
Calculate the equivalent resistance between point M and N.

A) $5\Omega $
B) $6\Omega $
C) $3\Omega $
D) $7\Omega $
Answer
133.5k+ views
Hint: To solve this connection you should know the series connection and parallel connection of resistances. In series connection, the equivalent resistance is simply the sum of the individual resistances connected in series.
In parallel connection, the reciprocal of equivalent resistance is equal to the sum of the reciprocal of individual resistances connected parallelly.
Complete step by step solution:
From the figure resistance ${R_1}$ and ${R_2}$ in series so the equivalent of them will be the direct of sum of their individual resistances,
$R' = {R_1} + {R_2}$
Put the values of ${R_1}$ and ${R_2}$ in the above equation to get the value of resistance.
$R' = 6 + 6$
$\Rightarrow 12\,\Omega$
Similarly, resistance ${R_4}$ and ${R_5}$ are also in series so the equivalent of them will also be the direct of sum of their individual resistances,
$R'' = {R_4} + {R_5}$
Put the values of ${R_4}$ and ${R_5}$ in above equation,
$R'' = 6 + 6$
$ \Rightarrow 12\,\Omega $
Now, we can see that resistance $R',{R_3}$ and $R’’$ are connected by parallel connection so the equivalent resistance of these resistances can be determined as,
$\dfrac{1}{{R'''}} = \dfrac{1}{{R'}} + \dfrac{1}{{{R_3}}} + \dfrac{1}{{R''}}$
Substitute the values of resistances in above equation,
$\dfrac{1}{{R'''}} = \dfrac{1}{{12}} + \dfrac{1}{6} + \dfrac{1}{{12}}\\
\Rightarrow \dfrac{{1 + 2 + 1}}{{12}}\\
\Rightarrow \dfrac{1}{{R'''}} = \dfrac{4}{{12}} = \dfrac{1}{3}\\
\Rightarrow R''' = 3\,\Omega
$
Resistances ${R_6},R'''$ and ${R_7}$ are connected in series so the equivalent resistance of them determined as,
${R_{eq}} = {R_6} + R''' + {R_7}$
Put the values in the above equation to get equivalent resistance.
$ {R_{eq}} = 2 + 3 + 2\\
\Rightarrow 7\,\Omega
$
Thus the equivalent resistance of the circuit is $7\,\Omega $.
Note: In this type of questions equivalent resistance cannot calculate directly. You have to use proper steps to determine equivalent resistance.
Solve the question by breaking it into small circuits and then find step by step resistance. Make sure that resistance connected parallel should be solved by the parallel connection method of equivalent resistance and resistance connected in series connection should be determined by equivalent resistance of series connection method.
In parallel connection, the reciprocal of equivalent resistance is equal to the sum of the reciprocal of individual resistances connected parallelly.
Complete step by step solution:
From the figure resistance ${R_1}$ and ${R_2}$ in series so the equivalent of them will be the direct of sum of their individual resistances,
$R' = {R_1} + {R_2}$
Put the values of ${R_1}$ and ${R_2}$ in the above equation to get the value of resistance.
$R' = 6 + 6$
$\Rightarrow 12\,\Omega$
Similarly, resistance ${R_4}$ and ${R_5}$ are also in series so the equivalent of them will also be the direct of sum of their individual resistances,
$R'' = {R_4} + {R_5}$
Put the values of ${R_4}$ and ${R_5}$ in above equation,
$R'' = 6 + 6$
$ \Rightarrow 12\,\Omega $
Now, we can see that resistance $R',{R_3}$ and $R’’$ are connected by parallel connection so the equivalent resistance of these resistances can be determined as,
$\dfrac{1}{{R'''}} = \dfrac{1}{{R'}} + \dfrac{1}{{{R_3}}} + \dfrac{1}{{R''}}$
Substitute the values of resistances in above equation,
$\dfrac{1}{{R'''}} = \dfrac{1}{{12}} + \dfrac{1}{6} + \dfrac{1}{{12}}\\
\Rightarrow \dfrac{{1 + 2 + 1}}{{12}}\\
\Rightarrow \dfrac{1}{{R'''}} = \dfrac{4}{{12}} = \dfrac{1}{3}\\
\Rightarrow R''' = 3\,\Omega
$
Resistances ${R_6},R'''$ and ${R_7}$ are connected in series so the equivalent resistance of them determined as,
${R_{eq}} = {R_6} + R''' + {R_7}$
Put the values in the above equation to get equivalent resistance.
$ {R_{eq}} = 2 + 3 + 2\\
\Rightarrow 7\,\Omega
$
Thus the equivalent resistance of the circuit is $7\,\Omega $.
Note: In this type of questions equivalent resistance cannot calculate directly. You have to use proper steps to determine equivalent resistance.
Solve the question by breaking it into small circuits and then find step by step resistance. Make sure that resistance connected parallel should be solved by the parallel connection method of equivalent resistance and resistance connected in series connection should be determined by equivalent resistance of series connection method.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main Books 2023-24: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 13 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Wheatstone Bridge for JEE Main Physics 2025

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

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

Diffraction of Light - Young’s Single Slit Experiment

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

JEE Advanced 2024 Syllabus Weightage

Current Loop as Magnetic Dipole and Its Derivation for JEE
