Two cells of emf 1V and 2V and internal resistance 2 $$\Omega$$ and 1 $$\Omega$$, respectively, are connected in series with an external resistance of 6 $$\Omega$$. The total current in the circuit is $$I_1$$. Now the same two cells in parallel configuration are connected to same external resistance. In this case, the total current drawn is $$I_2$$. The value of $$\left(\dfrac{I_1}{I_2}\right)$$ is $$\dfrac{x}{3}$$. The value of x is ________.

Case 1: Cells in series

For series combination, the net emf is the algebraic sum  $$E_s = E_1 + E_2 = 1 + 2 = 3\;{\rm V}$$
The net internal resistance is the arithmetic sum  $$r_s = r_1 + r_2 = 2 + 1 = 3\;\Omega$$
Total resistance in the external circuit is  $$R_{\text{tot}} = r_s + R = 3 + 6 = 9\;\Omega$$
Hence the current in the series circuit is

$$I_1 = \frac{E_s}{R_{\text{tot}}} = \frac{3}{9} = \frac{1}{3}\;{\rm A}$$

Case 2: Cells in parallel

For two unequal cells in parallel, use the general formulae:
Equivalent internal resistance: $$\frac{1}{r_p} = \frac{1}{r_1} + \frac{1}{r_2}$$
Equivalent emf: $$E_p = \frac{\dfrac{E_1}{r_1} + \dfrac{E_2}{r_2}}{\dfrac{1}{r_1} + \dfrac{1}{r_2}}$$

Compute $$r_p$$: $$\frac{1}{r_p} = \frac{1}{2} + \frac{1}{1} = 0.5 + 1 = 1.5$$
$$r_p = \frac{1}{1.5} = \frac{2}{3}\;\Omega$$

Compute $$E_p$$: $$E_p = \frac{\dfrac{1}{2} + \dfrac{2}{1}}{1.5} = \frac{0.5 + 2}{1.5} = \frac{2.5}{1.5} = \frac{5}{3}\;{\rm V}$$

Total resistance now is  $$R_{\text{tot}} = R + r_p = 6 + \frac{2}{3} = \frac{20}{3}\;\Omega$$
Therefore the current in the parallel configuration is

$$I_2 = \frac{E_p}{R_{\text{tot}}} = \frac{\dfrac{5}{3}}{\dfrac{20}{3}} = \frac{5}{20} = \frac{1}{4}\;{\rm A}$$

Required ratio

$$\frac{I_1}{I_2} = \frac{\dfrac{1}{3}}{\dfrac{1}{4}} = \frac{1}{3} \times 4 = \frac{4}{3}$$

The problem states $$\dfrac{I_1}{I_2} = \dfrac{x}{3}$$, so $$x = 4$$.

Hence, the value of $$x$$ is 4.