The current through a 4/3 Ω external resistance connected to a parallel combination of two cells of 2 V and 1 V emf and internal resistances of 1 Ω and 2 Ω respectively is ______.

The correct option is 5A/6.

To find the current through the external resistance, we first find the equivalent electromotive force (emf) and the equivalent internal resistance of the parallel combination of the two cells.
Let the two cells have:
Emf of the first cell, _E1 = 2 V, and its internal resistance, _r1 = 1 Ω.
Emf of the second cell, _E2 = 1 V, and its internal resistance, _r2 = 2 Ω.
External resistance, R = 4/3 Ω.

For a parallel combination of two cells, the equivalent emf (_Eeq) is given by the formula:
$$E_{\mathrm{eq}}=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}}$$
Substituting the given values:
$$E_{\mathrm{eq}}=\frac{\frac{2}{1}+\frac{1}{2}}{\frac{1}{1}+\frac{1}{2}}=\frac{2+0.5}{1+0.5}=\frac{2.5}{1.5}=\frac{5}{3}\text{V}$$

The equivalent internal resistance (_req) of the parallel combination is given by:
$$\frac{1}{r_{\mathrm{eq}}}=\frac{1}{r_1}+\frac{1}{r_2}$$
$$\frac{1}{r_{\mathrm{eq}}}=\frac{1}{1}+\frac{1}{2}=\frac{3}{2}{\Omega }^{-1}$$
Taking the reciprocal, we get:
$$r_{\mathrm{eq}}=\frac{2}{3}\Omega$$

Now, the current (I) through the external resistance R connected across this equivalent cell is:
$$I=\frac{E_{\mathrm{eq}}}{R+r_{\mathrm{eq}}}$$
Substituting the values of _Eeq, _req, and R:
$$I=\frac{\frac{5}{3}}{\frac{4}{3}+\frac{2}{3}}$$
Simplify the denominator:
$$\frac{4}{3}+\frac{2}{3}=\frac{6}{3}=2\Omega$$
Now calculate the current:
$$I=\frac{\frac{5}{3}}{2}=\frac{5}{3\times 2}=\frac{5}{6}\text{A}$$

Thus, the current through the external resistance is 5A/6.