Question

$24$ cells, each having the same e.m.f. and $2$ ohm internal resistance, are used to draw maximum current through an external resistance of $3$ ohm. The cells should be connected :

• A. In series
• B. in parallel
• C. In $4$ rows, each row having $6$ cells
• D. In $6$ rows, each row having $4$ cells

Answer 1

The correct option is A In series

(A) Option in series

Net EMF = $n\epsilon$

= 24$ϵ$

Net resistance =nr

= 24$\times$2

=48$\Omega$

Current = $\frac{24\epsilon }{nr+R}$

$=\frac{24\epsilon }{nr+R}$

$=\frac{24\epsilon }{48+3}$

$=\frac{24\epsilon }{51}=0.47\epsilon$

(B) Net EMF = $\epsilon$ = (In parallel EMF is same )

Net resistance $=\frac{r}{n}=\frac{1\Omega }{12}$

Current $=\frac{\epsilon }{\frac{1}{12}+3}$

$=\frac{12\epsilon }{37}=0.32\epsilon$

(C) Option

net EMF in each row$=n\epsilon$

$=6\epsilon$

net resistance in each row $=n\epsilon$

$=12\Omega$

net EMF will be $6\epsilon$ ( this will be in parallel)

Net resistance $\frac{r}{n}$

$=\frac{12}{4}=3\Omega$

Net current $=\frac{6\epsilon }{3+3}$

$=1\epsilon A$

(D) Option

Net EMF in each row $n\epsilon$

=$4\epsilon$

Net resistance in each row = nr

= 4$\times$2

= 8$\Omega$

Net EMF will be $4\epsilon$ ( this will be in parallel)

Net resistance $\frac{r}{n}=\frac{8}{6}=\frac{4}{3}\Omega$

Current $\frac{4\epsilon }{\frac{4}{3}+3}=\frac{12\epsilon }{13}=0.92\epsilon$

Hence current will be maximum in (C) option