1. **State the problem:** Find the Thevenin and Norton equivalent circuits seen from terminals a-b for the given circuit. 2. **Identify variables and components:** The circuit has a 20 V voltage source, a 6\Omega resistor, a dependent current source of value 2i, a 10\Omega resistor, and another 6\Omega resistor connected between the node and terminal B. 3. **Define currents:** Let current through the dependent source be $i$. The current through the 6\Omega resistor connected to terminal B is $i_1$. 4. **Write the mesh equation:** The voltage source and the 6\Omega resistor are in series, so voltage at node between resistors and current source is $$ V_{node} = 20 - 6i. $$ 5. **Apply KCL at the node between the 2i current source, 10\Omega resistor and 6\Omega resistor:** $$ 2i = \frac{V_{node} - V_a}{10} + \frac{V_{node} - V_b}{6} $$ Terminals a-b have voltages $V_a$ and $V_b$, but for Thevenin equivalent, we find open-circuit voltage $V_{oc} = V_{a} - V_{b}$. 6. **The voltage at node connected to 10\Omega resistor and dependent source is $V_{node}$, terminal b is reference (0 V), so $V_b=0$, and with $V_a$ at the terminal of the 10\Omega resistor:** $$ I_{10\Omega} = \frac{V_{node} - V_a}{10} $$ $$ I_{6\Omega} = \frac{V_{node} - 0}{6} = \frac{V_{node}}{6} $$ 7. **Since terminals are open (no load), current through terminals is zero, so $i$ (the current controlling dependent source) is the current through the 6\Omega resistor on the left branch and equals current through voltage source branch:** $$ i = \frac{20 - V_{node}}{6} $$ 8. **Substitute $i$ into equation for dependent current source:** $$ 2i = 2 \times \frac{20 - V_{node}}{6} = \frac{40 - 2V_{node}}{6} $$ 9. **Write KCL at node:** $$ \frac{40 - 2V_{node}}{6} = \frac{V_{node} - V_a}{10} + \frac{V_{node}}{6} $$ Since $V_a$ is voltage at node connected to 10\Omega and terminal a, and terminal a is the node at the end of 10\Omega resistor, $$ V_a = V_{node} - I_{10\Omega} \times 10 = V_{node} - 10 \times \frac{V_{node} - V_a}{10} = V_{node} - (V_{node} - V_a) = V_a $$ So $V_a$ is voltage at terminal, so the open circuit voltage $V_{oc} = V_a - V_b = V_a - 0 = V_a$. But this is complicated; instead, the current through 10\Omega resistor is: $$ I_{10\Omega} = \frac{V_{node} - V_a}{10}. $$ On open circuit terminals, no current flows out, so current through 10\Omega resistor equals current $i_{10}\Omega = 0$? This means terminal a has no load, so $I_{10\Omega} = 0$. Then $$ \frac{V_{node} - V_a}{10} = 0 \implies V_{node} = V_a. $$ 10. **Use KCL with this:** $$ \frac{40 - 2V_{node}}{6} = 0 + \frac{V_{node}}{6} $$ $$ \frac{40 - 2V_{node}}{6} = \frac{V_{node}}{6} $$ Multiply both sides by 6: $$ 40 - 2V_{node} = V_{node} $$ $$ 40 = 3V_{node} $$ $$ V_{node} = \frac{40}{3} \approx 13.33 V. $$ 11. **Therefore open circuit voltage $V_{oc} = V_a = V_{node} = 13.33 V$.** 12. **Find Thevenin resistance $R_{th}$ by zeroing independent source and turning off dependent source effect:** - Zero voltage source \to replace by short circuit. - To find $R_{th}$ with dependent source, apply a test source at terminals a-b and find voltage/current ratio. 13. **Apply a test voltage $V_{test}$ at terminals a-b, measure current $I_{test}$, then** $$ R_{th} = \frac{V_{test}}{I_{test}}. $$ 14. **Let $V_{test} = V_a - V_b$, terminal b=0, $V_a = V_{test}$** Write KCL at node (same as earlier), with dependent source present. Use $i$ current through left 6\Omega resistor is: $$ i = \frac{0 - 0}{6} = 0 $$ Zero voltage source replaced by short, so node left is 0 V, no independent source. Dependent source current $=2i=0$. Then at node: $$ 0 = \frac{V_{node} - V_a}{10} + \frac{V_{node}}{6} $$ Since $V_a=V_{test}$, and node is connected to $V_a$ via 10\Omega, $$ \frac{V_{node} - V_{test}}{10} + \frac{V_{node}}{6} = 0 $$ Multiply both sides by 30: $$ 3(V_{node} - V_{test}) + 5 V_{node} = 0 $$ $$ 3V_{node} - 3V_{test} + 5V_{node} = 0 $$ $$ 8V_{node} = 3V_{test} $$ $$ V_{node} = \frac{3}{8} V_{test} $$ Current $I_{test}$ flowing into terminal a through 10\Omega resistor is: $$ I_{test} = \frac{V_{node} - V_{test}}{10} = \frac{\frac{3}{8}V_{test} - V_{test}}{10} = \frac{-\frac{5}{8} V_{test}}{10} = -\frac{5}{80} V_{test} = -\frac{1}{16} V_{test} $$ Taking magnitude, $$ I_{test} = \frac{1}{16} V_{test} $$ 15. **Therefore the Thevenin resistance is:** $$ R_{th} = \frac{V_{test}}{I_{test}} = \frac{V_{test}}{\frac{1}{16} V_{test}} = 16\ \Omega. $$ 16. **Norton equivalent current $I_N$ is the short circuit current across terminals a-b:** $$ I_N = \frac{V_{oc}}{R_{th}} = \frac{13.33}{16} = 0.833 A. $$ **Final answers:** - Thevenin voltage $V_{th} = 13.33 V$ - Thevenin resistance $R_{th} = 16 \Omega$ - Norton current $I_N = 0.833 A$