$A$ battery of $24$ cells,each of emf $1.5\,V$ and internal resistance $2\,\Omega$,is to be connected in order to send the maximum current through a $12\,\Omega$ resistor. The correct arrangement of cells will be:

  • A
    $2$ rows of $12$ cells connected in series
  • B
    $3$ rows of $8$ cells connected in series
  • C
    $4$ rows of $6$ cells connected in series
  • D
    All of these

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Similar Questions

In how many ways can a combination of cells be done? Explain.

There are a large number of cells available,each marked $(6 \,V, 0.5 \,\Omega)$,to be used to supply current to a device of resistance $0.75 \,\Omega$,requiring $24 \,A$ current. How should the cells be arranged so that power is transmitted to the load using the minimum number of cells?

Ten identical cells each emf $2 \, V$ and internal resistance $1 \, \Omega$ are connected in series with two cells wrongly connected. $A$ resistor of $10 \, \Omega$ is connected to the combination. What is the current through the resistor (in $ \, A$)?

When two cells of emfs $E_1$ and $E_2$ and different internal resistances $r_1$ and $r_2$ are connected in series with an external load resistor $R$,the current through the load is $5 \ A$. If the polarity of the cell of emf $E_2$ is reversed,then the current through the load is $2 \ A$. Then $\frac{E_1}{E_2} = $

When can the internal resistance of a cell be neglected? Explain.

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