1. **Problem 1: Find the current through the 4 Ω resistor in Fig. 01 using the superposition principle.** 2. The superposition principle states that in a linear circuit with multiple independent sources, the total current or voltage is the algebraic sum of the currents or voltages caused by each independent source acting alone while all other independent sources are replaced by their internal impedances (voltage sources replaced by short circuits and current sources by open circuits). 3. **Step 1: Consider the 54 V source alone (replace 48 V source by a short circuit).** - The circuit reduces to 54 V source, 24 Ω resistor, 12 Ω resistor, and 4 Ω resistor with the 48 V source shorted. - Calculate equivalent resistance and current through 4 Ω resistor. 4. **Step 2: Consider the 48 V source alone (replace 54 V source by a short circuit).** - The circuit reduces to 48 V source, 4 Ω resistor, 12 Ω resistor, and 24 Ω resistor with the 54 V source shorted. - Calculate equivalent resistance and current through 4 Ω resistor. 5. **Step 3: Add the currents from both steps algebraically to find total current through 4 Ω resistor.** 6. **Problem 2: Use superposition theorem to calculate the voltage drop across the 2 Ω resistor in Fig. 02.** 7. The circuit has two current sources: 24 A and 6 A. 8. **Step 1: Consider 24 A source alone (replace 6 A source by open circuit).** - Calculate voltage across 2 Ω resistor. 9. **Step 2: Consider 6 A source alone (replace 24 A source by open circuit).** - Calculate voltage across 2 Ω resistor. 10. **Step 3: Add the voltages from both steps algebraically to find total voltage across 2 Ω resistor.** --- ### Detailed calculations for Problem 1: - When 48 V source is replaced by a short circuit: - The 12 Ω and 4 Ω resistors are in series: $12 + 4 = 16\ \Omega$ - This 16 Ω is in parallel with 24 Ω resistor: $$R_{eq1} = \frac{24 \times 16}{24 + 16} = \frac{384}{40} = 9.6\ \Omega$$ - Total current from 54 V source: $$I_1 = \frac{54}{9.6} = 5.625\ A$$ - Current divides between 24 Ω and 16 Ω branches: $$I_{4\Omega,1} = I_1 \times \frac{24}{24 + 16} = 5.625 \times \frac{24}{40} = 3.375\ A$$ - Current through 4 Ω resistor is same as through 16 Ω branch, so $I_{4\Omega,1} = 3.375\ A$ - When 54 V source is replaced by a short circuit: - The 24 Ω and 12 Ω resistors are in series: $24 + 12 = 36\ \Omega$ - This 36 Ω is in parallel with 4 Ω resistor: $$R_{eq2} = \frac{36 \times 4}{36 + 4} = \frac{144}{40} = 3.6\ \Omega$$ - Total current from 48 V source: $$I_2 = \frac{48}{3.6} = 13.333\ A$$ - Current divides between 36 Ω and 4 Ω branches: $$I_{4\Omega,2} = I_2 \times \frac{36}{36 + 4} = 13.333 \times \frac{36}{40} = 12\ A$$ - Total current through 4 Ω resistor: $$I_{4\Omega} = I_{4\Omega,1} + I_{4\Omega,2} = 3.375 + 12 = 15.375\ A$$ --- ### Detailed calculations for Problem 2: - When 6 A source is replaced by open circuit: - The 24 A current source feeds three parallel branches: 4 Ω, 2 Ω, and 6 Ω. - Voltage across 2 Ω resistor: $$V_{2\Omega,1} = I \times R = 24 \times 2 = 48\ V$$ - When 24 A source is replaced by open circuit: - The 6 A current source feeds the 6 Ω resistor only. - No current flows through 2 Ω resistor, so voltage across 2 Ω resistor is 0 V. - Total voltage across 2 Ω resistor: $$V_{2\Omega} = V_{2\Omega,1} + 0 = 48\ V$$ --- **Final answers:** - Current through 4 Ω resistor in Fig. 01: $15.375\ A$ - Voltage drop across 2 Ω resistor in Fig. 02: $48\ V$