Question

24 cells each of internal resistance 0.5 ohm and e.m.f 1.5v are used to send current to send external current through an external circuit of 3 ohm resistance the best method of grouping them is. ans:m=2,n=12​

Answer 1

Explanation:

if you know it's continution please tell me

Answer 2

Answer:

There will be 12 cells in series and 2 in parallel.

Explanation:

Let the number of cells in series be "n" and the number of cells in parallel be "m".So,

$mn=24$     (1)

The resistance of the cells in series=0.5n

And the total resistance of the circuit is,

$R=\frac{0.5n}{m}$      (2)

The maximum current passes through the circuit when the internal resistance of the battery of the cells is equal to the external resistance. So, we can write equation (2) as,

$3=\frac{0.5n}{m}$

$m=\frac{0.5n}{3}$       (3)

By putting the value of "m" in equation (1) we get;

$n\times \frac{0.5n}{3}=24$

$n^2=\frac{24\times 3}{0.5}$

$n^2=144$

$n=\sqrt{144}$

$n=12$      (4)

By putting the value of "n" in equation (1) we get;

$12\times m=24$

$m=2$       (5)

Hence, the correct answer among the given option is (4).i.e.n=12,m=2.