1. **Problem statement:** Find the currents $I_1$, $I_2$, and $I_3$ in the given circuit using Kirchhoff's laws. 2. **Identify the loops and junctions:** - Junction at point $b$: currents satisfy $I_1 = I_2 + I_3$ (Kirchhoff's Current Law). - Loop 1: $f \to a \to b \to e \to f$ (left loop with 32V battery and 2\Omega and 8\Omega resistors). - Loop 2: $b \to c \to d \to e \to b$ (right loop with 4\Omega resistor and 20V battery). 3. **Write Kirchhoff's Voltage Law (KVL) equations for each loop:** - Loop 1 (clockwise): $$32 - 2I_1 - 8I_3 = 0$$ - Loop 2 (clockwise): $$4I_2 - 20 + 8I_3 = 0$$ 4. **Use the junction equation:** $$I_1 = I_2 + I_3$$ 5. **Substitute $I_1$ from junction into Loop 1 equation:** $$32 - 2(I_2 + I_3) - 8I_3 = 0$$ $$32 - 2I_2 - 2I_3 - 8I_3 = 0$$ $$32 - 2I_2 - 10I_3 = 0$$ 6. **Rewrite Loop 2 equation:** $$4I_2 - 20 + 8I_3 = 0$$ $$4I_2 + 8I_3 = 20$$ 7. **Now solve the system:** From Loop 1: $$2I_2 + 10I_3 = 32$$ From Loop 2: $$4I_2 + 8I_3 = 20$$ 8. **Multiply Loop 1 equation by 2 to align $I_2$ coefficients:** $$4I_2 + 20I_3 = 64$$ 9. **Subtract Loop 2 equation from this:** $$ (4I_2 + 20I_3) - (4I_2 + 8I_3) = 64 - 20$$ $$ 12I_3 = 44$$ $$ I_3 = \frac{44}{12} = \frac{11}{3} \approx 3.67 \text{ A}$$ 10. **Substitute $I_3$ back into Loop 2 equation:** $$4I_2 + 8 \times \frac{11}{3} = 20$$ $$4I_2 + \frac{88}{3} = 20$$ $$4I_2 = 20 - \frac{88}{3} = \frac{60}{3} - \frac{88}{3} = -\frac{28}{3}$$ $$I_2 = -\frac{28}{12} = -\frac{7}{3} \approx -2.33 \text{ A}$$ 11. **Find $I_1$ using junction equation:** $$I_1 = I_2 + I_3 = -\frac{7}{3} + \frac{11}{3} = \frac{4}{3} \approx 1.33 \text{ A}$$ 12. **Interpretation:** - $I_1 \approx 1.33$ A flows from left to right. - $I_2 \approx -2.33$ A means current $I_2$ flows opposite to assumed direction. - $I_3 \approx 3.67$ A flows downward as assumed. **Final answers:** $$I_1 = \frac{4}{3} \text{ A}, \quad I_2 = -\frac{7}{3} \text{ A}, \quad I_3 = \frac{11}{3} \text{ A}$$