1. **State the problem:** Find the current $I_{A\to B}$ flowing from node A to node B in the given circuit with resistors and voltage sources. 2. **Analyze the circuit:** The circuit has three resistors: 1\Omega between 5V and A, 2\Omega between A and 10V, and 3\Omega between A and B. 3. **Apply Kirchhoff's Voltage Law (KVL) and Ohm's Law:** Let the current from A to B be $I$. The voltage at node A is unknown; denote it as $V_A$. 4. **Write equations for currents at node A:** - Current from 5V source through 1\Omega resistor to A: $I_1 = \frac{5 - V_A}{1} = 5 - V_A$ - Current from A to 10V source through 2\Omega resistor: $I_2 = \frac{V_A - 10}{2}$ - Current from A to B through 3\Omega resistor: $I = \frac{V_A - V_B}{3}$ 5. **Note:** Node B is connected only through the 3\Omega resistor to A, and no other elements are connected to B, so we can assume $V_B = 0$ (reference node). 6. **Apply Kirchhoff's Current Law (KCL) at node A:** Sum of currents leaving A is zero: $$I_1 + I_2 + I = 0$$ Substitute: $$ (5 - V_A) + \frac{V_A - 10}{2} + \frac{V_A - 0}{3} = 0$$ 7. **Multiply through by 6 (LCM of denominators 1, 2, 3) to clear fractions:** $$6(5 - V_A) + 3(V_A - 10) + 2V_A = 0$$ $$30 - 6V_A + 3V_A - 30 + 2V_A = 0$$ 8. **Simplify:** $$30 - 6V_A + 3V_A - 30 + 2V_A = 0$$ $$(-6V_A + 3V_A + 2V_A) + (30 - 30) = 0$$ $$(-6V_A + 5V_A) + 0 = 0$$ $$-V_A = 0$$ 9. **Solve for $V_A$:** $$V_A = 0$$ 10. **Calculate current $I_{A\to B}$:** $$I = \frac{V_A - V_B}{3} = \frac{0 - 0}{3} = 0$$ **Final answer:** $$I_{A\to B} = 0$$ This means no current flows from A to B under the given conditions.