Question

A number of n cells each of emf e and internal resistance r are connected over m rows and external resistance R . Derive the condition for which the current in the resistance R is maximum.

Answer 1

The expression for which the current in the resistance R is maximum is I = Ne / nr +mR

Explanation:

Current "I" =  ne / (  nr/m  + R)  = ne x m/nr + mR

I  = nme / nr +mR

But mn = N

So

I = Ne / nr +mR

The expression for which the current in the resistance R is maximum is I = Ne / nr +mR

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Answer 2

The current in the resistance R is maximum When n cells each of emf and internal resistance are connected in series with external resistance.

Explanation:

Given that,

A number of n cells each of emf e and internal resistance r are connected in series and m identical cells are connected in parallel with external resistance.

(I). When n cells each of emf e and internal resistance r are connected in series with external resistance

The net internal resistance = nr

Total resistance = R+nr

We need to calculate the current

Using formula of ohm's law

$I=\dfrac{ne}{R+r}$

The current is $\dfrac{ne}{R+r}$.

(II). When m identical cells are connected in parallel with external resistance

The net internal resistance is  $\dfrac{r}{m}$

The total resistance is $R+\dfrac{r}{m}$

We need to calculate the current

Using ohm's law

$I=\dfrac{e}{R+\dfrac{r}{m}}$

Hence, The current in the resistance R is maximum When n cells each of emf and internal resistance are connected in series with external resistance.

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Topic : internal resistance

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