In the circuit shown, current (in A) through 50 V and 30 V batteries are, respectively:

$$I_{20\Omega} = \frac{50\text{ V}}{20\ \Omega} = 2.5\text{ A (downward)}$$

$$I_{10\Omega} = \frac{30\text{ V}}{10\ \Omega} = 3\text{ A (downward)}$$

$$I_{horizontal} = \frac{50 - 30}{5 + 5} = \frac{20}{10} = 2\text{ A (left to right)}$$

$$I_{50V} = I_{20\Omega} + I_{horizontal} = 2.5\text{ A} + 2\text{ A} = 4.5\text{ A}$$

$$2\text{ A} + I_{30V} = 3\text{ A} \implies I_{30V} = 1\text{ A}$$

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