Question

$n$ Identical cells are joined in series with two cells A and B with reversed polarities. Emf of each cell is $E$ and the internal resistance is $R$. Potential difference across the cell A and B is : $(n > 4)$
(A) $\dfrac{{2E}}{n}$
(B) $2E(1 - \dfrac{1}{n})$
(C ) $\dfrac{{4E}}{n}$
(D) $2E(1 - \dfrac{2}{n})$

Answer 1

In the question, it is given that two cells are connected such that they have opposite polarities. This means they cancel some cells having opposite polarity. So, first calculate the net emf of the connection, then find the current flowing through them.

Complete Step By Step Solution:
Since cells A and B are connected in opposite polarities, they cancel emf of cells of opposite polarity.
The $n$cells are connected in series, hence their emf adds up.
Now, let us find, the net emf in the circuit:
Net emf $= (n - 2E) - 2E$
Where, $2E$ corresponds to emf of A and B having opposite polarity.
Now, let us calculate the current flowing through the circuit.
Since $n$cells are connected each having $R$internal resistance, total resistance in the circuit, is:

$r = nR$
$r =$Total internal resistances.

Now, applying ohm’s law, we know:
Potential difference across two ends of a conductor is directly proportional to the current through the conductor, resistance being the constant of proportionality.

$V = IR$

Where,
$V =$Potential Difference $I =$ Current
$R =$Resistance
Therefore,

$NetEMF = \dfrac{I}{r}$

Putting the values:

$(n - 2E) - 2E = \dfrac{I}{{nR}}$
$I = \dfrac{{(n - 4)E}}{{nR}}$

Where, $I$is the current flowing through the circuit
Now, potential difference across A and B :

$E + IR$

Let $\Delta v =$ potential difference across A and B
So,

$\Delta v = E + IR$
Putting the value of $I$, we get:

$\Delta v = E + \dfrac{{(n - 4)E}}{{nR}}R$

Thus, on solving the equation we get:

$\Delta v = 2E\left[ {\dfrac{{n - 2}}{n}} \right]$

So, finally we arrive at:

$\Delta v = 2E\left[ {1 - \dfrac{2}{n}} \right]$

This is the required solution.

SO, option (D) IS CORRECT.

Note: When cells are connected in series, the same amount of current flows through them.
Here, since the cells A and B connected had opposite polarity, the net Emf was obtained after subtraction, if they were of same polarity, the emf’s would have been added. To increase the supply voltage across a circuit the cells are connected in series with the same polarity.