1. **State the problem:** We need to find the current $I_{AB}$ in the given circuit using the Superposition theorem. The circuit has a 6A current source, a 5\,\Omega resistor, a 2\,\Omega resistor, a 3\,\Omega resistor, and a short circuit (SC). All resistances are in ohms. 2. **Superposition theorem:** We analyze the circuit by considering one independent source at a time while replacing the other sources with their internal resistances (current sources replaced by open circuits, voltage sources by short circuits). 3. **Step 1: Consider only the 6A current source** - Replace the d.c. voltage source with a short circuit. - The 5\,\Omega resistor and the 3\,\Omega resistor with SC form a path to node B. - Calculate the current $I_{AB}^{(1)}$ due to the 6A source. 4. **Step 2: Consider only the d.c. voltage source** - Replace the 6A current source with an open circuit. - Calculate the current $I_{AB}^{(2)}$ due to the voltage source. 5. **Calculate $I_{AB}^{(1)}$:** - With the voltage source shorted, the 2\,\Omega resistor is connected to the 6A current source. - The 3\,\Omega resistor is shorted to node B, so it effectively bypasses the 3\,\Omega resistor. - The current from the 6A source splits between the 2\,\Omega resistor and the short circuit. - Since the short circuit has zero resistance, all current flows through the short circuit, so no current flows through the 2\,\Omega resistor. - The 5\,\Omega resistor is connected to node A and B, but with the voltage source shorted, the current $I_{AB}^{(1)}$ is 0. 6. **Calculate $I_{AB}^{(2)}$:** - With the 6A current source open circuited, the circuit reduces to the voltage source connected to the 5\,\Omega resistor in parallel with the series combination of 2\,\Omega and 3\,\Omega resistors. - The short circuit bypasses the 3\,\Omega resistor, so effectively the 3\,\Omega resistor is shorted out. - The total resistance between A and B is the parallel of 5\,\Omega and 2\,\Omega. - Calculate equivalent resistance: $$R_{eq} = \frac{5 \times 2}{5 + 2} = \frac{10}{7} \approx 1.4286\,\Omega$$ - The current $I_{AB}^{(2)}$ is the voltage divided by $R_{eq}$, but voltage value is not given, so assume voltage source $V$. 7. **Combine currents:** - Total current $I_{AB} = I_{AB}^{(1)} + I_{AB}^{(2)} = 0 + I_{AB}^{(2)} = I_{AB}^{(2)}$ **Note:** The problem does not provide the voltage value of the d.c. source, so the exact numeric value of $I_{AB}$ cannot be computed without it. **Final answer:** $$I_{AB} = \frac{V}{\frac{5 \times 2}{5 + 2}} = \frac{7V}{10}$$ where $V$ is the voltage of the d.c. source.