Answer

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**Hint**So for solving this question, we will first calculate the effective voltages and effective resistance in both the series and parallel cells connection. And then we will find the current separately for both. After that equating both the current we will get the resistance.

Formula used:

Current,

$I = \dfrac{E}{r}$,

$I$, will be the current

$E$, will be the effective voltage

$r$, will be the effective resistance

**Complete Step By Step Solution**

Let the voltage of each cell is equal to $E$, the internal resistance of each cell is equal to $r$

When cells are in series-

The effective voltage of the circuit will be$ = E + E = 2E$

And effective resistance of the circuit will be$ = r + r + 3 = 2r + 3$

Therefore the current will be

${I_1} = \dfrac{{2E}}{{2r + 3}}$

When cells are in parallel, then

The effective voltage of the circuit will be $ = E$

And effective resistance of the circuit will be $ = \dfrac{r}{3} + r$

Therefore the current will be

${I_2} = \dfrac{{2E}}{{r + 6}}$

Since the current for both of them is the same that is we can write it as

$ \Rightarrow {I_1} = {I_2}$

Now equating both of the equations, we get

$ \Rightarrow \dfrac{{2E}}{{2r + 3}} = \dfrac{{2E}}{{r + 6}}$

Since the numerator is the same so both of them will cancel each other.

Now,

$ \Rightarrow 2r + 3 = r + 6$

Now on solving the above equation, we will get

$ \Rightarrow r = 3\Omega $

Hence, $3\Omega $ will be the resistance of the cell.

**Therefore, the option $B$ will be the correct answer.**

**Note**So if the resistors are in parallel then we can find it by using the formula ${R_{eq}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$and so on. And if it is in series then we will use the formula ${R_{eq}} = {R_1} + {R_2}$and so on. Now if there is the case of the capacitor then it will become reverse of the above. That is parallel we will simply add the capacitor but if it is in series then we will add like this ${C_{eq}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$.

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