1. **State the problem:** Find the currents $I_1$, $I_2$, and $I_3$ in the given circuit using Kirchhoff's laws. 2. **Identify the circuit elements and nodes:** - Resistors: $2\Omega$ (between $a$ and $b$), $4\Omega$ (between $b$ and $c$), $8\Omega$ (between $b$ and $e$). - Voltage sources: $32$ V (between $f$ and $a$), $20$ V (between $c$ and $d$). - Currents: $I_1$ from $a$ to $b$, $I_2$ from $b$ to $c$, $I_3$ from $b$ to $e$. 3. **Apply Kirchhoff's Current Law (KCL) at node $b$:** $$I_1 = I_2 + I_3$$ 4. **Apply Kirchhoff's Voltage Law (KVL) to the left loop $f \to a \to b \to e \to f$:** $$32 - 2I_1 - 8I_3 = 0$$ 5. **Apply KVL to the right loop $b \to c \to d \to e \to b$:** $$4I_2 + 20 - 8I_3 = 0$$ 6. **Rewrite the equations:** - From KCL: $$I_1 - I_2 - I_3 = 0$$ - Left loop: $$32 = 2I_1 + 8I_3$$ - Right loop: $$4I_2 + 20 = 8I_3$$ 7. **Express $I_2$ from the right loop:** $$4I_2 = 8I_3 - 20 \implies I_2 = \frac{8I_3 - 20}{4} = 2I_3 - 5$$ 8. **Substitute $I_2$ into KCL:** $$I_1 - (2I_3 - 5) - I_3 = 0 \implies I_1 - 2I_3 + 5 - I_3 = 0 \implies I_1 - 3I_3 + 5 = 0$$ 9. **Express $I_1$ from above:** $$I_1 = 3I_3 - 5$$ 10. **Substitute $I_1$ into left loop equation:** $$32 = 2(3I_3 - 5) + 8I_3 = 6I_3 - 10 + 8I_3 = 14I_3 - 10$$ 11. **Solve for $I_3$:** $$14I_3 = 42 \implies I_3 = 3$$ 12. **Find $I_1$:** $$I_1 = 3(3) - 5 = 9 - 5 = 4$$ 13. **Find $I_2$:** $$I_2 = 2(3) - 5 = 6 - 5 = 1$$ **Final answer:** $$I_1 = 4\,A, \quad I_2 = 1\,A, \quad I_3 = 3\,A$$