1. **State the problem:** We need to find the Thevenin equivalent circuit to the left of terminals (a) and (b) in the given circuit, then find the current $i$ flowing downward through the 1 $\Omega$ resistor. 2. **Identify the portion for Thevenin equivalent:** Thevenin equivalent is found by looking back into the circuit from terminals (a) and (b), replacing the load (the 4 $\Omega$ and 1 $\Omega$ resistors) with an open circuit. 3. **Find Thevenin voltage $V_{th}$:** - Remove the load resistors (4 $\Omega$ and 1 $\Omega$) from terminals (a) and (b). - Calculate the open-circuit voltage across terminals (a) and (b). 4. **Find Thevenin resistance $R_{th}$:** - Turn off all independent sources: voltage sources replaced by short circuits, current sources by open circuits. - Calculate equivalent resistance seen from terminals (a) and (b). 5. **Analyze the circuit:** - The circuit has a 12 V voltage source in series with a 6 $\Omega$ resistor. - Then a 2 A current source downward. - Then another 6 $\Omega$ resistor before terminal (a). 6. **Calculate $V_{th}$:** - The current source forces 2 A downward between the two 6 $\Omega$ resistors. - Voltage across the lower 6 $\Omega$ resistor (between node after current source and terminal a) is $V = IR = 2 \times 6 = 12$ V downward. - Voltage at node (a) relative to the node after the current source is $-12$ V (since current flows downward). - Voltage at node after current source relative to ground: - Voltage drop across first 6 $\Omega$ resistor: $I = ?$ (unknown), but since current source fixes current, the current through first 6 $\Omega$ resistor is not fixed. 7. **Use mesh or node analysis:** - Let node after first 6 $\Omega$ resistor be $V_1$. - Ground is at negative terminal of 12 V source. - Voltage at node before first 6 $\Omega$ resistor is 12 V (positive terminal of source). - Current through first 6 $\Omega$ resistor: $I_1 = \frac{12 - V_1}{6}$. - Current source forces 2 A downward from $V_1$ to $V_2$ (node after current source). - Current through second 6 $\Omega$ resistor: $I_2 = \frac{V_2 - V_a}{6}$, but $V_a$ is terminal a voltage. 8. **Since terminals (a) and (b) are open for $V_{th}$ calculation, no current flows through load, so $V_a = V_b$ (open circuit).** - Let $V_b = 0$ (reference). - Then $V_a = V_b = 0$. 9. **Apply KCL at node $V_2$:** - Current from $V_1$ to $V_2$ is 2 A downward (current source). - Current through 6 $\Omega$ resistor from $V_2$ to $V_a$ is $I_2 = \frac{V_2 - 0}{6} = \frac{V_2}{6}$. - KCL: $2 = I_2 = \frac{V_2}{6} \Rightarrow V_2 = 12$ V. 10. **Apply KCL at node $V_1$:** - Current through first 6 $\Omega$ resistor: $I_1 = \frac{12 - V_1}{6}$. - Current source current downward from $V_1$ to $V_2$ is 2 A. - No other branches, so KCL at $V_1$: $I_1 = 2$ A. - Solve: $\frac{12 - V_1}{6} = 2 \Rightarrow 12 - V_1 = 12 \Rightarrow V_1 = 0$ V. 11. **Calculate $V_{th}$:** - Voltage across terminals (a) and (b) is $V_a - V_b = 0 - 0 = 0$ V. - But this contradicts previous step; re-examine terminal voltages. 12. **Reconsider terminal voltages:** - Terminal (a) is node after second 6 $\Omega$ resistor, which is $V_a$. - Terminal (b) is ground reference. - From step 9, $V_2 = 12$ V. - Voltage drop across second 6 $\Omega$ resistor is $V_2 - V_a$. - Since load is removed, no current flows through second 6 $\Omega$ resistor, so $V_a = V_2 = 12$ V. - Therefore, $V_{th} = V_a - V_b = 12 - 0 = 12$ V. 13. **Calculate $R_{th}$:** - Turn off sources: voltage source replaced by short circuit, current source replaced by open circuit. - Circuit reduces to two 6 $\Omega$ resistors in series between terminals (a) and (b). - So, $R_{th} = 6 + 6 = 12$ $\Omega$. 14. **Thevenin equivalent circuit:** - Voltage source $V_{th} = 12$ V in series with $R_{th} = 12$ $\Omega$ connected to terminals (a) and (b). 15. **Find current $i$ through 1 $\Omega$ resistor:** - Load resistors 4 $\Omega$ and 1 $\Omega$ are in parallel. - Equivalent load resistance $R_L = \frac{4 \times 1}{4 + 1} = \frac{4}{5} = 0.8$ $\Omega$. 16. **Total resistance in circuit:** - $R_{total} = R_{th} + R_L = 12 + 0.8 = 12.8$ $\Omega$. 17. **Current from Thevenin source:** - $I = \frac{V_{th}}{R_{total}} = \frac{12}{12.8} = 0.9375$ A. 18. **Current $i$ through 1 $\Omega$ resistor:** - Current divides in parallel: $i = I \times \frac{4}{4 + 1} = 0.9375 \times \frac{4}{5} = 0.75$ A downward. **Final answer:** $$i = 0.75 \text{ A downward through the 1 } \Omega \text{ resistor}.$$