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(B) Let $n$ be the number of cells connected in series and $m$ be the number of parallel rows,each containing $n$ cells. The total number of cells is

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(B) Let $n$ be the number of cells connected in series and $m$ be the number of parallel rows,each containing $n$ cells. The total number of cells is $N = nm$.The total emf of the combination is $E_{eq} = nE = n(1.5)$.The total internal resistance is $r_{eq} = \frac{nr}{m} = \frac{n(1)}{m} = \frac{n}{m}$.The current $I$ in the external resistance $R = 30 \ \Omega$ is given by $I = \frac{nE}{R + \frac{nr}{m}}$.Given $I = 1.5 \ A$,$E = 1.5 \ V$,$R = 30 \ \Omega$,and $r = 1 \ \Omega$:$1.5 = \frac{n(1.5)}{30 + \frac{n}{m}} \implies 30 + \frac{n}{m} = n \implies 30 = n(1 - \frac{1}{m}) = n(\frac{m-1}{m})$.Since $N = nm$,we have $n = \frac{N}{m}$. Substituting this: $30 = \frac{N}{m} \cdot \frac{m-1}{m} = N \frac{m-1}{m^2}$.To minimize $N$,we need to maximize the function $f(m) = \frac{m-1}{m^2}$. Setting $f'(m) = 0$ gives $m=2$.For $m=2$,$30 = N \frac{2-1}{2^2} = N \frac{1}{4} \implies N = 120$.

Each cell has emf $1.5 \ V$ and internal resistance $1 \ \Omega$. The minimum number of such cells required to produce a maximum current of $1.5 \ A$ in an external load resistance of $30 \ \Omega$ is

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