1. **State the problem:** Find the current through the 15 Ω resistor using the Superposition principle in the given circuit with voltage sources 30 V and 60 V and resistors 30 Ω, 10 Ω, and 15 Ω. 2. **Superposition principle:** Analyze the circuit twice, each time considering only one voltage source active and replacing the other with its internal resistance (for ideal voltage sources, replace with a short circuit). --- ### Step 1: Consider only the 30 V source active (60 V replaced by a short circuit) - The 60 V source is replaced by a wire (short circuit). - The circuit now has the 30 V source, and resistors 30 Ω, 10 Ω, and 15 Ω arranged accordingly. - The 10 Ω and 15 Ω resistors are in parallel because the bottom branch is shorted by the 60 V source replacement. Calculate the equivalent resistance of 10 Ω and 15 Ω in parallel: $$ R_{10||15} = \frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6\,\Omega $$ - This 6 Ω is in series with the 30 Ω resistor: $$ R_{total1} = 30 + 6 = 36\,\Omega $$ - Current from the 30 V source: $$ I_1 = \frac{30}{36} = \frac{5}{6} = 0.8333\,A $$ - Voltage across the parallel combination (10 Ω and 15 Ω): $$ V_{parallel} = I_1 \times 6 = 0.8333 \times 6 = 5\,V $$ - Current through the 15 Ω resistor: $$ I_{15,1} = \frac{V_{parallel}}{15} = \frac{5}{15} = \frac{1}{3} = 0.3333\,A $$ --- ### Step 2: Consider only the 60 V source active (30 V replaced by a short circuit) - The 30 V source is replaced by a wire (short circuit). - The 30 Ω resistor is now in parallel with the 10 Ω resistor because the left branch is shorted. Calculate the equivalent resistance of 30 Ω and 10 Ω in parallel: $$ R_{30||10} = \frac{30 \times 10}{30 + 10} = \frac{300}{40} = 7.5\,\Omega $$ - This 7.5 Ω is in series with the 15 Ω resistor: $$ R_{total2} = 7.5 + 15 = 22.5\,\Omega $$ - Current from the 60 V source: $$ I_2 = \frac{60}{22.5} = \frac{8}{3} = 2.6667\,A $$ - Voltage across the 15 Ω resistor: $$ V_{15} = I_2 \times 15 = 2.6667 \times 15 = 40\,V $$ - Current through the 15 Ω resistor: $$ I_{15,2} = \frac{V_{15}}{15} = \frac{40}{15} = 2.6667\,A $$ --- ### Step 3: Combine the currents from both sources - The direction of currents must be considered; assuming both currents flow in the same direction through the 15 Ω resistor, total current is: $$ I_{15} = I_{15,1} + I_{15,2} = 0.3333 + 2.6667 = 3\,A $$ **Final answer:** The current through the 15 Ω resistor is **3 A**.