}}_{{\rm{2 }}}}}} + \frac{1}{{{{\rm{R}}_{{\rm{3 }}}}}} + .....\]
Complete answer:
We have been provided in the given data that,
All wires have the same resistance and equivalent resistance between A and B is Ry.
Let us restructure the given circuit by breaking the circuit to simpler circuits for easy understanding, we have


Now, we have to calculate the equivalent resistance, for that we can write as
\[\frac{{{\rm{7}}{{\rm{R}}_{{\rm{0 }}}}}}{3} = {{\rm{R}}_{{\rm{eq }}}}\]
As, from the given circuit it is understood that,
\[{{\rm{R}}_{{\rm{0 }}}} = \frac{{\rm{R}}}{3}\]
Now, we have substitute the value obtained before, we get
\[{{\rm{R}}_{{\rm{eq }}}} = \frac{{7{\rm{R}}}}{9}\]
Therefore, when keys are closed, then the equivalent resistance will become \[{{\rm{R}}_{{\rm{eq }}}} = \frac{{7{\rm{R}}}}{9}\]
Hence, option B is the correct answer
Note:All wires have the same resistance and equivalent resistance between A and B is \[Ry\], then when keys are closed, the equivalent resistance will become \[\frac{{7{R_0}}}{9}\].
To generate the same impact on the circuit by allowing the same amount of current, one resistor can be used in place of two or more. This is referred to as efficient resistance.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding Uniform Acceleration in Physics

Understanding the Electric Field of a Uniformly Charged Ring

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

