There are 45 number of cells with internal resistance of each cell is 0.5Ω . To get the maximum current through a resistance of 2.5Ω , one can use m rows of cells, each row having n cells. The values of m and n are

Question

There are 45 number of cells with internal resistance of each cell is 0.5Ω . To get the maximum current through a resistance of 2.5Ω , one can use m rows of cells, each row having n cells. The values of m and n are (A) m=3,n=15 (B) m=5,n=9 (C) m=9,n=5 (D) m=15,n=3

Tardigrade Admin · Accepted Answer

No. of cells in series =n No. of rows =m Total internal resistance r'=(n r/m) To have max. current (n r/m)=R (n (. 0 . 5 .)/m)=2.5⇒ (n/m)=5 5 n× m=45⇒ n=15,m=3 3

Q. There are $45$ number of cells with internal resistance of each cell is $0.5Ω$ . To get the maximum current through a resistance of $2.5Ω$ , one can use $m$ rows of cells, each row having $n$ cells. The values of $m$ and $n$ are

$m = 3, n = 15$

$m = 5, n = 9$

$m = 9, n = 5$

$m = 15, n = 3$

No. of cells in series $= n$
No. of rows $= m$
Total internal resistance $r^{^{'}} = \frac{n r}{m}$
To have max. current $\frac{n r}{m} = R$
$\frac{n ( 0.5 )}{m} = 2.5 \Rightarrow \frac{n}{m} = 5$ 5 $< b r / >$ n\times m=45\Rightarrow n=15,m=3 $3$

Q. There are 45 number of cells with internal resistance of each cell is 0.5Ω . To get the maximum current through a resistance of 2.5Ω , one can use m rows of cells, each row having n cells. The values of m and n are

m=3,n=15

m=5,n=9

m=9,n=5

m=15,n=3