Q.11.    Show how you would connect three resistors, each of resistance6\Omega , so that the combination has a resistance of (i) 9\Omega, (ii) 4\Omega .
 

Answers (1)

(i) R_1=R_2=R_3=6

    R=Equivalent  resistance

    \frac{1}{R_1_2}=\frac{1}{R_1}+\frac{1}{R_2} 

\Rightarrow \frac{1}{R_1_2}=\frac{1}{6}+\frac{1}{6}

\Rightarrow \frac{1}{R_1_2}=\frac{1+1}{6}

\Rightarrow R_1_2=\frac{6}{2}=3

R=R_1_2+R_3=3+6=9\Omega

(ii)

.R_1=R_2=R_3=6

  R=Equivalent  resistance

R_1_2=R_1+R_2=6+6=12\Omega

 

  \frac{1}{R}=\frac{1}{R_1_2}+\frac{1}{R_3} 

\Rightarrow \frac{1}{R}=\frac{1}{12}+\frac{1}{6}

\Rightarrow \frac{1}{R}=\frac{1+2}{12}

\Rightarrow R=\frac{12}{3}=4

 

 

 

 

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout NEET May 2022

An exhaustive E-learning program for the complete preparation of NEET..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions