Determine the current passing through the 10 resistor in the...

Let the common wire on the right side of the circuit be the reference node with a potential of $0\text{ V}$.

The $4.5\text{ V}$ battery has its positive terminal (longer line) on the left, so the potential at the node to the left of the $3\text{ }\Omega$ resistor is $4.5\text{ V}$.

The $3\text{ V}$ battery also has its positive terminal on the left, so the potential at the node between the $10\text{ }\Omega$ resistor and the $3\text{ V}$ battery is $3\text{ V}$.

Let the potential at the central junction connecting the $3\text{ }\Omega$, $6\text{ }\Omega$, and $10\text{ }\Omega$ resistors be $V$.

Applying Kirchhoff's Current Law (KCL) at this central junction, the sum of currents leaving the node is zero:

$\dfrac{V - 4.5}{3} + \dfrac{V - 0}{6} + \dfrac{V - 3}{10} = 0$

To solve for $V$, multiply the entire equation by the least common multiple of the denominators, which is $30$:

$10(V - 4.5) + 5(V) + 3(V - 3) = 0$

$10V - 45 + 5V + 3V - 9 = 0$

$18V - 54 = 0$

$18V = 54 \Rightarrow V = 3\text{ V}$

The current passing through the $10\text{ }\Omega$ resistor is given by the potential difference across it divided by its resistance:

$I_{10} = \dfrac{V - 3}{10} = \dfrac{3 - 3}{10} = 0\text{ A}$

Answer: $0\text{ A}$