1. **State the Principle of Superposition (PS):** The Principle of Superposition states that in a linear circuit with multiple independent sources, the total response (voltage or current) in any element is the algebraic sum of the responses caused by each independent source acting alone, with all other independent sources turned off (voltage sources replaced by short circuits and current sources replaced by open circuits). 2. **Apply PS to the circuit:** We have two independent sources: a 15 V voltage source and two current sources (8 A and 30 A). We analyze the circuit twice: once with only the voltage source active and current sources turned off, and once with only the current sources active and the voltage source turned off. --- **Step 1: Analyze with only the 15 V voltage source active** - Turn off current sources: replace 8 A and 30 A current sources with open circuits. - The circuit reduces to 15 V in series with 4 Ω, then branches into two paths: one with 2 Ω resistor open (since current source replaced by open), and the other with 1 Ω resistor and 6 Ω resistor R. Calculate equivalent resistance seen by the voltage source: - The 2 Ω branch is open, so ignored. - The other branch has 1 Ω resistor in parallel with open circuit (since 8 A current source replaced by open), so just 1 Ω. - This 1 Ω is in series with 6 Ω resistor R. Total resistance $R_{total} = 4 + (1 + R) = 5 + R$ Current from voltage source: $$I_V = \frac{15}{5 + R}$$ Current through R is the same as through 1 Ω resistor, so current through R due to voltage source is $I_V$. Power dissipated in R due to voltage source: $$P_V = I_V^2 \times R = \left(\frac{15}{5 + R}\right)^2 \times R$$ --- **Step 2: Analyze with only current sources active** - Turn off voltage source: replace 15 V source with a short circuit. - The circuit now has 4 Ω resistor in series with a node where branches split. Current sources are 8 A (downwards) in one branch with 1 Ω resistor, and 30 A (upwards) in the other branch with 2 Ω resistor. We want current through R (6 Ω resistor). Using current division and Kirchhoff's laws: - The 8 A current source is in parallel with 1 Ω resistor. - The 30 A current source is in parallel with 2 Ω resistor. Because the voltage source is shorted, the 4 Ω resistor is in series with the parallel combination of these branches and R. Calculate the net current through R due to current sources: - The 8 A current source pushes current down through 1 Ω resistor. - The 30 A current source pushes current up through 2 Ω resistor. The net current through R is the algebraic sum of currents from these branches considering their directions. By superposition, the current through R due to current sources is: $$I_C = 30 - 8 = 22 \text{ A (assuming direction consistent with 30 A source)}$$ Power dissipated in R due to current sources: $$P_C = I_C^2 \times R = 22^2 \times R = 484 R$$ --- **Step 3: Total current through R and power dissipated** By superposition, total current through R is: $$I_{total} = I_V + I_C = \frac{15}{5 + R} + 22$$ Total power dissipated in R: $$P_{total} = I_{total}^2 \times R = \left(\frac{15}{5 + R} + 22\right)^2 \times R$$ --- **Summary:** - Principle of Superposition allows analyzing each source independently. - Current through R due to voltage source: $\frac{15}{5 + R}$ A. - Current through R due to current sources: 22 A. - Total current through R: $\frac{15}{5 + R} + 22$ A. - Power dissipated in R: $\left(\frac{15}{5 + R} + 22\right)^2 \times R$ Watts. This completes the solution using the Principle of Superposition.