1. **State the problem:** We have a circuit with resistors and voltage sources. We need to find the magnitude and direction of currents $I_1$ (through 12 ohm resistor), $I_2$ (through 3 ohm resistor), and current through 6 ohm resistor using the Superposition theorem. 2. **Superposition theorem:** Analyze the circuit by considering one voltage source at a time while replacing the other voltage source with a short circuit. --- ### Step 1: Consider only the 24 V source (replace 18 V source with a short) - The 18 V source between B and D is replaced by a wire. - The circuit reduces to 24 V source between A and D, resistors 12 ohm (A-C), 3 ohm (B-C), and 6 ohm (C-D). - Points B and D are shorted, so B and D are at the same potential. Calculate equivalent resistance and currents: - Resistors 3 ohm and 6 ohm are in series between B and D (shorted), so combined resistance $R_{BD} = 3 + 6 = 9$ ohm. - This 9 ohm is in parallel with 12 ohm resistor between A and D. Equivalent resistance seen by 24 V source: $$ R_{eq1} = \frac{12 \times 9}{12 + 9} = \frac{108}{21} = \frac{36}{7} \approx 5.14 \text{ ohm} $$ Current from 24 V source: $$ I_{total1} = \frac{24}{R_{eq1}} = \frac{24}{5.14} \approx 4.67 \text{ A} $$ Current division: - Current through 12 ohm resistor: $$ I_{1(24V)} = I_{total1} \times \frac{9}{12 + 9} = 4.67 \times \frac{9}{21} = 2.0 \text{ A} $$ - Current through 3 and 6 ohm branch: $$ I_{2(24V)} = I_{total1} \times \frac{12}{21} = 4.67 \times \frac{12}{21} = 2.67 \text{ A} $$ Current through 6 ohm resistor is same as $I_{2(24V)} = 2.67$ A. Direction: - $I_1$ flows from A to C. - $I_2$ flows from B to C. - Current through 6 ohm flows from C to D. --- ### Step 2: Consider only the 18 V source (replace 24 V source with a short) - The 24 V source between A and D is replaced by a wire. - Now, A and D are shorted. Calculate equivalent resistance and currents: - Resistors 12 ohm (A-C) and 6 ohm (C-D) are in series between A and D (shorted), so combined resistance $R_{AD} = 12 + 6 = 18$ ohm. - This 18 ohm is in parallel with 3 ohm resistor between B and D. Equivalent resistance seen by 18 V source: $$ R_{eq2} = \frac{18 \times 3}{18 + 3} = \frac{54}{21} = \frac{18}{7} \approx 2.57 \text{ ohm} $$ Current from 18 V source: $$ I_{total2} = \frac{18}{R_{eq2}} = \frac{18}{2.57} \approx 7.0 \text{ A} $$ Current division: - Current through 3 ohm resistor: $$ I_{2(18V)} = I_{total2} \times \frac{18}{18 + 3} = 7.0 \times \frac{18}{21} = 6.0 \text{ A} $$ - Current through 12 and 6 ohm branch: $$ I_{AD(18V)} = I_{total2} \times \frac{3}{21} = 7.0 \times \frac{3}{21} = 1.0 \text{ A} $$ Current through 12 ohm resistor and 6 ohm resistor is $1.0$ A (same current in series). Direction: - $I_2$ flows from B to C. - Current through 12 ohm resistor flows from A to C. - Current through 6 ohm resistor flows from C to D. --- ### Step 3: Combine currents from both sources - Total current through 12 ohm resistor: $$ I_1 = I_{1(24V)} + I_{AD(18V)} = 2.0 + 1.0 = 3.0 \text{ A} $$ - Total current through 3 ohm resistor: $$ I_2 = I_{2(24V)} + I_{2(18V)} = 2.67 + 6.0 = 8.67 \text{ A} $$ - Total current through 6 ohm resistor: $$ I_6 = I_{2(24V)} + I_{AD(18V)} = 2.67 + 1.0 = 3.67 \text{ A} $$ ### Final answer: - Current through 12 ohm resistor $I_1 = 3.0$ A from A to C. - Current through 3 ohm resistor $I_2 = 8.67$ A from B to C. - Current through 6 ohm resistor $I_6 = 3.67$ A from C to D. These directions are consistent with the assumed directions in the problem.