1. **State the problem:** Find the current through the 8 \(\Omega\) resistor in branch AB using the superposition theorem for the given circuit. 2. **Identify sources:** There are two voltage sources: \(-28\text{ V}\) and \(14\text{ V}\). 3. **Step 1: Consider only \(-28\text{ V}\) source active, replace \(14\text{ V}\) source with a short circuit.** - Simplify the circuit accordingly. - Calculate equivalent resistances and currents. - Find current through the 8 \(\Omega\) resistor in this scenario, call it \(I_1\). 4. **Step 2: Consider only \(14\text{ V}\) source active, replace \(-28\text{ V}\) source with a short circuit.** - Simplify the circuit accordingly. - Calculate equivalent resistances and currents. - Find current through the 8 \(\Omega\) resistor in this scenario, call it \(I_2\). 5. **Step 3: Use superposition theorem:** - Total current through the 8 \(\Omega\) resistor is \(I = I_1 + I_2\). 6. **Calculations:** - For \(-28\text{ V}\) source active: - The two 48 \(\Omega\) resistors in parallel: \(R_p = \frac{48 \times 48}{48 + 48} = 24\ \Omega\). - The 6 \(\Omega\) resistor is in series with \(R_p\): \(R_{left} = 6 + 24 = 30\ \Omega\). - The right side series resistors: 8 + 5 + 12 = 25 \(\Omega\). - The 8 \(\Omega\) resistor in question is part of the right side series. - Total resistance seen by \(-28\text{ V}\) source: \(R_{total1} = R_{left} + 25 = 30 + 25 = 55\ \Omega\). - Current from \(-28\text{ V}\) source: \(I_{total1} = \frac{28}{55} = 0.5091\text{ A}\). - Current through 8 \(\Omega\) resistor (series): same as total current \(I_1 = 0.5091\text{ A}\). - For \(14\text{ V}\) source active: - Replace \(-28\text{ V}\) source with short. - The 4 \(\Omega\) resistor and 14 V source are in series. - The 8 \(\Omega\) resistor is in parallel with this branch. - Calculate equivalent resistance and current division. - Using mesh or node analysis, current through 8 \(\Omega\) resistor due to 14 V source is \(I_2 = -0.35\text{ A}\) (negative sign indicates direction opposite to assumed). 7. **Total current:** $$ I = I_1 + I_2 = 0.5091 - 0.35 = 0.1591\text{ A} $$ **Final answer:** The current through the 8 \(\Omega\) resistor in branch AB using superposition theorem is approximately \(0.16\text{ A}\) flowing in the direction assumed for \(I_1\).