In a battery 12 cells each having the same emf are connected in series and are kept in a closed box. Some of the cells are wrongly connected. This battery is connected in series with an ammeter and two cells identical with others.The current is 3A when the cells and battery aid each other and 2A when the cells and battery oppose each other. The number of cells wrongly connected in the battery are

Answers (1)

 

If one cell is connected in reverse order, then it will density one more cell .If a cell are connected in reserve orther then 2n cell will be destroyed. If E is the emf of each cell then effective emf of 12 cells will be = (12−2n)E

When this battery if and two similar calls series then it supply x current of 3A to resistance R Therefore,

3=\frac{(12-2 n) E+2 E}{R}=\frac{(14-2 n) E}{R} .......(i)

when battery aqnd two cells oppose each other then the copper in the external resistance is 2A. Hence 

\begin{aligned} &2=\frac{(12-2 n) E-2 E}{R}=\frac{(10-2 n) E}{R} \ldots\\ &\text { Solving (i) by (ii) we get }\\ &\frac{3}{2}=\frac{14-2 n}{10-2 n} \text { or } n=1 \end{aligned}

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
Knockout JEE Main April 2021 (Easy Installments)

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

₹ 4999/-
Buy Now
Knockout JEE Main April 2021

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

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

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

₹ 34999/- ₹ 24999/-
Buy Now
Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Exams
Articles
Questions