1. **Problem Statement:** Calculate the current flowing through resistor $R_3$ in the given circuit with two loops and two voltage sources $V_1=12$ V and $V_2=9$ V. 2. **Identify the loops and currents:** Let $I_1$ be the current in the left loop and $I_2$ be the current in the right loop. The resistor $R_3$ is shared between the two loops, so the current through $R_3$ is $I_1 - I_2$ (assuming directions as shown). 3. **Write Kirchhoff's Voltage Law (KVL) equations:** - Left loop (with $V_1$, $R_1$, $R_2$, and $R_3$): $$12 - 10I_1 - 2I_1 - 6(I_1 - I_2) = 0$$ - Right loop (with $V_2$, $R_3$, and $R_4$): $$9 - 22I_2 - 6(I_2 - I_1) = 0$$ 4. **Simplify the equations:** Left loop: $$12 - 12I_1 - 6I_1 + 6I_2 = 0$$ $$12 - 18I_1 + 6I_2 = 0$$ Right loop: $$9 - 22I_2 - 6I_2 + 6I_1 = 0$$ $$9 - 28I_2 + 6I_1 = 0$$ 5. **Rewrite as system of linear equations:** $$\begin{cases} -18I_1 + 6I_2 = -12 \\ 6I_1 - 28I_2 = -9 \end{cases}$$ 6. **Divide first equation by 6:** $$-3I_1 + I_2 = -2$$ 7. **Divide second equation by 2:** $$3I_1 - 14I_2 = -4.5$$ 8. **Solve the system:** From first equation: $$I_2 = -2 + 3I_1$$ Substitute into second: $$3I_1 - 14(-2 + 3I_1) = -4.5$$ $$3I_1 + 28 - 42I_1 = -4.5$$ $$-39I_1 = -32.5$$ $$I_1 = \frac{32.5}{39} \approx 0.8333\, \text{A}$$ 9. **Find $I_2$:** $$I_2 = -2 + 3(0.8333) = -2 + 2.5 = 0.5\, \text{A}$$ 10. **Calculate current through $R_3$:** $$I_{R_3} = I_1 - I_2 = 0.8333 - 0.5 = 0.3333\, \text{A}$$ **Final answer:** The current flowing through resistor $R_3$ is approximately $0.33$ A.