Twelve cells,each having emf $E$ volts,are connected in series and are kept in a closed box. Some of these cells are wrongly connected with positive and negative terminals reversed. This $12$-cell battery is connected in series with an ammeter,an external resistance $R$ ohms,and a two-cell battery (two cells of the same type used earlier,connected perfectly in series). The current in the circuit when the $12$-cell battery and $2$-cell battery aid each other is $3 \text{ A}$,and it is $2 \text{ A}$ when they oppose each other. Then,the number of cells in the $12$-cell battery that are connected wrongly is:

  • A
    $4$
  • B
    $3$
  • C
    $2$
  • D
    $1$

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The circuit in the figure shows two cells connected in opposition to each other. Cell $E_1$ has an $emf$ of $6 \ V$ and an internal resistance of $2 \ \Omega$. Cell $E_2$ has an $emf$ of $4 \ V$ and an internal resistance of $8 \ \Omega$. Find the potential difference between the points $A$ and $B$.

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