1. **State the problem:** Find the current through the 3 \(\Omega\) resistor connected between points C and D in the given circuit. 2. **Analyze the circuit:** The circuit is a rectangular loop with resistors and two voltage sources (9 V on the left, 4.5 V on the right). The resistor between C and D is 3 \(\Omega\). 3. **Label currents:** Assume currents \(I_1\) flows through the left vertical branch, \(I_2\) through the middle vertical 3 \(\Omega\) resistor between C and D, and \(I_3\) through the right vertical branch. 4. **Apply Kirchhoff's Voltage Law (KVL):** Write loop equations for the two loops. - Left loop (including 9 V battery, 3 \(\Omega\), 1.5 \(\Omega\), and vertical 3 \(\Omega\) resistor): $$9 - 3I_1 - 1.5I_1 - 3(I_1 - I_2) = 0$$ Simplify: $$9 - 4.5I_1 - 3I_1 + 3I_2 = 0$$ $$9 - 7.5I_1 + 3I_2 = 0$$ - Right loop (including 4.5 V battery, 3 \(\Omega\), and vertical 3 \(\Omega\) resistor): $$-4.5 + 3I_3 + 3(I_3 - I_2) = 0$$ Simplify: $$-4.5 + 3I_3 + 3I_3 - 3I_2 = 0$$ $$-4.5 + 6I_3 - 3I_2 = 0$$ 5. **Apply Kirchhoff's Current Law (KCL) at node C:** $$I_1 = I_2 + I_3$$ 6. **Solve the system of equations:** From KCL: $$I_1 = I_2 + I_3$$ From left loop: $$9 - 7.5I_1 + 3I_2 = 0 \Rightarrow 7.5I_1 - 3I_2 = 9$$ From right loop: $$-4.5 + 6I_3 - 3I_2 = 0 \Rightarrow 6I_3 - 3I_2 = 4.5$$ Substitute \(I_1 = I_2 + I_3\) into left loop: $$7.5(I_2 + I_3) - 3I_2 = 9$$ $$7.5I_2 + 7.5I_3 - 3I_2 = 9$$ $$4.5I_2 + 7.5I_3 = 9$$ Right loop equation: $$6I_3 - 3I_2 = 4.5$$ Rewrite system: $$4.5I_2 + 7.5I_3 = 9$$ $$-3I_2 + 6I_3 = 4.5$$ Multiply second equation by 1.5 to align \(I_2\) coefficients: $$-4.5I_2 + 9I_3 = 6.75$$ Add to first equation: $$(4.5I_2 + 7.5I_3) + (-4.5I_2 + 9I_3) = 9 + 6.75$$ $$16.5I_3 = 15.75$$ $$I_3 = \frac{15.75}{16.5} = 0.9545\text{ A}$$ Substitute \(I_3\) back into first equation: $$4.5I_2 + 7.5(0.9545) = 9$$ $$4.5I_2 + 7.158 = 9$$ $$4.5I_2 = 1.842$$ $$I_2 = \frac{1.842}{4.5} = 0.4093\text{ A}$$ 7. **Final answer:** The current through the 3 \(\Omega\) resistor between C and D is approximately \(\boxed{0.41\text{ A}}\).